Tüm alıştırma soruları

5556 soru

Soru 2121Soru

In the standard (x,y)(x,y) coordinate plane, the midpoint of a line segment is (2,5)(2, 5). If one of the endpoints of the segment is (2,1)(-2, 1), what is the xx-coordinate of the other endpoint?

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Cevap: 6

Cevap

The xx-coordinate of the other endpoint is 6.
According to the midpoint formula, the xx-coordinate of the midpoint is the average of the xx-coordinates of the two endpoints. Substituting the given values yields the equation 2=2+x222 = \frac{-2 + x_2}{2}. Multiplying both sides by 2 gives 4=2+x24 = -2 + x_2, and adding 2 to both sides results in x2=6x_2 = 6.

Adım Adım Çözüm

1
Set up the midpoint equation for the xx-coordinate using the midpoint formula xm=x1+x22x_m = \frac{x_1 + x_2}{2}.
2=2+x222 = \frac{-2 + x_2}{2}
The xx-coordinate of the midpoint is the average of the xx-coordinates of the endpoints.
2
Multiply both sides of the equation by 2 to solve for the numerator.
4=2+x24 = -2 + x_2
Multiplying by 2 eliminates the denominator on the right side.
3
Add 2 to both sides of the equation to isolate x2x_2.
x2=6x_2 = 6
Adding 2 to both sides isolates the variable x2x_2.

Anahtar Kavram

Midpoint Formula
Soru 2122Soru

An online streaming service offers a basic monthly plan that costs 1212 dollars plus 1.501.50 dollars for each premium movie rented. A premium monthly plan costs 2020 dollars and includes 33 free premium movie rentals, with each additional rental costing 1.001.00 dollar. If a subscriber rents xx premium movies in a month, where x>3x > 3, which of the following expressions represents the savings, in dollars, of the premium plan compared to the basic plan?

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Cevap: 0.50x50.50x - 5

Cevap

0.50x50.50x - 5
The cost of the basic plan for xx rentals is 12+1.50x12 + 1.50x dollars. Under the premium plan, the first 33 rentals are free, so the subscriber only pays for x3x - 3 rentals. The cost of the premium plan is 20+1.00(x3)20 + 1.00(x - 3) dollars, which simplifies to 17+x17 + x dollars. The savings of the premium plan compared to the basic plan is the difference between the basic cost and the premium cost: (12+1.50x)(17+x)=0.50x5(12 + 1.50x) - (17 + x) = 0.50x - 5 dollars.

Adım Adım Çözüm

1
Write the expression for the cost of the basic monthly plan.
12+1.50x12 + 1.50x
The basic plan has a flat fee of 1212 dollars and a rate of 1.501.50 dollars per movie for all xx movies rented.
2
Write the expression for the cost of the premium monthly plan.
20+1.00(x3)20 + 1.00(x - 3), which simplifies to 17+x17 + x.
The premium plan has a flat fee of 2020 dollars. Since 33 rentals are free, the subscriber only pays 1.001.00 dollar each for the remaining x3x - 3 rentals.
3
Subtract the premium plan cost from the basic plan cost to find the savings.
(12+1.50x)(17+x)=0.50x5(12 + 1.50x) - (17 + x) = 0.50x - 5
Savings represents the difference in cost: Basic Cost minus Premium Cost.

Anahtar Kavram

Translating verbal descriptions of multi-step costs into linear algebraic expressions and simplifying them.
Tahmini Süre:1m 30s
Soru 2123Soru

In the standard (x,y)(x, y) coordinate plane, a line L1L_1 with a positive slope mm and a negative yy-intercept bb passes through the point (4,3)(4, 3). The region in the fourth quadrant bounded by the line L1L_1, the xx-axis, and the yy-axis has an area of exactly 88 square units. What is the yy-coordinate of the yy-intercept of line L1L_1?

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Cevap: -6

Cevap

The y-coordinate of the y-intercept of line L1L_1 is 6-6.
The correct answer is 6-6. Substituting (4,3)(4, 3) into the slope-intercept equation y=mx+by = mx + b gives 3=4m+b3 = 4m + b, or m=3b4m = \frac{3-b}{4}. The area of the right triangle in the fourth quadrant is 12×base×height=12(bm)(b)=b22m=8\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (-\frac{b}{m})(-b) = \frac{b^2}{2m} = 8. Substituting mm yields b2=16(3b4)b^2 = 16\left(\frac{3-b}{4}\right), which simplifies to the quadratic equation b2+4b12=0b^2 + 4b - 12 = 0. Factoring gives (b+6)(b2)=0(b+6)(b-2) = 0. Since the y-intercept bb must be negative, we have b=6b = -6.

Adım Adım Çözüm

1
Substitute the given point into the slope-intercept equation
m=3b4m = \frac{3 - b}{4}
Since the line passes through (4,3)(4, 3), substituting these coordinates into y=mx+by = mx + b allows us to express the slope mm in terms of the y-intercept bb.
2
Determine the intercepts and the dimensions of the bounded region
Base =bm= -\frac{b}{m} and Height =b= -b
The boundary of the region in the fourth quadrant is a right triangle formed by the origin, the x-intercept (bm,0)(-\frac{b}{m}, 0), and the y-intercept (0,b)(0, b).
3
Set up the area of the triangle and equate it to 8
b2=16mb^2 = 16m
The area of a right triangle is 12×base×height\frac{1}{2} \times \text{base} \times \text{height}, so 12(bm)(b)=8\frac{1}{2} \left(-\frac{b}{m}\right)(-b) = 8 simplifies to b2=16mb^2 = 16m.
4
Substitute mm into the area equation and solve the resulting quadratic equation
b=6b = -6 (discarding b=2b = 2)
Substituting m=3b4m = \frac{3 - b}{4} yields b2+4b12=0b^2 + 4b - 12 = 0, which factors into (b+6)(b2)=0(b + 6)(b - 2) = 0. Since the region is in the fourth quadrant, the y-intercept must be negative (b<0b < 0).

Anahtar Kavram

Using linear equation intercepts to calculate bounded areas on the coordinate plane and relating variables using point substitution.
Soru 2124Soru

In the standard (x,y)(x, y) coordinate plane, a triangle has vertices A(1,2)A(1, 2), B(7,2)B(7, 2), and CC. The midpoint of side ACAC lies on the line y=3x3y = 3x - 3, and the midpoint of side BCBC lies on the line y=2x+13y = -2x + 13. What is the distance between point CC and the midpoint of side ABAB?

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Cevap: 5\sqrt{5}

Cevap

The distance between point CC and the midpoint of side ABAB is 5\sqrt{5}.
The coordinates of point C(3,4)C(3, 4) are determined by setting up the midpoint coordinates for sides ACAC and BCBC and substituting them into their respective line equations. The midpoint of ABAB is calculated to be (4,2)(4, 2). Using the distance formula between C(3,4)C(3, 4) and (4,2)(4, 2) yields (43)2+(24)2=5\sqrt{(4-3)^2 + (2-4)^2} = \sqrt{5}.

Adım Adım Çözüm

1
Express the midpoint of side ACAC in terms of the unknown coordinates of point C(x,y)C(x, y) and substitute it into the given line equation.
The midpoint of ACAC is MAC=(x+12,y+22)M_{AC} = \left(\frac{x+1}{2}, \frac{y+2}{2}\right). Substituting this into y=3x3y = 3x - 3 yields: y+22=3(x+12)3\frac{y+2}{2} = 3\left(\frac{x+1}{2}\right) - 3, which simplifies to y=3x5y = 3x - 5.
Since the midpoint of ACAC lies on the line y=3x3y = 3x - 3, its coordinates must satisfy the equation of the line.
2
Express the midpoint of side BCBC in terms of the unknown coordinates of point C(x,y)C(x, y) and substitute it into the second given line equation.
The midpoint of BCBC is MBC=(x+72,y+22)M_{BC} = \left(\frac{x+7}{2}, \frac{y+2}{2}\right). Substituting this into y=2x+13y = -2x + 13 yields: y+22=2(x+72)+13\frac{y+2}{2} = -2\left(\frac{x+7}{2}\right) + 13, which simplifies to y=2x+10y = -2x + 10.
Since the midpoint of BCBC lies on the line y=2x+13y = -2x + 13, its coordinates must satisfy this equation.
3
Solve the system of two linear equations to find the coordinates of point C(x,y)C(x, y).
Equating the two expressions for yy: 3x5=2x+105x=15x=33x - 5 = -2x + 10 \Rightarrow 5x = 15 \Rightarrow x = 3. Substituting x=3x = 3 back into the first equation: y=3(3)5=4y = 3(3) - 5 = 4. Thus, C=(3,4)C = (3, 4).
Point CC must simultaneously satisfy the midpoint constraints on both sides ACAC and BCBC.
4
Find the coordinates of the midpoint of side ABAB.
The midpoint of segment ABAB with endpoints A(1,2)A(1, 2) and B(7,2)B(7, 2) is MAB=(1+72,2+22)=(4,2)M_{AB} = \left(\frac{1+7}{2}, \frac{2+2}{2}\right) = (4, 2).
The question asks for the distance between point CC and the midpoint of ABAB, so we need to determine the coordinates of this midpoint first.
5
Calculate the distance between point C(3,4)C(3, 4) and the midpoint MAB(4,2)M_{AB}(4, 2) using the distance formula.
The distance dd is: d=(43)2+(24)2=12+(2)2=1+4=5d = \sqrt{(4-3)^2 + (2-4)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}.
The distance formula d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} is used to find the straight-line distance between two points in a coordinate plane.

Anahtar Kavram

Applying the midpoint and distance formulas within coordinate geometry constraint systems.

Alternatif Yöntem

Instead of algebraically solving for the lines of midpoints, one can translate the lines using vectors. The set of possible points CC when the midpoint of ACAC lies on line L1L_1 is a line L1L'_1 obtained by dilating L1L_1 by a factor of 2 with respect to center AA. Dilating y=3x3y = 3x - 3 from A(1,2)A(1, 2) gives the line y=3x5y = 3x - 5. Similarly, dilating y=2x+13y = -2x + 13 from B(7,2)B(7, 2) by a factor of 2 gives y=2x+10y = -2x + 10. The intersection of these two dilated lines is point C(3,4)C(3, 4).
Tahmini Süre:2m 30s
Soru 2125Soru

In the standard (x,y)(x,y) coordinate plane, a line with a negative slope passes through the point (3,2)(3, 2) and has positive integer intercepts (a,0)(a, 0) and (0,b)(0, b). What is the sum of all possible values of aa?

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Cevap: 24

Cevap

The sum of all possible values of aa is 24.
The correct answer is the sum 24. Writing the line's equation in intercept form gives xa+yb=1\frac{x}{a} + \frac{y}{b} = 1. Substituting the point (3,2)(3, 2) yields 3a+2b=1\frac{3}{a} + \frac{2}{b} = 1. Solving for bb results in b=2+6a3b = 2 + \frac{6}{a-3}. Because both aa and bb must be positive integers, the expression a3a-3 must be a positive divisor of 66. The positive divisors of 66 are 1,2,3,1, 2, 3, and 66, which correspond to the aa values of 4,5,6,4, 5, 6, and 99. Summing these values gives 4+5+6+9=244 + 5 + 6 + 9 = 24.

Adım Adım Çözüm

1
Set up the intercept form of the linear equation.
xa+yb=1\frac{x}{a} + \frac{y}{b} = 1
A line with xx-intercept (a,0)(a,0) and yy-intercept (0,b)(0,b) can be written in intercept form.
2
Substitute the given point (3,2)(3, 2) into the equation.
3a+2b=1\frac{3}{a} + \frac{2}{b} = 1
Since the line passes through (3,2)(3,2), these coordinates must satisfy the equation.
3
Solve the equation for bb in terms of aa.
2b=13a    2b=a3a    b=2aa3\frac{2}{b} = 1 - \frac{3}{a} \implies \frac{2}{b} = \frac{a-3}{a} \implies b = \frac{2a}{a-3}
Expressing bb in terms of aa helps analyze the integer constraints.
4
Rewrite the expression for bb to isolate the fractional part.
b=2(a3)+6a3=2+6a3b = \frac{2(a-3) + 6}{a-3} = 2 + \frac{6}{a-3}
This form allows us to see when bb will be an integer based on the divisors of the numerator.
5
Determine the positive integer solutions for aa and bb.
a3a-3 must be a positive divisor of 66. The positive divisors of 66 are 1,2,3,1, 2, 3, and 66. This yields:
- If a3=1    a=4,b=8a-3=1 \implies a=4, b=8
- If a3=2    a=5,b=5a-3=2 \implies a=5, b=5
- If a3=3    a=6,b=4a-3=3 \implies a=6, b=4
- If a3=6    a=9,b=3a-3=6 \implies a=9, b=3
Since aa and bb must be positive integers, a3a-3 must be positive and divide 6 evenly. Negative divisors (like 1-1 or 2-2) would make bb negative or zero.
6
Sum the possible values of aa.
4+5+6+9=244 + 5 + 6 + 9 = 24
To find the final answer, we sum all the valid xx-intercept values.

Anahtar Kavram

Using the intercept form of a linear equation and applying integer constraints to find coordinates.
Soru 2126Soru

If xx and yy are positive real numbers, what is the simplified form of the expression below?

4x1/2y3x1/3y12x1y\frac{4 x^{1/2} y^3 \cdot x^{1/3} y^{-1}}{2 x^{-1} y}
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Cevap: 2x11/6y2x^{11/6}y

Cevap

The simplified form of the expression is 2x11/6y2x^{11/6}y.
The correct answer is 2x11/6y2x^{11/6}y. First, simplify the constant coefficients to obtain 42=2\frac{4}{2} = 2. In the numerator, combine the bases by adding exponents: x1/2x1/3=x1/2+1/3=x5/6x^{1/2} \cdot x^{1/3} = x^{1/2 + 1/3} = x^{5/6}, and y3y1=y31=y2y^3 \cdot y^{-1} = y^{3 - 1} = y^2. Next, divide by the terms in the denominator using the quotient rule: x5/6x1=x5/6(1)=x11/6\frac{x^{5/6}}{x^{-1}} = x^{5/6 - (-1)} = x^{11/6}, and y2y1=y21=y\frac{y^2}{y^1} = y^{2 - 1} = y. This yields the fully simplified expression 2x11/6y2x^{11/6}y.

Adım Adım Çözüm

1
Simplify the coefficients of the fraction.
42=2\frac{4}{2} = 2
Dividing the numerical constants in the numerator and denominator simplifies the constant multiplier of the expression.
2
Combine the xx terms in the numerator using the product rule.
x1/2x1/3=x1/2+1/3=x5/6x^{1/2} \cdot x^{1/3} = x^{1/2 + 1/3} = x^{5/6}
When multiplying terms with the same base, add their exponents: 12+13=36+26=56\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}.
3
Combine the yy terms in the numerator using the product rule.
y3y1=y3+(1)=y2y^3 \cdot y^{-1} = y^{3 + (-1)} = y^2
When multiplying terms with the same base, add their exponents: 31=23 - 1 = 2.
4
Simplify the xx terms in the fraction using the quotient rule.
x5/6x1=x5/6(1)=x5/6+1=x11/6\frac{x^{5/6}}{x^{-1}} = x^{5/6 - (-1)} = x^{5/6 + 1} = x^{11/6}
When dividing terms with the same base, subtract the denominator's exponent from the numerator's exponent.
5
Simplify the yy terms in the fraction using the quotient rule.
y2y1=y21=y\frac{y^2}{y^1} = y^{2 - 1} = y
When dividing terms with the same base, subtract the denominator's exponent from the numerator's exponent.
6
Combine the simplified components.
2x11/6y2x^{11/6}y
Multiply the simplified coefficient, xx term, and yy term together to get the final simplified expression.

Anahtar Kavram

Properties of Exponents in Algebraic Expressions
Tahmini Süre:1m 30s
Soru 2127Soru

A parabola in the standard (x,y)(x, y) coordinate plane is defined by the equation x26x8y+25=0x^2 - 6x - 8y + 25 = 0. What is the distance, in coordinate units, between the focus and the directrix of this parabola?

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Cevap: 4

Cevap

The distance between the focus and the directrix of the parabola is 4.
By completing the square on the equation x26x8y+25=0x^2 - 6x - 8y + 25 = 0, we get (x3)2=8(y2)(x-3)^2 = 8(y-2). Since the coefficient of the linear factor is 88, we set 4p=84p = 8, which yields p=2p = 2. The distance from the focus to the directrix is 2p=2(2)=42p = 2(2) = 4.

Adım Adım Çözüm

1
Isolate the terms containing xx on one side of the equation.
x26x=8y25x^2 - 6x = 8y - 25
To set up the equation for completing the square on the xx terms.
2
Complete the square for the quadratic expression in xx by adding 99 to both sides.
x26x+9=8y16    (x3)2=8y16x^2 - 6x + 9 = 8y - 16 \implies (x-3)^2 = 8y - 16
Adding (6/2)2=9( -6/2 )^2 = 9 creates a perfect square trinomial on the left side.
3
Factor out the coefficient of yy on the right side to write the equation in standard form.
(x3)2=8(y2)(x-3)^2 = 8(y-2)
This matches the standard form equation (xh)2=4p(yk)(x-h)^2 = 4p(y-k) for a vertical parabola.
4
Determine the value of the focal parameter pp from the standard form.
4p=8    p=24p = 8 \implies p = 2
Comparing the standard form coefficient 4p4p with the value 88 gives p=2p = 2.
5
Calculate the total distance between the focus and the directrix.
2p=2(2)=42p = 2(2) = 4
The vertex is situated halfway between the focus and the directrix, making the distance between them 2p2p.

Anahtar Kavram

Finding the geometric properties of a parabola by completing the square to convert its general equation to standard form.
Soru 2128Soru

In the standard (x,y)(x,y) coordinate plane, a line is defined by the equation 3x2y=123x - 2y = 12. What is the sum of the xx-intercept and the yy-intercept of this line?

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Cevap: -2

Cevap

The sum of the xx-intercept and the yy-intercept of the line is 2-2.
To find the xx-intercept, set y=0y = 0 in the equation 3x2y=123x - 2y = 12, which gives 3x=123x = 12, so x=4x = 4. To find the yy-intercept, set x=0x = 0, which gives 2y=12-2y = 12, so y=6y = -6. Adding these two values together gives 4+(6)=24 + (-6) = -2.

Adım Adım Çözüm

1
Find the xx-intercept by setting y=0y = 0 in the equation.
Substitute y=0y = 0 into 3x2y=123x - 2y = 12 to get 3x2(0)=123x - 2(0) = 12, which simplifies to 3x=123x = 12, yielding x=4x = 4.
The xx-intercept of a line is the point where the line crosses the xx-axis, which occurs when the yy-coordinate is 00.
2
Find the yy-intercept by setting x=0x = 0 in the equation.
Substitute x=0x = 0 into 3x2y=123x - 2y = 12 to get 3(0)2y=123(0) - 2y = 12, which simplifies to 2y=12-2y = 12, yielding y=6y = -6.
The yy-intercept of a line is the point where the line crosses the yy-axis, which occurs when the xx-coordinate is 00.
3
Add the xx-intercept and yy-intercept together.
Calculate 4+(6)=24 + (-6) = -2.
The question asks for the sum of the xx-intercept and the yy-intercept.

Anahtar Kavram

Finding the xx- and yy-intercepts of a linear equation in standard form.
Tahmini Süre:1m 0s
Soru 2129Soru

If 82x1=(14)x38^{2x - 1} = \left(\frac{1}{4}\right)^{x - 3}, what is the value of xx?

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Cevap: 98\frac{9}{8}

Cevap

The correct answer is 98\frac{9}{8}.
By writing both sides of the equation with a common base of 22, we get (23)2x1=(22)x3(2^3)^{2x-1} = (2^{-2})^{x-3}. Applying the power-to-a-power exponent rule, this simplifies to 26x3=22x+62^{6x-3} = 2^{-2x+6}. Since the bases are now identical, their exponents must be equal: 6x3=2x+66x-3 = -2x+6. Adding 2x2x and 33 to both sides results in 8x=98x = 9, which gives x=98x = \frac{9}{8}.

Adım Adım Çözüm

1
Express both bases as powers of 22.
8=238 = 2^3 and 14=22\frac{1}{4} = 2^{-2}, so the equation becomes (23)2x1=(22)x3(2^3)^{2x-1} = (2^{-2})^{x-3}.
Finding a common base allows us to equate the exponents directly.
2
Apply the power-to-a-power rule (am)n=amn(a^m)^n = a^{mn} to simplify the exponents.
23(2x1)=22(x3)26x3=22x+62^{3(2x-1)} = 2^{-2(x-3)} \Rightarrow 2^{6x-3} = 2^{-2x+6}.
This simplifies the exponential expressions on both sides of the equation.
3
Equate the exponents and solve for xx.
6x3=2x+68x=9x=986x - 3 = -2x + 6 \Rightarrow 8x = 9 \Rightarrow x = \frac{9}{8}.
Since the bases are equal, their exponents must be equal.

Anahtar Kavram

Solving exponential equations by finding a common base and applying exponent properties.
Tahmini Süre:1m 30s
Soru 2130Soru

An ellipse in the standard (x,y)(x, y) coordinate plane is defined by the equation x2+4y2+6x8y+9=0x^2 + 4y^2 + 6x - 8y + 9 = 0. What is the length of the major axis of this ellipse?

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Cevap: 4

Cevap

4
To find the length of the major axis, rewrite the general equation of the ellipse in standard form by completing the square. Grouping the terms yields (x2+6x)+4(y22y)=9(x^2 + 6x) + 4(y^2 - 2y) = -9. Completing the square for both variables gives (x+3)29+4[(y1)21]=9(x+3)^2 - 9 + 4[(y-1)^2 - 1] = -9, which simplifies to (x+3)2+4(y1)2=4(x+3)^2 + 4(y-1)^2 = 4. Dividing both sides by 4 gives the standard form (x+3)24+(y1)21=1\frac{(x+3)^2}{4} + \frac{(y-1)^2}{1} = 1. In this form, the horizontal axis is the major axis because the denominator under the xx-term (a2=4a^2 = 4) is larger than the denominator under the yy-term (b2=1b^2 = 1). Since a2=4a^2 = 4, the semi-major axis is a=2a = 2. Therefore, the total length of the major axis is 2a=2(2)=42a = 2(2) = 4.

Adım Adım Çözüm

1
Group the xx-terms and yy-terms and move the constant to the right side of the equation.
(x2+6x)+(4y28y)=9(x^2 + 6x) + (4y^2 - 8y) = -9
Grouping like terms allows completing the square for each variable independently.
2
Factor out the coefficient of y2y^2 from the yy-terms.
(x2+6x)+4(y22y)=9(x^2 + 6x) + 4(y^2 - 2y) = -9
Before completing the square, the leading coefficient of the squared terms inside the parentheses must be 1.
3
Complete the square for both the xx and yy expressions by adding and subtracting the square of half of the linear coefficients.
((x+3)29)+4((y1)21)=9((x+3)^2 - 9) + 4((y-1)^2 - 1) = -9
This rewrites the quadratic expressions into perfect square trinomial form.
4
Distribute the coefficients and simplify the constant terms.
(x+3)29+4(y1)24=9    (x+3)2+4(y1)213=9    (x+3)2+4(y1)2=4(x+3)^2 - 9 + 4(y-1)^2 - 4 = -9 \implies (x+3)^2 + 4(y-1)^2 - 13 = -9 \implies (x+3)^2 + 4(y-1)^2 = 4
Isolating the squared terms on one side helps convert the equation to the standard form of an ellipse.
5
Divide both sides of the equation by 4 to set the right side equal to 1.
(x+3)24+(y1)21=1\frac{(x+3)^2}{4} + \frac{(y-1)^2}{1} = 1
The standard form of a horizontal ellipse is (xh)2a2+(yk)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1.
6
Identify the values of a2a^2 and b2b^2 and calculate the length of the major axis.
a2=4    a=2a^2 = 4 \implies a = 2. The major axis length is 2a=2(2)=42a = 2(2) = 4.
The length of the major axis is twice the length of the semi-major axis (aa).

Anahtar Kavram

Rewriting the general equation of an ellipse into standard form by completing the square to find its key features, such as the length of the major axis.

Alternatif Yöntem

Another way to find the length of the major axis is to find the vertices of the ellipse by finding the maximum and minimum x-values where the equation has real solutions for y, though completing the square is the standard and most direct method.
Tahmini Süre:1m 30s
Soru 2131Soru

In the standard (x,y)(x, y) coordinate plane, a parabola is defined by the equation y=x2+2x+7y = x^2 + 2x + 7 and a line is defined by the equation y=mx+3y = mx + 3, where mm is a constant. If the system of equations consisting of this parabola and line has exactly one real solution, and this solution lies in the first quadrant, what is the value of mm?

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Cevap: 6

Cevap

6
The correct answer is 66. When setting the equations of the parabola and line equal, we get x2+(2m)x+4=0x^2 + (2-m)x + 4 = 0. Setting the discriminant to zero yields (2m)216=0(2-m)^2 - 16 = 0, which gives m=6m = 6 or m=2m = -2. Substituting m=6m = 6 back gives a single intersection point of (2,15)(2, 15), which is in the first quadrant since both coordinates are positive. The other value, m=2m = -2, gives an intersection point of (2,7)(-2, 7), which is in the second quadrant.

Adım Adım Çözüm

1
Set the equations of the parabola and the line equal to each other to find their intersection points.
x2+2x+7=mx+3x^2 + 2x + 7 = mx + 3
To find the coordinates where the two graphs intersect.
2
Rearrange the equation into standard quadratic form: ax2+bx+c=0ax^2 + bx + c = 0.
x2+(2m)x+4=0x^2 + (2 - m)x + 4 = 0
To express the intersection condition as a single quadratic equation where we can analyze the number of solutions.
3
Set the discriminant of the quadratic equation to zero.
Δ=(2m)24(1)(4)=0    (2m)216=0\Delta = (2 - m)^2 - 4(1)(4) = 0 \implies (2 - m)^2 - 16 = 0
For the system to have exactly one real solution, the quadratic equation must have a discriminant of zero (tangency).
4
Solve for the possible values of mm.
(2m)2=16    2m=±4(2 - m)^2 = 16 \implies 2 - m = \pm 4, giving m=2m = -2 or m=6m = 6.
To find all values of mm that result in exactly one intersection point.
5
Find the intersection point for each value of mm and determine which lies in the first quadrant.
For m=2m = -2, (x+2)2=0    x=2(x+2)^2 = 0 \implies x = -2 and y=2(2)+3=7y = -2(-2)+3 = 7, yielding (2,7)(-2, 7) (Quadrant II). For m=6m = 6, (x2)2=0    x=2(x-2)^2 = 0 \implies x = 2 and y=6(2)+3=15y = 6(2)+3 = 15, yielding (2,15)(2, 15) (Quadrant I).
To satisfy the condition that the single solution must lie in the first quadrant (where both x>0x > 0 and y>0y > 0).

Anahtar Kavram

Solving systems of linear and quadratic equations and applying the discriminant to find conditions for tangency.
Tahmini Süre:2m 0s
Soru 2132Soru

A thermometer is considered accurate if its temperature reading, TT degrees Fahrenheit, differs from the actual temperature by less than 1.5F1.5^\circ\text{F}. If the actual temperature is 72.0F72.0^\circ\text{F}, which of the following inequalities represents the range of reading temperatures, TT, that are NOT considered accurate?

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Cevap: T72.01.5|T - 72.0| \ge 1.5

Cevap

T72.01.5|T - 72.0| \ge 1.5
The difference between the temperature reading, TT, and the actual temperature of 72.0F72.0^\circ\text{F} is represented by T72.0|T - 72.0|. A thermometer is accurate when this difference is less than 1.5F1.5^\circ\text{F} (T72.0<1.5|T - 72.0| < 1.5). Therefore, the thermometer is not accurate when the difference is greater than or equal to 1.5F1.5^\circ\text{F}, which is written as T72.01.5|T - 72.0| \ge 1.5.

Adım Adım Çözüm

1
Represent the difference between the reading and the actual temperature.
T72.0|T - 72.0|
The difference between the reading temperature and the actual temperature of 72.0F72.0^\circ\text{F} is given by the absolute value expression, representing the distance between the two temperatures on a thermometer.
2
Formulate the condition for an accurate reading.
T72.0<1.5|T - 72.0| < 1.5
An accurate reading differs from the actual temperature by less than 1.5F1.5^\circ\text{F}.
3
Determine the complement condition for a reading that is NOT accurate.
T72.01.5|T - 72.0| \ge 1.5
To find the range of readings that are NOT accurate, we take the complement of the accurate condition. The opposite of 'less than 1.51.5' is 'greater than or equal to 1.51.5'.

Anahtar Kavram

Absolute Value Inequalities in Real-World Contexts
Tahmini Süre:1m 0s
Soru 2133Soru

In the standard (x,y)(x, y) coordinate plane, a line LL passes through the point (2,3)(2, -3) and has a yy-intercept of (0,b)(0, b), where b>0b > 0. If the area of the triangular region bounded by the line LL, the xx-axis, and the yy-axis is 44 square units, what is the value of bb?

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Cevap: 6

Cevap

6
The correct answer is the option containing 6. Using the two points (2,3)(2, -3) and (0,b)(0, b), we find the slope of the line is m=b+32m = -\frac{b+3}{2}, which gives the equation y=b+32x+by = -\frac{b+3}{2}x + b. Setting y=0y=0 shows that the xx-intercept is at x=2bb+3x = \frac{2b}{b+3}. The area of the right triangle formed by the intercepts and the origin is 12×base×height=12×2bb+3×b=b2b+3\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{2b}{b+3} \times b = \frac{b^2}{b+3}. Setting this area equal to 44 yields b24b12=0b^2 - 4b - 12 = 0. Factoring gives (b6)(b+2)=0(b - 6)(b + 2) = 0. Since we are given b>0b > 0, bb must be 66.

Adım Adım Çözüm

1
Find the slope of the line LL in terms of bb.
The slope mm is given by 3b20=b+32\frac{-3 - b}{2 - 0} = -\frac{b+3}{2}.
The line passes through (2,3)(2, -3) and its yy-intercept (0,b)(0, b).
2
Write the equation of the line LL and determine its xx-intercept.
The equation is y=b+32x+by = -\frac{b+3}{2}x + b. Setting y=0y = 0 gives the xx-intercept x=2bb+3x = \frac{2b}{b+3}.
The xx-intercept is the point where the line crosses the xx-axis (y=0y = 0).
3
Set up the area equation for the triangle formed by the axes and the line.
The area is 12×base×height=12×2bb+3×b=b2b+3=4\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{2b}{b+3} \times b = \frac{b^2}{b+3} = 4.
The base of the triangle is the xx-intercept, and the height is the yy-intercept bb (since both are positive for b>0b > 0).
4
Solve the quadratic equation for bb.
b2=4(b+3)b24b12=0(b6)(b+2)=0b^2 = 4(b + 3) \Rightarrow b^2 - 4b - 12 = 0 \Rightarrow (b - 6)(b + 2) = 0. Since b>0b > 0, b=6b = 6.
We must solve the equation and select the positive solution because the problem specifies b>0b > 0.

Anahtar Kavram

Using the coordinates of a point and intercepts to write a linear equation, finding intercepts, and calculating the area of a coordinate triangle.

Alternatif Yöntem

Instead of setting up the area algebraically first, you can test the answer choices. For example, testing the correct value 6: the y-intercept is (0,6)(0, 6). The slope of the line passing through (0,6)(0, 6) and (2,3)(2, -3) is m=3620=4.5m = \frac{-3 - 6}{2 - 0} = -4.5. The equation of the line is y=4.5x+6y = -4.5x + 6. The x-intercept is found by setting y=0y = 0, giving x=64.5=43x = \frac{6}{4.5} = \frac{4}{3}. The area of the triangle is 12×43×6=4\frac{1}{2} \times \frac{4}{3} \times 6 = 4, which matches the given area of 4 square units.
Tahmini Süre:1m 30s
Soru 2134Soru

The equation of a parabola is given by (x4)2=12(y+1)(x - 4)^2 = 12(y + 1). What is the yy-coordinate of the focus of this parabola?

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Cevap: 2

Cevap

The correct answer is 2.
The equation (x4)2=12(y+1)(x - 4)^2 = 12(y + 1) is a parabola with a vertical axis of symmetry, vertex at (4,1)(4, -1), and focal length p=3p = 3. The focus is located pp units above the vertex, yielding a yy-coordinate of 1+3=2-1 + 3 = 2.

Adım Adım Çözüm

1
Identify the standard form of the parabola's equation.
The equation (x4)2=12(y+1)(x - 4)^2 = 12(y + 1) matches (xh)2=4p(yk)(x - h)^2 = 4p(y - k).
This form allows us to find the vertex and the focal distance pp directly.
2
Determine the vertex and focal distance pp.
The vertex is (4,1)(4, -1) and p=3p = 3 since 4p=124p = 12.
Matching the given equation terms to the standard form reveals these properties.
3
Find the coordinates of the focus.
The focus is at (4,2)(4, 2).
The focus is located pp units vertically above the vertex for a parabola opening upward.

Anahtar Kavram

Focus of a Parabola
Soru 2135Soru

The measures of the three interior angles of a triangle are in the ratio 2:3:52:3:5. What is the measure of the largest angle in the triangle?

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Cevap: 9090^\circ

Cevap

The correct answer is 9090^\circ.
Since the interior angles of a triangle sum to 180180^\circ and their ratio is 2:3:52:3:5, the sum of the ratio parts is 2+3+5=102 + 3 + 5 = 10. Dividing the total 180180^\circ by 10 parts yields 1818^\circ per ratio unit. The largest angle corresponds to the largest part of the ratio, which is 5. Therefore, the measure of the largest angle is 5×18=905 \times 18^\circ = 90^\circ.

Adım Adım Çözüm

1
Find the total number of parts in the ratio by adding the terms together.
The total number of parts is 2+3+5=102 + 3 + 5 = 10 parts.
This determines how many equal units the total angle measure is divided into.
2
Divide the total sum of the interior angles of a triangle by the total number of parts to find the degree measure of one part.
The sum of the interior angles of a triangle is 180180^\circ. Dividing by the total parts gives 18010=18\frac{180^\circ}{10} = 18^\circ per part.
The Triangle Angle Sum Theorem states that the interior angles of a triangle always sum to 180180^\circ.
3
Multiply the value of one part by the ratio term representing the largest angle.
The largest angle is represented by the term 5, so its measure is 5×18=905 \times 18^\circ = 90^\circ.
This gives the measure of the largest angle in the triangle.

Anahtar Kavram

The interior angles of a triangle always sum to 180180^\circ, and individual angle measures can be found from a given ratio by dividing the total degrees by the sum of the ratio parts.
Soru 2136Soru

What is the product of all real values of xx that satisfy the equation log3(x)6logx(3)=1\log_3(x) - 6\log_x(3) = 1?

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Cevap: 3

Cevap

The product of all real values of xx that satisfy the equation is 3.
By applying the change-of-base formula, the equation becomes log3(x)6log3(x)=1\log_3(x) - \frac{6}{\log_3(x)} = 1. Substituting y=log3(x)y = \log_3(x) leads to y2y6=0y^2 - y - 6 = 0, which has roots y=3y = 3 and y=2y = -2. These roots correspond to x=33=27x = 3^3 = 27 and x=32=19x = 3^{-2} = \frac{1}{9}. Both solutions are valid because they are positive and do not equal 1. The product of these solutions is 27×19=327 \times \frac{1}{9} = 3. Alternatively, using Vieta's formulas, the sum of the roots of the quadratic equation is y1+y2=1y_1 + y_2 = 1. The product of the solutions is x1x2=3y13y2=3y1+y2=31=3x_1 x_2 = 3^{y_1} \cdot 3^{y_2} = 3^{y_1 + y_2} = 3^1 = 3.

Adım Adım Çözüm

1
Apply the change-of-base formula to rewrite the variable base term.
log3(x)6log3(x)=1\log_3(x) - \frac{6}{\log_3(x)} = 1
This expresses the equation in terms of logarithms with the same base.
2
Use substitution to convert the equation into a quadratic form.
y6y=1y2y6=0y - \frac{6}{y} = 1 \Rightarrow y^2 - y - 6 = 0 where y=log3(x)y = \log_3(x)
Substitution simplifies the logarithmic equation into a polynomial equation.
3
Solve the quadratic equation by factoring.
(y3)(y+2)=0y=3(y-3)(y+2) = 0 \Rightarrow y = 3 or y=2y = -2
Finding the roots for yy is the intermediate step to solving for xx.
4
Back-substitute to find the values of xx.
x=33=27x = 3^3 = 27 and x=32=19x = 3^{-2} = \frac{1}{9}
Converting from logarithmic form back to exponential form yields the values of xx.
5
Multiply the solutions together.
27×19=327 \times \frac{1}{9} = 3
The question asks for the product of all real solutions.

Anahtar Kavram

Solving logarithmic equations using the change-of-base formula and quadratic substitution.

Alternatif Yöntem

Instead of solving for individual values of xx, note that if y1y_1 and y2y_2 are the roots of the quadratic equation y2y6=0y^2 - y - 6 = 0, then y1+y2=1y_1 + y_2 = 1 by Vieta's formulas. Since x1=3y1x_1 = 3^{y_1} and x2=3y2x_2 = 3^{y_2}, the product of the solutions is x1x2=3y13y2=3y1+y2=31=3x_1 x_2 = 3^{y_1} \cdot 3^{y_2} = 3^{y_1 + y_2} = 3^1 = 3.
Tahmini Süre:2m 0s
Soru 2137Soru

In ABC\triangle ABC, the measure of exterior angle BCD\angle BCD is 115115^\circ. If the measure of interior angle A\angle A is 4545^\circ, what is the measure, in degrees, of interior angle B\angle B?

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Cevap: 70

Cevap

The measure of interior angle B\angle B is 7070 degrees.
By the Exterior Angle Theorem, the measure of exterior angle BCD\angle BCD is equal to the sum of the two remote interior angles, A\angle A and B\angle B. We can write this relationship as mBCD=mA+mBm\angle BCD = m\angle A + m\angle B. Substituting 115115^\circ for mBCDm\angle BCD and 4545^\circ for mAm\angle A gives 115=45+mB115 = 45 + m\angle B. Solving for mBm\angle B yields 7070^\circ.

Adım Adım Çözüm

1
Set up the equation using the Exterior Angle Theorem.
mBCD=mA+mBm\angle BCD = m\angle A + m\angle B
The measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles.
2
Substitute the given measurements into the equation.
115=45+mB115 = 45 + m\angle B
The exterior angle BCD\angle BCD measures 115115^\circ and the remote interior angle A\angle A measures 4545^\circ.
3
Solve for the unknown angle measure by subtraction.
mB=70m\angle B = 70
Subtracting 4545 from both sides isolates mBm\angle B.

Anahtar Kavram

Exterior Angle Theorem
Soru 2138Soru

In the standard (x,y)(x, y) coordinate plane, a line passes through the point (3,2)(3, 2) and has a yy-intercept of 4-4. If the point (k,5k+2)(k, 5k + 2) also lies on this line, what is the value of kk?

Cevabı ve açıklamayı göster

Cevap: -2

Cevap

The value of kk is 2-2.
The line has a slope of 22 and a y-intercept of 4-4, giving the equation y=2x4y = 2x - 4. Substituting the coordinates of (k,5k+2)(k, 5k + 2) results in 5k+2=2k45k + 2 = 2k - 4, which simplifies to 3k=63k = -6, yielding k=2k = -2.

Adım Adım Çözüm

1
Identify the coordinates of the y-intercept.
The y-intercept of 4-4 corresponds to the point (0,4)(0, -4).
The y-intercept is the point where the line crosses the y-axis, meaning the x-coordinate is 0.
2
Calculate the slope of the line.
The slope mm is 22.
Using the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} with the points (3,2)(3, 2) and (0,4)(0, -4) gives m=2(4)30=63=2m = \frac{2 - (-4)}{3 - 0} = \frac{6}{3} = 2.
3
Write the equation of the line.
The equation of the line is y=2x4y = 2x - 4.
Using the slope-intercept form y=mx+by = mx + b, where the slope m=2m = 2 and the y-intercept b=4b = -4.
4
Substitute the point (k,5k+2)(k, 5k + 2) into the line's equation.
The equation becomes 5k+2=2k45k + 2 = 2k - 4.
Since the point lies on the line, its coordinates must satisfy the line's equation.
5
Solve the linear equation for kk.
k=2k = -2.
Subtracting 2k2k from both sides gives 3k+2=43k + 2 = -4. Subtracting 22 from both sides gives 3k=63k = -6. Dividing by 33 gives k=2k = -2.

Anahtar Kavram

Finding the equation of a line from a point and an intercept, and solving for parameters of points on that line.
Soru 2139Soru

Two cyclists start at opposite ends of a 9090-mile trail at the same time and ride toward each other. One cyclist rides at a constant speed that is 33 miles per hour faster than the other cyclist. If the two cyclists meet after exactly 33 hours, what is the constant speed, in miles per hour, of the faster cyclist?

Cevabı ve açıklamayı göster

Cevap: 16.5

Cevap

The speed of the faster cyclist is 16.516.5 miles per hour.
The correct answer of 16.516.5 is found by setting the speed of the slower cyclist to ss and the faster cyclist to s+3s + 3. Since both cyclists ride toward each other for 33 hours, their combined distance is 3s+3(s+3)=903s + 3(s + 3) = 90. Solving for ss yields 6s+9=906s + 9 = 90, which simplifies to 6s=816s = 81, or s=13.5s = 13.5. Adding 33 to 13.513.5 gives the speed of the faster cyclist, which is 16.516.5 miles per hour.

Adım Adım Çözüm

1
Define variables for the speeds of both cyclists in terms of a single variable.
Let ss be the speed of the slower cyclist. The speed of the faster cyclist is s+3s + 3.
Using a single variable simplifies the setup of a solvable linear equation.
2
Write a linear equation using the relationship Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}.
The equation is 3s+3(s+3)=903s + 3(s + 3) = 90.
The sum of the distances traveled by both cyclists when they meet must equal the total length of the trail, which is 9090 miles.
3
Solve the equation for ss.
6s+9=90    6s=81    s=13.56s + 9 = 90 \implies 6s = 81 \implies s = 13.5.
This yields the speed of the slower cyclist.
4
Calculate the speed of the faster cyclist.
13.5+3=16.513.5 + 3 = 16.5.
The question specifically asks for the speed of the faster cyclist, which is represented by s+3s + 3.

Anahtar Kavram

Translating distance-rate-time relationships from word problems into solvable linear equations.

Alternatif Yöntem

An alternative approach is to use the concept of relative speed. Since the two cyclists are moving directly toward each other, their relative speed of approach is the sum of their individual speeds. They cover a total of 9090 miles in 33 hours, which means their combined speed is 903=30\frac{90}{3} = 30 miles per hour. If the speed of the faster cyclist is ff and the slower is ss, then f+s=30f + s = 30 and fs=3f - s = 3. Adding these two equations gives 2f=332f = 33, which yields f=16.5f = 16.5 miles per hour.
Tahmini Süre:1m 15s
Soru 2140Soru

In the standard (x,y)(x, y) coordinate plane, an ellipse is defined by the equation 7x2+16y242x32y33=07x^2 + 16y^2 - 42x - 32y - 33 = 0. A parabola has its vertex at the focus of the ellipse with the smaller xx-coordinate, and its focus at the focus of the ellipse with the larger xx-coordinate. What is the larger of the two yy-coordinates of the points on the parabola that have an xx-coordinate of 6?

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Cevap: 13

Cevap

The larger of the two yy-coordinates of the points on the parabola is 13.
By completing the square on the general ellipse equation, we get (x3)216+(y1)27=1\frac{(x-3)^2}{16} + \frac{(y-1)^2}{7} = 1. The center is (3,1)(3, 1) and the focal distance is c=167=3c = \sqrt{16-7} = 3, meaning the foci are at (0,1)(0, 1) and (6,1)(6, 1). The parabola has its vertex at (0,1)(0, 1) and focus at (6,1)(6, 1), which means it opens to the right with p=6p = 6. Its equation is (y1)2=24x(y - 1)^2 = 24x. Substituting x=6x = 6 yields (y1)2=144(y - 1)^2 = 144, so y1=±12y - 1 = \pm 12. The two possible yy-coordinates are 1313 and 11-11, of which 1313 is the larger value.

Adım Adım Çözüm

1
Complete the square for the given ellipse equation to rewrite it in standard form.
(x3)216+(y1)27=1\frac{(x-3)^2}{16} + \frac{(y-1)^2}{7} = 1
Converting the equation to standard form is necessary to determine the center and semi-axis lengths of the ellipse.
2
Find the focal distance cc and calculate the coordinates of the foci.
Focal distance c=3c = 3; Foci at (0,1)(0, 1) and (6,1)(6, 1)
For an ellipse, the distance cc from the center (h,k)(h, k) to the foci is a2b2\sqrt{a^2 - b^2}. Since the major axis is horizontal, the foci are located at (h±c,k)(h \pm c, k).
3
Use the foci coordinates to identify the vertex and focus of the parabola.
Vertex: (0,1)(0, 1); Focus: (6,1)(6, 1)
The problem defines the parabola's vertex as the ellipse focus with the smaller xx-coordinate, and the parabola's focus as the ellipse focus with the larger xx-coordinate.
4
Determine the equation of the parabola using its vertex and focus.
(y1)2=24x(y - 1)^2 = 24x
The parabola is horizontal and opens to the right with focal distance p=6p = 6. The standard form is (yk)2=4p(xh)(y - k)^2 = 4p(x - h).
5
Substitute x=6x = 6 into the parabola equation and solve for the larger yy-value.
y=13y = 13
Substituting x=6x = 6 yields (y1)2=144(y - 1)^2 = 144, which gives y=1+12=13y = 1 + 12 = 13 or y=112=11y = 1 - 12 = -11. The larger value is 13.

Anahtar Kavram

Determining the equations and key features (foci, vertices, focal parameters) of ellipses and parabolas by rewriting equations into standard forms.

Alternatif Yöntem

Once the equation (y1)2=24x(y - 1)^2 = 24x is established, recognize that at x=6x = 6 (which is the xx-coordinate of the focus), the points on the parabola form the endpoints of the latus rectum. The length of the latus rectum is 4p=244p = 24, so the points lie at distance 2p=122p = 12 vertically above and below the focus (6,1)(6, 1). Thus, the yy-coordinates are 1±121 \pm 12, immediately yielding the larger coordinate as 13.
Tahmini Süre:3m 0s
ÖncekiSayfa 107 / 278Sonraki
Tüm alıştırma soruları — ACT | Examkin