Advanced Math

438 soru

Soru 241Soru

In the xyxy-plane, the graph of the quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c, where aa, bb, and cc are constants, has a vertex (h,k)(h, k) that lies on the line y=4x+8y = 4x + 8. If the graph of y=f(x)y = f(x) has xx-intercepts at x=2x = -2 and x=6x = 6, what is the value of cc?

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Cevap: 12

Cevap

The value of c is 12.
The correct answer is 12. The axis of symmetry of the quadratic function lies midway between its x-intercepts at x=2x = -2 and x=6x = 6, giving an x-coordinate of h=2h = 2 for the vertex. Since the vertex lies on the line y=4x+8y = 4x + 8, its y-coordinate is k=4(2)+8=16k = 4(2) + 8 = 16. Substituting the vertex (2,16)(2, 16) into the vertex form of a quadratic function gives f(x)=a(x2)2+16f(x) = a(x - 2)^2 + 16. Using the x-intercept at (6,0)(6, 0) to solve for aa gives 0=a(62)2+160 = a(6 - 2)^2 + 16, which yields a=1a = -1. Evaluating the function at x=0x = 0 to find the constant term cc gives c=f(0)=(02)2+16=12c = f(0) = -(0 - 2)^2 + 16 = 12.

Adım Adım Çözüm

1
Determine the x-coordinate of the vertex of the quadratic function.
h=2h = 2
The axis of symmetry of a parabola lies midway between its x-intercepts. The midpoint of the intercepts x=2x = -2 and x=6x = 6 is calculated as 2+62=2\frac{-2 + 6}{2} = 2.
2
Determine the y-coordinate of the vertex using the given linear equation.
k=16k = 16
Since the vertex (h,k)(h, k) lies on the line y=4x+8y = 4x + 8, substituting x=h=2x = h = 2 into the line equation gives k=4(2)+8=16k = 4(2) + 8 = 16.
3
Write the quadratic function in vertex form and solve for the leading coefficient aa.
a=1a = -1
The vertex form is f(x)=a(xh)2+kf(x) = a(x - h)^2 + k. Substituting the vertex (2,16)(2, 16) yields f(x)=a(x2)2+16f(x) = a(x - 2)^2 + 16. Using the x-intercept (6,0)(6, 0), substitute x=6x = 6 and f(x)=0f(x) = 0 into the equation: 0=a(62)2+160 = a(6 - 2)^2 + 16, which simplifies to 16a=1616a = -16, resulting in a=1a = -1.
4
Expand the quadratic function or evaluate it at x=0x = 0 to find the value of cc.
c=12c = 12
In standard form f(x)=ax2+bx+cf(x) = ax^2 + bx + c, the constant term cc is equal to f(0)f(0). Substituting x=0x = 0 into f(x)=(x2)2+16f(x) = -(x - 2)^2 + 16 yields f(0)=(02)2+16=4+16=12f(0) = -(0 - 2)^2 + 16 = -4 + 16 = 12.

Anahtar Kavram

Using symmetry and the vertex form of a quadratic function to determine its standard form coefficients.

Alternatif Yöntem

Alternatively, since the x-intercepts are 2-2 and 66, the quadratic function can be written in factored form as f(x)=a(x+2)(x6)f(x) = a(x + 2)(x - 6). Expanding this gives f(x)=a(x24x12)=ax24ax12af(x) = a(x^2 - 4x - 12) = ax^2 - 4ax - 12a. Comparing this to the standard form f(x)=ax2+bx+cf(x) = ax^2 + bx + c, we see that the x-coordinate of the vertex is h=b2a=4a2a=2h = -\frac{b}{2a} = -\frac{-4a}{2a} = 2. Using the line equation y=4x+8y = 4x + 8 at x=2x = 2, we find the vertex y-coordinate is k=16k = 16. Since the vertex is (2,16)(2, 16), we evaluate the factored form at x=2x = 2: 16=a(2+2)(26)16=16aa=116 = a(2 + 2)(2 - 6) \Rightarrow 16 = -16a \Rightarrow a = -1. The constant term cc is 12a-12a, so c=12(1)=12c = -12(-1) = 12.
Tahmini Süre:2m 30s
Soru 242Soru

If (x,y)(x, y) is a solution to the system of equations below and x>1x > 1, what is the value of xx?

y2x=3y25x2=20\begin{aligned} y - 2x &= 3 \\ y^2 - 5x^2 &= 20 \end{aligned}
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Cevap: 11

Cevap

The value of xx is 11.
The system can be solved by substituting the first equation into the second. Rearranging the first equation gives y=2x+3y = 2x + 3. Substituting this expression for yy in the second equation gives (2x+3)25x2=20(2x + 3)^2 - 5x^2 = 20. Expanding the binomial yields 4x2+12x+95x2=204x^2 + 12x + 9 - 5x^2 = 20, which simplifies to x2+12x+9=20-x^2 + 12x + 9 = 20. Rearranging into standard quadratic form gives x212x+11=0x^2 - 12x + 11 = 0. Factoring the quadratic equation results in (x11)(x1)=0(x - 11)(x - 1) = 0. Thus, x=11x = 11 or x=1x = 1. Since the problem specifies x>1x > 1, the value of xx must be 11.

Adım Adım Çözüm

1
Rearrange the first equation to express yy in terms of xx.
y=2x+3y = 2x + 3
This allows for substitution into the second equation.
2
Substitute y=2x+3y = 2x + 3 into the second equation.
(2x+3)25x2=20(2x + 3)^2 - 5x^2 = 20
To eliminate yy and obtain an equation in terms of xx only.
3
Expand and simplify the equation.
x212x+11=0x^2 - 12x + 11 = 0
Expanding (2x+3)2(2x + 3)^2 gives 4x2+12x+94x^2 + 12x + 9. Subtracting 5x25x^2 yields x2+12x+9=20-x^2 + 12x + 9 = 20. Subtracting 20 from both sides gives x2+12x11=0-x^2 + 12x - 11 = 0. Multiplying the entire equation by 1-1 results in standard form.
4
Factor the quadratic equation.
(x11)(x1)=0(x - 11)(x - 1) = 0
To find the possible values of xx.
5
Apply the given constraint x>1x > 1.
x=11x = 11
The solutions are x=11x = 11 and x=1x = 1. The constraint x>1x > 1 excludes x=1x = 1, leaving x=11x = 11 as the only valid solution.

Anahtar Kavram

Solving a system of nonlinear equations by substitution and factoring a quadratic equation.
Soru 243Soru

If xx is a real number that satisfies the equation xx36x23x=2x\frac{x}{x-3} - \frac{6}{x^2-3x} = \frac{2}{x}, what is the value of xx?

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Cevap: 2

Cevap

2
The correct answer is 2. Multiplying the entire equation by the lowest common denominator, x(x3)x(x-3), yields x26=2(x3)x^2 - 6 = 2(x - 3). Distributing the right side gives x26=2x6x^2 - 6 = 2x - 6. Adding 6 to both sides and subtracting 2x2x results in x22x=0x^2 - 2x = 0. Factoring this equation gives x(x2)=0x(x-2) = 0, which has solutions x=0x = 0 and x=2x = 2. However, x=0x = 0 is extraneous because it results in division by zero in the original equation. Thus, the only valid real solution is 2.

Adım Adım Çözüm

1
Identify the common denominator and multiply the equation by it.
x26=2(x3)x^2 - 6 = 2(x - 3)
To clear the fractions and simplify the rational equation.
2
Distribute and simplify the equation.
x22x=0x^2 - 2x = 0
To set up the quadratic equation in standard form.
3
Factor and solve for xx.
x=0x = 0 or x=2x = 2
To find potential solutions to the equation.
4
Check for extraneous solutions by substituting the values back into the original denominators.
x=2x = 2
x=0x = 0 makes the denominators equal to zero and is therefore extraneous.

Anahtar Kavram

Solving rational equations and identifying extraneous solutions.
Soru 244Soru

If x34=9x3\frac{x - 3}{4} = \frac{9}{x - 3} and x>3x > 3, what is the value of xx?

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Cevap: 9

Cevap

9
Cross-multiplying the equation x34=9x3\frac{x - 3}{4} = \frac{9}{x - 3} gives (x3)2=36(x - 3)^2 = 36. Taking the square root of both sides gives x3=6x - 3 = 6 or x3=6x - 3 = -6. Solving for xx yields x=9x = 9 or x=3x = -3. Since the question specifies the constraint x>3x > 3, the only valid solution is 99.

Adım Adım Çözüm

1
Cross-multiply the equation to eliminate the denominators.
(x3)2=36(x - 3)^2 = 36
Multiplying both sides of the equation by 4(x3)4(x - 3) simplifies the rational equation into a quadratic form.
2
Take the square root of both sides of the equation.
x3=6x - 3 = 6 or x3=6x - 3 = -6
Taking the square root of a squared term yields both positive and negative root options.
3
Solve each linear equation for xx and apply the constraint x>3x > 3.
x=9x = 9
Adding 33 to both sides gives x=9x = 9 or x=3x = -3. The constraint x>3x > 3 excludes the negative solution, leaving 99 as the only valid value.

Anahtar Kavram

Solving rational equations by cross-multiplication and factoring the resulting quadratic equation while adhering to domain constraints.
Soru 245Soru

A parabola defined by the equation y=ax2+bx+cy = ax^2 + bx + c, where aa, bb, and cc are constants, has its vertex in the second quadrant of the coordinate plane. If the parabola passes through the point (0,0)(0,0), which of the following inequalities must be true?

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Cevap: ab>0ab > 0

Cevap

The inequality ab>0ab > 0 must be true.
Since the parabola passes through the point (0,0)(0,0) and its vertex (h,k)(h, k) lies in the second quadrant where k>0k > 0, the vertex must represent a maximum value. Therefore, the parabola opens downward, so the leading coefficient aa is negative. The x-coordinate of the vertex, hh, is also negative because it lies in the second quadrant. The vertex x-coordinate is defined by h=b2ah = -\frac{b}{2a}, which means b=2ahb = -2ah. Because both aa and hh are negative, their product ahah is positive, which makes bb negative when multiplied by 2-2. Finally, since both aa and bb are negative, their product abab must be positive.

Adım Adım Çözüm

1
Determine the signs of the vertex coordinates from its quadrant.
The vertex (h,k)(h, k) lies in the second quadrant, which means h<0h < 0 and k>0k > 0.
Points in the second quadrant have negative x-coordinates and positive y-coordinates.
2
Determine the value of the constant cc.
c=0c = 0
Since the parabola passes through (0,0)(0,0), substituting x=0x = 0 must result in y=0y = 0.
3
Determine the sign of the leading coefficient aa.
a<0a < 0
The vertex is at (h,k)(h, k) with k>0k > 0, and the graph passes through (0,0)(0, 0) where the y-value is 00. Since the maximum value must be at least the value at any other point, the vertex (h,k)(h,k) is a maximum. A parabola with a maximum opens downward, so a<0a < 0.
4
Determine the sign of the coefficient bb using the vertex formula.
b<0b < 0
The x-coordinate of the vertex is h=b2ah = -\frac{b}{2a}, which can be rewritten as b=2ahb = -2ah. Since a<0a < 0 and h<0h < 0, their product ahah is positive. Multiplying this positive product by 2-2 yields a negative value for bb.
5
Find the sign of the product abab.
ab>0ab > 0
Since both aa and bb are negative, their product abab must be positive.

Anahtar Kavram

Analyzing quadratic coefficients and vertex properties in the coordinate plane.
Soru 246Soru

The graph of the function ff in the xyxy-plane is translated 44 units to the right and 33 units down to create the graph of the function gg. Which of the following equations defines g(x)g(x) in terms of f(x)f(x)?

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Cevap: g(x)=f(x4)3g(x) = f(x - 4) - 3

Cevap

The equation that defines g(x)g(x) in terms of f(x)f(x) is g(x)=f(x4)3g(x) = f(x - 4) - 3.
The equation g(x)=f(x4)3g(x) = f(x - 4) - 3 correctly represents the translated function. A translation of 44 units to the right is represented by replacing xx with x4x - 4 within the function's input, and a translation of 33 units down is represented by subtracting 33 from the outer function.

Adım Adım Çözüm

1
Apply the horizontal translation to the function input.
f(x4)f(x - 4)
Translating a function f(x)f(x) horizontally to the right by hh units is represented by substituting xx with xhx - h. For a shift of 44 units to the right, we replace xx with x4x - 4.
2
Apply the vertical translation to the function output.
f(x4)3f(x - 4) - 3
Translating a function vertically down by kk units is represented by subtracting kk from the entire function. For a shift of 33 units down, we subtract 33.
3
Write the final equation for g(x)g(x).
g(x)=f(x4)3g(x) = f(x - 4) - 3
Combining the horizontal and vertical transformations yields the function g(x)g(x).

Anahtar Kavram

Graph transformations of functions
Soru 247Soru

A polynomial p(x)p(x) has a remainder of 33 when divided by x5x - 5. Which of the following equations must be true?

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Cevap: p(5)=3p(5) = 3

Cevap

The equation p(5)=3p(5) = 3 must be true.
According to the Remainder Theorem, when a polynomial p(x)p(x) is divided by a linear expression of the form xcx - c, the remainder is equal to p(c)p(c). Here, the divisor is x5x - 5, which gives c=5c = 5. The remainder is given as 33. Therefore, evaluating the polynomial at x=5x = 5 must yield 33, which is written as p(5)=3p(5) = 3.

Adım Adım Çözüm

1
Identify the divisor and the remainder from the given problem statement.
The divisor is x5x - 5 and the remainder is 33.
Applying the Remainder Theorem requires identifying the value of cc in the divisor form xcx - c and the remainder value.
2
Apply the Remainder Theorem to relate the divisor and the remainder to the polynomial function.
Since the divisor is x5x - 5, we have c=5c = 5. The Remainder Theorem states that the remainder when p(x)p(x) is divided by xcx - c is p(c)p(c). Therefore, p(5)=3p(5) = 3.
This establishes the mathematical relationship directly showing which equation must be true.

Anahtar Kavram

The Remainder Theorem
Tahmini Süre:45s
Soru 248Soru

In the polynomial function p(x)=x35x2+2x+kp(x) = x^3 - 5x^2 + 2x + k, the constant kk is chosen such that p(x)p(x) is divisible by x4x - 4. What is the value of kk?

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Cevap: 8

Cevap

The value of kk is 8.
According to the Factor Theorem, a polynomial p(x)p(x) is divisible by xcx - c if and only if p(c)=0p(c) = 0. Since p(x)p(x) is divisible by x4x - 4, we must have p(4)=0p(4) = 0. Substituting 44 for xx in the equation for p(x)p(x) yields 435(4)2+2(4)+k=04^3 - 5(4)^2 + 2(4) + k = 0. Simplifying the numerical terms gives 6480+8+k=064 - 80 + 8 + k = 0, which simplifies further to 8+k=0-8 + k = 0. Therefore, the value of the constant kk is 88.

Adım Adım Çözüm

1
Apply the Factor Theorem to relate the factor x4x - 4 to the value of the polynomial.
Since x4x - 4 is a factor of p(x)p(x), p(4)=0p(4) = 0.
By the Factor Theorem, if xcx - c is a factor of a polynomial p(x)p(x), then p(c)=0p(c) = 0.
2
Substitute x=4x = 4 into the polynomial expression.
435(4)2+2(4)+k=04^3 - 5(4)^2 + 2(4) + k = 0
We substitute x=4x = 4 into p(x)=x35x2+2x+kp(x) = x^3 - 5x^2 + 2x + k and set the expression to 00.
3
Simplify the numerical expression to solve for kk.
6480+8+k=0    8+k=0    k=864 - 80 + 8 + k = 0 \implies -8 + k = 0 \implies k = 8
Evaluate powers, multiply, and solve the resulting linear equation for kk.

Anahtar Kavram

Factor Theorem
Tahmini Süre:1m 0s
Soru 249Soru

For all x2x \neq 2, the expression 3x2+kx8x2\frac{3x^2 + kx - 8}{x - 2} is equivalent to 3x+10+12x23x + 10 + \frac{12}{x - 2}, where kk is a constant. What is the value of kk?

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Cevap: 4

Cevap

The value of the constant kk is 44.
The correct answer is 44. By rewriting the right-hand side of the equation with a common denominator of x2x - 2, the expression becomes (3x+10)(x2)+12x2\frac{(3x + 10)(x - 2) + 12}{x - 2}. Expanding and simplifying the numerator yields 3x2+4x83x^2 + 4x - 8. Comparing this to the numerator of the left-hand side, 3x2+kx83x^2 + kx - 8, shows that the coefficient of the xx term, kk, must equal 44.

Adım Adım Çözüm

1
Set up the equation representing the equivalence of the two expressions.
3x2+kx8x2=3x+10+12x2\frac{3x^2 + kx - 8}{x - 2} = 3x + 10 + \frac{12}{x - 2}
This establishes the relationship that must hold for all x2x \neq 2.
2
Find a common denominator for the terms on the right-hand side.
3x+10+12x2=(3x+10)(x2)x2+12x23x + 10 + \frac{12}{x - 2} = \frac{(3x + 10)(x - 2)}{x - 2} + \frac{12}{x - 2}
To combine the terms into a single rational expression, they must have the same denominator.
3
Expand and simplify the numerator on the right-hand side.
(3x+10)(x2)+12=3x2+4x8(3x + 10)(x - 2) + 12 = 3x^2 + 4x - 8
Expanding the product of binomials and combining like terms yields the simplified numerator.
4
Equate the numerators of the simplified expressions to solve for kk.
3x2+kx8=3x2+4x83x^2 + kx - 8 = 3x^2 + 4x - 8, which means k=4k = 4.
Since the denominators are equal, the numerators must be identical for all values of x2x \neq 2, meaning their corresponding coefficients must match.

Anahtar Kavram

Equivalent Algebraic Expressions

Alternatif Yöntem

An alternative method is to substitute a convenient value for xx that is not equal to 22. Substituting x=1x = 1 into both sides of the equivalence gives: 3(1)2+k(1)812=3(1)+10+1212\frac{3(1)^2 + k(1) - 8}{1 - 2} = 3(1) + 10 + \frac{12}{1 - 2}. Simplifying this yields k51=1312\frac{k - 5}{-1} = 13 - 12, which simplifies to k+5=1-k + 5 = 1. Solving for kk results in k=4k = 4.
Tahmini Süre:1m 30s
Soru 250Soru

For which of the following values of kk does the equation x2kx+9x3=0\frac{x^2 - kx + 9}{x - 3} = 0 have exactly one real solution?

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Cevap: 6-6

Cevap

The correct value is 6-6.
For the equation x2kx+9x3=0\frac{x^2 - kx + 9}{x - 3} = 0 to have a solution, we must have x3x \neq 3. Multiplying both sides by the denominator yields the quadratic equation x2kx+9=0x^2 - kx + 9 = 0. For a quadratic equation to have exactly one real solution, its discriminant must equal zero: (k)24(1)(9)=k236=0(-k)^2 - 4(1)(9) = k^2 - 36 = 0, which gives k=6k = 6 or k=6k = -6. If k=6k = 6, the equation is (x3)2=0(x-3)^2 = 0, which has the single solution x=3x = 3. Since x=3x = 3 is excluded by the domain restriction, the equation has zero real solutions. If k=6k = -6, the equation is (x+3)2=0(x+3)^2 = 0, which has the single solution x=3x = -3. Since 3-3 is a valid value, this value of kk yields exactly one real solution.

Adım Adım Çözüm

1
Identify the domain restriction of the rational equation.
The denominator x3x - 3 cannot be zero, which means x3x \neq 3.
Division by zero is undefined, so x=3x = 3 cannot be a solution to the equation.
2
Simplify the equation by multiplying both sides by the denominator.
x2kx+9=0x^2 - kx + 9 = 0 for x3x \neq 3.
A rational expression equals zero when its numerator is zero and its denominator is non-zero.
3
Set the discriminant of the quadratic equation to zero to find the values of kk that yield exactly one real root.
Δ=(k)24(1)(9)=k236=0    k=6\Delta = (-k)^2 - 4(1)(9) = k^2 - 36 = 0 \implies k = 6 or k=6k = -6.
A quadratic equation has exactly one real solution (a double root) when its discriminant is zero.
4
Verify if the roots for each value of kk are valid under the domain restriction x3x \neq 3.
If k=6k = 6, x26x+9=(x3)2=0    x=3x^2 - 6x + 9 = (x - 3)^2 = 0 \implies x = 3 (invalid/extraneous). If k=6k = -6, x2+6x+9=(x+3)2=0    x=3x^2 + 6x + 9 = (x + 3)^2 = 0 \implies x = -3 (valid).
The value k=6k = 6 makes x=3x = 3 the only solution, which is excluded. Therefore, only k=6k = -6 yields exactly one valid real solution (x=3x = -3).

Anahtar Kavram

Solving rational equations that simplify to quadratic equations, using the discriminant to find the number of solutions, and identifying extraneous solutions due to domain restrictions.
Tahmini Süre:2m 0s
Soru 251Soru
If xx is a real solution to the equation
2x5x3+x5x24x+3=1x1\frac{2x - 5}{x - 3} + \frac{x - 5}{x^2 - 4x + 3} = -\frac{1}{x - 1}
what is the value of 12x1 - 2x?
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Cevap: 2

Cevap

2
The correct answer is 2. Multiplying the entire equation by the least common denominator (x3)(x1)(x - 3)(x - 1) simplifies the rational equation to the quadratic equation 2x25x3=02x^2 - 5x - 3 = 0. This quadratic factors into (2x+1)(x3)=0(2x + 1)(x - 3) = 0, yielding two potential solutions: x=1/2x = -1/2 and x=3x = 3. However, x=3x = 3 is an extraneous solution because it results in a denominator of zero in the original equation. Thus, the only valid real solution is x=1/2x = -1/2. Substituting this into the expression 12x1 - 2x yields 12(1/2)=21 - 2(-1/2) = 2.

Adım Adım Çözüm

1
Factor the quadratic expression in the denominator to identify the least common denominator.
The quadratic in the denominator of the second term factors as x24x+3=(x3)(x1)x^2 - 4x + 3 = (x - 3)(x - 1). Thus, the least common denominator for all terms in the equation is (x3)(x1)(x - 3)(x - 1), with the restrictions that x3x \neq 3 and x1x \neq 1.
Finding a common denominator allows us to multiply both sides of the equation to clear all rational expressions.
2
Multiply both sides of the equation by the least common denominator (x3)(x1)(x - 3)(x - 1).
(2x5)(x1)+(x5)=1(x3)(2x - 5)(x - 1) + (x - 5) = -1(x - 3)
This clears the fractions and converts the rational equation into a polynomial equation.
3
Expand and simplify both sides of the equation.
(2x27x+5)+(x5)=x+3(2x^2 - 7x + 5) + (x - 5) = -x + 3
2x26x=x+32x^2 - 6x = -x + 3
Expanding the terms allows us to combine like terms and set up a quadratic equation.
4
Rearrange the terms to set the quadratic equation equal to zero.
2x25x3=02x^2 - 5x - 3 = 0
Quadratic equations must be set to zero to be solved by factoring or using the quadratic formula.
5
Solve the quadratic equation by factoring.
(2x+1)(x3)=0(2x + 1)(x - 3) = 0
This gives two potential solutions: x=1/2x = -1/2 and x=3x = 3.
Factoring is the most direct algebraic method to find the roots of the quadratic equation.
6
Check the potential solutions against the original domain restrictions to identify any extraneous solutions.
For x=3x = 3, the denominators x3x - 3 and x24x+3x^2 - 4x + 3 equal zero, so x=3x = 3 is an extraneous solution. For x=1/2x = -1/2, the denominators are non-zero, so x=1/2x = -1/2 is the only valid solution.
Squaring or multiplying by variable expressions can introduce solutions that are undefined in the original equation.
7
Substitute the valid solution x=1/2x = -1/2 into the expression 12x1 - 2x.
12(12)=1+1=21 - 2\left(-\frac{1}{2}\right) = 1 + 1 = 2
The question asks for the value of the expression 12x1 - 2x, not the value of xx itself.

Anahtar Kavram

Solving rational equations by clearing denominators and checking for extraneous solutions.
Soru 252Soru

Two different cultures of bacteria, culture AA and culture BB, begin growing at the same time. The population of culture AA increases by 300%300\% every 88 hours, and the population of culture BB increases by 700%700\% every 99 hours. If both cultures start with the same initial population, after how many hours will the population of culture BB be 44 times the population of culture AA?

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Cevap: 24

Cevap

The population of culture B will be 4 times the population of culture A after 24 hours.
The correct answer is the option stating 24 hours. The growth model for culture A is PA(t)=P04t/8P_A(t) = P_0 \cdot 4^{t/8} and for culture B is PB(t)=P08t/9P_B(t) = P_0 \cdot 8^{t/9}. Setting PB(t)=4PA(t)P_B(t) = 4 P_A(t) and converting all terms to base 2 yields 2t/3=22+t/42^{t/3} = 2^{2 + t/4}. Equating exponents gives t/3=2+t/4t/3 = 2 + t/4, which simplifies to t/12=2t/12 = 2 and solves to t=24t = 24 hours.

Adım Adım Çözüm

1
Determine the growth factors and set up the population equations for both cultures.
For culture AA, a 300%300\% increase means the population becomes 1+3.00=41 + 3.00 = 4 times its previous value every 88 hours, so PA(t)=P04t/8P_A(t) = P_0 \cdot 4^{t/8}. For culture BB, a 700%700\% increase means the population becomes 1+7.00=81 + 7.00 = 8 times its previous value every 99 hours, so PB(t)=P08t/9P_B(t) = P_0 \cdot 8^{t/9}.
Establishing the correct exponential growth models with the appropriate growth factors and time constants is necessary to relate the populations.
2
Set up the equation to find when the population of culture BB is 44 times that of culture AA.
PB(t)=4PA(t)    P08t/9=4P04t/8    8t/9=44t/8P_B(t) = 4 P_A(t) \implies P_0 \cdot 8^{t/9} = 4 \cdot P_0 \cdot 4^{t/8} \implies 8^{t/9} = 4 \cdot 4^{t/8}
This sets up the equation that must be solved for tt by dividing both sides by the non-zero initial population P0P_0.
3
Express all terms with a common base of 22.
(23)t/9=22(22)t/8    2t/3=222t/4(2^3)^{t/9} = 2^2 \cdot (2^2)^{t/8} \implies 2^{t/3} = 2^2 \cdot 2^{t/4}
Expressing bases as powers of 22 allows the use of exponent rules to simplify the equation.
4
Apply the product rule of exponents to combine the terms on the right side.
2^{t/3} = 2^{2 + t/4}
The rule 2a2b=2a+b2^a \cdot 2^b = 2^{a+b} simplifies the multiplication of exponential terms with the same base.
5
Equate the exponents and solve the resulting linear equation for tt.
t/3 = 2 + t/4 \implies t/3 - t/4 = 2 \implies t/12 = 2 \implies t = 24
Since the bases are equal, their exponents must be equal. Solving the linear equation yields the time in hours.

Anahtar Kavram

Solving exponential equations by converting to a common base and applying exponent laws.
Tahmini Süre:3m 0s
Soru 253Soru

If xx is the real solution to the equation below, what is the value of xx?

2x230x+100x5+140x+2=10\frac{2x^2 - 30x + 100}{x - 5} + \frac{140}{x + 2} = 10
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Cevap: 8

Cevap

The correct answer is 8.
To solve the rational equation, we first state the domain restrictions as x5x \neq 5 and x2x \neq -2 because these values make the denominators zero. Next, we factor the numerator of the first term: 2x230x+100=2(x215x+50)=2(x5)(x10)2x^2 - 30x + 100 = 2(x^2 - 15x + 50) = 2(x - 5)(x - 10). Since x5x \neq 5, we can cancel the common factor of x5x - 5 from the numerator and denominator, which simplifies the first term to 2(x10)=2x202(x - 10) = 2x - 20. The equation can then be rewritten as 2x20+140x+2=102x - 20 + \frac{140}{x + 2} = 10. Subtracting 10 from both sides yields 2x30+140x+2=02x - 30 + \frac{140}{x + 2} = 0. Dividing the entire equation by 2 simplifies the coefficients, giving x15+70x+2=0x - 15 + \frac{70}{x + 2} = 0. To clear the fraction, we multiply the entire equation by x+2x + 2, leading to (x15)(x+2)+70=0(x - 15)(x + 2) + 70 = 0. Expanding the product gives x213x30+70=0x^2 - 13x - 30 + 70 = 0, which simplifies to x213x+40=0x^2 - 13x + 40 = 0. Factoring this quadratic equation yields (x5)(x8)=0(x - 5)(x - 8) = 0, giving the potential solutions x=5x = 5 and x=8x = 8. However, x=5x = 5 is extraneous because it is restricted from the domain of the original rational expression. Therefore, the only real solution is 8.

Adım Adım Çözüm

1
Find the values of xx that make any denominator equal to zero to identify domain restrictions.
x5x \neq 5 and x2x \neq -2
Dividing by zero is undefined in real numbers, so these values are excluded from the domain.
2
Factor the numerator of the first term, 2x230x+1002x^2 - 30x + 100.
2(x5)(x10)2(x - 5)(x - 10)
Factoring allows us to check for common factors in the numerator and denominator.
3
Simplify the first rational term by dividing the numerator and denominator by the common factor x5x - 5, under the assumption that x5x \neq 5.
2(x10)=2x202(x - 10) = 2x - 20
Reducing the expression simplifies the algebraic structure of the equation.
4
Substitute the simplified expression back into the original equation and subtract 10 from both sides.
2x30+140x+2=02x - 30 + \frac{140}{x + 2} = 0
To consolidate the constant terms and prepare the equation for clearing the remaining fraction.
5
Divide the entire equation by 2.
x15+70x+2=0x - 15 + \frac{70}{x + 2} = 0
Dividing by the common factor simplifies the coefficients before expanding.
6
Multiply the entire equation by x+2x + 2 to clear the denominator, then expand and combine like terms.
x213x+40=0x^2 - 13x + 40 = 0
This transforms the rational equation into a standard quadratic equation.
7
Factor the quadratic equation.
(x5)(x8)=0(x - 5)(x - 8) = 0
Factoring is the most direct method to find the roots of the quadratic equation.
8
Verify the candidate solutions against the domain restrictions identified in Step 1.
The candidate solution x=5x = 5 is extraneous because it is excluded from the domain of the original equation. The candidate solution x=8x = 8 is valid.
Extraneous solutions introduced during algebraic steps must be excluded from the final solution set.

Anahtar Kavram

Solving rational equations by simplifying terms, finding common denominators, and identifying extraneous solutions.
Soru 254Soru

In the xyxy-plane, the system of equations below has exactly one real solution.

y=x2+6x+2y = -x^2 + 6x + 2
y+2x=ky + 2x = k

If kk is a constant, what is the value of kk?

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Cevap: 18

Cevap

18
To find the value of the constant kk for which the system of equations has exactly one real solution, we can solve the system by substitution. Substituting y=k2xy = k - 2x from the second equation into the first equation gives k2x=x2+6x+2k - 2x = -x^2 + 6x + 2. Rearranging the terms to write this quadratic equation in standard form, ax2+bx+c=0ax^2 + bx + c = 0, yields x28x+(k2)=0x^2 - 8x + (k - 2) = 0. A quadratic equation has exactly one real solution when its discriminant, b24acb^2 - 4ac, is equal to zero. Substituting a=1a = 1, b=8b = -8, and c=k2c = k - 2 into the discriminant formula gives (8)24(1)(k2)=0(-8)^2 - 4(1)(k - 2) = 0. Simplifying this expression results in 644k+8=064 - 4k + 8 = 0, which simplifies further to 724k=072 - 4k = 0. Solving for kk yields k=18k = 18.

Adım Adım Çözüm

1
Rewrite the linear equation to express yy in terms of xx and kk.
y=k2xy = k - 2x
This allows substituting the expression for yy into the quadratic equation to eliminate yy.
2
Substitute the expression for yy into the quadratic equation.
k2x=x2+6x+2k - 2x = -x^2 + 6x + 2
To construct a single quadratic equation in terms of xx whose solutions correspond to the system's intersection points.
3
Rearrange the equation into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x28x+(k2)=0x^2 - 8x + (k - 2) = 0
To identify the coefficients a=1a = 1, b=8b = -8, and c=k2c = k - 2.
4
Set the discriminant of the quadratic equation to zero.
(8)24(1)(k2)=0(-8)^2 - 4(1)(k - 2) = 0
A quadratic equation has exactly one real solution if and only if its discriminant is zero.
5
Solve the linear equation for kk.
k=18k = 18
Simplify the equation: 644k+8=0    724k=0    4k=72    k=1864 - 4k + 8 = 0 \implies 72 - 4k = 0 \implies 4k = 72 \implies k = 18.

Anahtar Kavram

Nonlinear Systems of Equations
Soru 255Soru

The table below shows the population of a colony of bacteria, P(t)P(t), at various times tt, in hours.

Time tt (hours)Population P(t)P(t)
0120
3960
67,680

If the population can be modeled by an exponential function of the form P(t)=abtP(t) = a \cdot b^t, where aa and bb are constants, what is the value of bb?

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Cevap: 2

Cevap

2
The correct answer is the value that represents the hourly growth factor of the bacteria. Since the initial population at t=0t = 0 is 120, we have a=120a = 120. Using the population at t=3t = 3, we set up the equation 120b3=960120 \cdot b^3 = 960. Dividing both sides by 120 yields b3=8b^3 = 8. Taking the cube root of 8 gives b=2b = 2. We can verify this with the third data point: 12026=12064=7,680120 \cdot 2^6 = 120 \cdot 64 = 7,680, which matches the table.

Adım Adım Çözüm

1
Find the value of the constant aa using the initial value from the table.
a=120a = 120
When t=0t = 0, the population P(0)=120P(0) = 120. Substituting these values into P(t)=abtP(t) = a \cdot b^t gives 120=ab0120 = a \cdot b^0, and since b0=1b^0 = 1, we get a=120a = 120.
2
Set up an equation to solve for bb using another data point from the table.
120b3=960120 \cdot b^3 = 960
When t=3t = 3, the population P(3)=960P(3) = 960. Substituting a=120a = 120, t=3t = 3, and P(3)=960P(3) = 960 into the function gives 960=120b3960 = 120 \cdot b^3.
3
Isolate b3b^3 and solve for bb by taking the cube root.
b=2b = 2
Dividing both sides of 120b3=960120 \cdot b^3 = 960 by 120120 gives b3=8b^3 = 8. Taking the cube root of both sides gives b=83=2b = \sqrt[3]{8} = 2.

Anahtar Kavram

Determining parameters of exponential functions from a table of values.
Soru 256Soru

If (14)2x583x=(23)x5\frac{\left(\frac{1}{4}\right)^{2x-5}}{8^{3-x}} = \left(\sqrt[3]{2}\right)^{x-5}, what is the value of xx?

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Cevap: 2

Cevap

2
By writing all terms with a common base of 2, the equation simplifies to 2x+1=2x532^{-x+1} = 2^{\frac{x-5}{3}}. Equating the exponents yields x+1=x53-x + 1 = \frac{x-5}{3}. Solving this linear equation gives x=2x = 2.

Adım Adım Çözüm

1
Rewrite each base in terms of base 2.
14=22\frac{1}{4} = 2^{-2}, 8=238 = 2^3, and 23=213\sqrt[3]{2} = 2^{\frac{1}{3}}
Expressing all terms with a common base allows the exponents to be directly compared.
2
Substitute the bases back into the left side of the equation and simplify using exponent rules.
(22)2x5(23)3x=24x+10293x=2x+1\frac{(2^{-2})^{2x-5}}{(2^3)^{3-x}} = \frac{2^{-4x+10}}{2^{9-3x}} = 2^{-x+1}
Applying the power rule (am)n=amn(a^m)^n = a^{mn} and quotient rule aman=amn\frac{a^m}{a^n} = a^{m-n} simplifies the fraction.
3
Rewrite the right side of the equation using base 2.
(23)x5=(213)x5=2x53(\sqrt[3]{2})^{x-5} = (2^{\frac{1}{3}})^{x-5} = 2^{\frac{x-5}{3}}
Applying the power rule to the fractional exponent converts the radical expression.
4
Equate the simplified left and right sides, then set the exponents equal to each other.
x+1=x53-x + 1 = \frac{x-5}{3}
If two exponential expressions with the same base are equal, their exponents must be equal.
5
Solve the linear equation for xx.
3x+3=x5    8=4x    x=2-3x + 3 = x - 5 \implies 8 = 4x \implies x = 2
Isolating the variable xx yields the final solution.

Anahtar Kavram

Solving exponential equations by expressing all terms with a common base and applying exponent laws.
Soru 257Soru

If one of the solutions to the equation 2x29x+c=02x^2 - 9x + c = 0, where cc is a constant, is x=2x = 2, what is the other solution to the equation?

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Cevap: 52\frac{5}{2}

Cevap

The other solution to the equation is 52\frac{5}{2}.
The correct answer is found by using the relationship between the coefficients of a quadratic equation and the sum of its roots. For any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of the roots is ba-\frac{b}{a}. Substituting the values a=2a = 2 and b=9b = -9, we find that the sum of the roots is 92=92-\frac{-9}{2} = \frac{9}{2}. Since one of the solutions is given as 22, the other solution must be 922=52\frac{9}{2} - 2 = \frac{5}{2}. Alternatively, one can find the constant cc by substituting x=2x = 2 into the equation to get 2(2)29(2)+c=02(2)^2 - 9(2) + c = 0, which yields c=10c = 10. Solving the resulting equation 2x29x+10=02x^2 - 9x + 10 = 0 by factoring gives (2x5)(x2)=0(2x - 5)(x - 2) = 0, which confirms the other solution is 52\frac{5}{2}.

Adım Adım Çözüm

1
Identify the relationship between the coefficients and the sum of the roots for a quadratic equation.
For any quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0, the sum of its roots, x1x_1 and x2x_2, is given by the formula x1+x2=bax_1 + x_2 = -\frac{b}{a}.
This relationship allows us to find the second root directly without needing to calculate the value of the constant cc first.
2
Substitute the given root and the coefficients of the quadratic equation into the sum of the roots formula.
Here, a=2a = 2 and b=9b = -9. The given root is x1=2x_1 = 2. Substituting these values gives 2+x2=922 + x_2 = -\frac{-9}{2}, which simplifies to 2+x2=922 + x_2 = \frac{9}{2}.
This sets up a linear equation to solve for the unknown root, x2x_2.
3
Solve for the second root x2x_2.
x2=922=9242=52x_2 = \frac{9}{2} - 2 = \frac{9}{2} - \frac{4}{2} = \frac{5}{2}.
This isolates the unknown variable to find the final value.

Anahtar Kavram

Using the sum of roots formula (x1+x2=bax_1 + x_2 = -\frac{b}{a}) to solve for an unknown solution of a quadratic equation.
Soru 258Soru
If xx is a real solution to the equation
3xx1+2xx2=x+2x23x+2\frac{3x}{x - 1} + \frac{2x}{x - 2} = \frac{x + 2}{x^2 - 3x + 2}
what is the value of 5x45x - 4?
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Cevap: -5

Cevap

-5
The correct answer is 5-5. Solving the equation by multiplying both sides by the least common denominator (x1)(x2)(x-1)(x-2) yields the quadratic equation 5x29x2=05x^2 - 9x - 2 = 0. Factoring this equation gives the potential solutions x=1/5x = -1/5 and x=2x = 2. However, substituting x=2x = 2 into the original equation results in division by zero, meaning x=2x = 2 is extraneous and must be discarded. The only valid solution is x=1/5x = -1/5. Substituting this value into the expression 5x45x - 4 yields 5(1/5)4=55(-1/5) - 4 = -5.

Adım Adım Çözüm

1
Identify the least common denominator (LCD) and multiply both sides of the equation to clear the denominators.
The LCD is (x1)(x2)=x23x+2(x - 1)(x - 2) = x^2 - 3x + 2. Multiplying both sides by the LCD gives:
3x(x2)+2x(x1)=x+23x(x - 2) + 2x(x - 1) = x + 2
Multiplying by the LCD simplifies the rational equation into a polynomial equation.
2
Expand the terms and collect like terms to form a quadratic equation.
3x26x+2x22x=x+23x^2 - 6x + 2x^2 - 2x = x + 2
5x28x=x+25x^2 - 8x = x + 2
5x29x2=05x^2 - 9x - 2 = 0
This puts the equation in the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
3
Factor the quadratic equation to find potential solutions.
The quadratic factors as:
(5x+1)(x2)=0(5x + 1)(x - 2) = 0
This yields potential solutions x=15x = -\frac{1}{5} and x=2x = 2.
Factoring allows us to isolate the candidate values of xx.
4
Check the potential solutions in the original equation to identify any extraneous solutions.
For x=2x = 2, the denominators x2x - 2 and x23x+2x^2 - 3x + 2 become zero, which is undefined. Therefore, x=2x = 2 is an extraneous solution. For x=15x = -\frac{1}{5}, all denominators are non-zero, so it is the only valid solution.
Values that make any denominator in the original equation equal to zero must be excluded.
5
Substitute the valid solution into the target expression 5x45x - 4.
5(15)4=14=55\left(-\frac{1}{5}\right) - 4 = -1 - 4 = -5
Evaluating the expression with the only valid solution gives the final answer.

Anahtar Kavram

Solving rational equations by clearing denominators and checking for extraneous solutions.
Tahmini Süre:2m 30s
Soru 259Soru

The function ff is defined by f(x)=x26x+cf(x) = x^2 - 6x + c, where cc is a constant. In the xyxy-plane, the graph of ff has vertex AA. The function gg is defined by g(x)=f(x6)g(x) = -f(x - 6), and its graph has vertex BB. If the distance between points AA and BB is 1010, and c>10c > 10, what is the value of cc?

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Cevap: 13

Cevap

The value of cc is 1313.
Completing the square for f(x)=x26x+cf(x) = x^2 - 6x + c gives f(x)=(x3)2+c9f(x) = (x - 3)^2 + c - 9, which shows that vertex AA is located at (3,c9)(3, c - 9). The transformation g(x)=f(x6)g(x) = -f(x - 6) translates the graph 66 units to the right and reflects it vertically, giving vertex BB the coordinates (3+6,(c9))=(9,9c)(3 + 6, -(c - 9)) = (9, 9 - c). Using the distance formula, the distance between AA and BB is (93)2+((9c)(c9))2=36+(182c)2\sqrt{(9 - 3)^2 + ((9 - c) - (c - 9))^2} = \sqrt{36 + (18 - 2c)^2}. Setting this distance equal to 1010 and squaring both sides gives 36+(182c)2=10036 + (18 - 2c)^2 = 100, which simplifies to (182c)2=64(18 - 2c)^2 = 64. Taking the square root of both sides gives 182c=818 - 2c = 8 or 182c=818 - 2c = -8, yielding solutions of c=5c = 5 or c=13c = 13. Since the question specifies that c>10c > 10, the correct value must be 1313.

Adım Adım Çözüm

1
Rewrite the function f(x)=x26x+cf(x) = x^2 - 6x + c in vertex form by completing the square.
f(x)=(x3)2+c9f(x) = (x - 3)^2 + c - 9, which gives the coordinates of vertex AA as (3,c9)(3, c - 9).
Completing the square reveals the vertex (h,k)(h, k) of a quadratic function in the form y=a(xh)2+ky = a(x - h)^2 + k.
2
Determine the vertex BB of the graph of g(x)=f(x6)g(x) = -f(x - 6) by applying transformations to vertex A(3,c9)A(3, c - 9).
The horizontal shift of f(x6)f(x - 6) moves the vertex to (3+6,c9)=(9,c9)(3 + 6, c - 9) = (9, c - 9). The reflection of f(x6)-f(x - 6) negates the yy-coordinate of the vertex, resulting in B(9,9c)B(9, 9 - c).
The transformation f(xh)f(x - h) shifts a graph right by hh units, and the transformation f(x)-f(x) reflects it across the xx-axis.
3
Use the distance formula to set up an equation for the distance between A(3,c9)A(3, c - 9) and B(9,9c)B(9, 9 - c).
(93)2+((9c)(c9))2=10    62+(182c)2=10\sqrt{(9 - 3)^2 + ((9 - c) - (c - 9))^2} = 10 \implies \sqrt{6^2 + (18 - 2c)^2} = 10.
The distance dd between points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
4
Solve the radical equation 36+(182c)2=10\sqrt{36 + (18 - 2c)^2} = 10 for cc.
36+(182c)2=100    (182c)2=64    182c=836 + (18 - 2c)^2 = 100 \implies (18 - 2c)^2 = 64 \implies 18 - 2c = 8 or 182c=818 - 2c = -8. This yields c=5c = 5 or c=13c = 13.
Squaring both sides eliminates the square root, allowing us to solve the resulting quadratic equation.
5
Apply the constraint c>10c > 10 to choose the correct value for cc.
c=13c = 13.
The question specifies that cc must be greater than 1010, which excludes c=5c = 5.

Anahtar Kavram

Using vertex form of quadratic equations to determine vertex coordinates and applying transformations (horizontal shifts and vertical reflections) to find key graphical points.
Soru 260Soru

The mass of a radioactive sample, in grams, is modeled by the function M(t)=M0(18)t15M(t) = M_0 \left(\frac{1}{8}\right)^{\frac{t}{15}}, where M0M_0 is the initial mass of the sample and tt is the time, in minutes, since the sample started decaying. Which of the following functions correctly models the mass of the sample, in grams, in terms of ss seconds after the sample started decaying?

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Cevap: M(s)=M0(12)s300M(s) = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}

Cevap

The correct function modeling the mass in terms of seconds is M(s)=M0(12)s300M(s) = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}.
To express the mass in terms of ss seconds, we substitute t=s60t = \frac{s}{60} into the original decay function because the time in minutes is equal to the number of seconds divided by 60. This gives M(s)=M0(18)s60×15=M0(18)s900M(s) = M_0 \left(\frac{1}{8}\right)^{\frac{s}{60 \times 15}} = M_0 \left(\frac{1}{8}\right)^{\frac{s}{900}}. Next, since the options have a base of 12\frac{1}{2}, we rewrite the base 18\frac{1}{8} as (12)3\left(\frac{1}{2}\right)^3. Applying the power of a power rule (xa)b=xab(x^a)^b = x^{ab}, we multiply the exponent s900\frac{s}{900} by 3, which yields M(s)=M0(12)3×s900=M0(12)s300M(s) = M_0 \left(\frac{1}{2}\right)^{3 \times \frac{s}{900}} = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}.

Adım Adım Çözüm

1
Relate the time variables tt (in minutes) and ss (in seconds).
Since 1 minute is equivalent to 60 seconds, the relationship is t=s60t = \frac{s}{60}.
This substitution allows the function to take time input in seconds instead of minutes.
2
Substitute t=s60t = \frac{s}{60} into the exponent of the original model.
M(s)=M0(18)s/6015=M0(18)s900M(s) = M_0 \left(\frac{1}{8}\right)^{\frac{s/60}{15}} = M_0 \left(\frac{1}{8}\right)^{\frac{s}{900}}
This updates the function variable to seconds and simplifies the fractional exponent.
3
Rewrite the base 18\frac{1}{8} as a power of 12\frac{1}{2} and apply exponent rules.
Since 18=(12)3\frac{1}{8} = \left(\frac{1}{2}\right)^3, we write M(s)=M0((12)3)s900=M0(12)3×s900=M0(12)s300M(s) = M_0 \left(\left(\frac{1}{2}\right)^3\right)^{\frac{s}{900}} = M_0 \left(\frac{1}{2}\right)^{3 \times \frac{s}{900}} = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}.
Applying the power of a power rule, (am)n=amn(a^m)^n = a^{mn}, simplifies the function to its final form with a base of 12\frac{1}{2}.

Anahtar Kavram

Applying exponent rules to manipulate exponential bases and performing variable substitutions in contextual models.
Tahmini Süre:2m 0s
ÖncekiSayfa 13 / 22Sonraki
Advanced Math Alıştırma Soruları — SAT — Sayfa 13 | Examkin