Algebra

356 soru

Soru 1Soru

If b>1b > 1, which of the following expressions are equivalent to b3b23b16\frac{\sqrt{b^3 \cdot \sqrt[3]{b^2}}}{b^{-\frac{1}{6}}}? Select all that apply.

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Cevap: (b3)6(\sqrt[3]{b})^6; b73b13\frac{b^{\frac{7}{3}}}{b^{\frac{1}{3}}}; (b12)4\left(b^{-\frac{1}{2}}\right)^{-4}

Cevap

The equivalent expressions are (b3)6(\sqrt[3]{b})^6, b73b13\frac{b^{\frac{7}{3}}}{b^{\frac{1}{3}}}, and (b12)4\left(b^{-\frac{1}{2}}\right)^{-4}.
Simplifying the original expression yields b3b2/3b1/6=b11/6b1/6=b11/6(1/6)=b12/6=b2\frac{\sqrt{b^3 \cdot b^{2/3}}}{b^{-1/6}} = \frac{b^{11/6}}{b^{-1/6}} = b^{11/6 - (-1/6)} = b^{12/6} = b^2. The expression (b3)6(\sqrt[3]{b})^6 equals b6/3=b2b^{6/3} = b^2. The expression b7/3b1/3\frac{b^{7/3}}{b^{1/3}} equals b7/31/3=b2b^{7/3 - 1/3} = b^2. The expression (b1/2)4\left(b^{-1/2}\right)^{-4} equals b(1/2)(4)=b2b^{(-1/2)(-4)} = b^2. All three equal b2b^2.

Adım Adım Çözüm

1
Simplify the expression inside the square root in the numerator.
b3b23=b3b23=b3+23=b113b^3 \cdot \sqrt[3]{b^2} = b^3 \cdot b^{\frac{2}{3}} = b^{3 + \frac{2}{3}} = b^{\frac{11}{3}}
Convert radical to rational exponent and use product rule for exponents.
2
Apply the square root to the numerator.
b113=(b113)12=b116\sqrt{b^{\frac{11}{3}}} = \left(b^{\frac{11}{3}}\right)^{\frac{1}{2}} = b^{\frac{11}{6}}
Taking the square root is equivalent to raising to the power of 12\frac{1}{2}.
3
Divide by the denominator.
\frac{b^{\frac{11}{6}}}{b^{-\frac{1}{6}}} = b^{\frac{11}{6} - \left(-\frac{1}{6}\right)} = b^{\frac{12}{6}} = b^2
Apply quotient rule for exponents: subtract the exponent of the denominator from the numerator.
4
Evaluate each given option to determine which simplify to b2b^2.
The expressions (b3)6=b2(\sqrt[3]{b})^6 = b^2, b73b13=b2\frac{b^{\frac{7}{3}}}{b^{\frac{1}{3}}} = b^2, and (b12)4=b2\left(b^{-\frac{1}{2}}\right)^{-4} = b^2 are all equivalent.
Matching each simplified candidate expression to the target simplified value b2b^2.

Anahtar Kavram

Simplification of nested algebraic radicals and rational exponents using exponent rules
Tahmini Süre:2m 0s
Soru 2Soru

If xx is a real number that satisfies the compound absolute value inequality 2x574||2x - 5| - 7| \le 4, which of the following values could be the value of xx? Select all such values.

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Cevap: 2-2; 00; 66

Cevap

The real numbers 2-2, 00, and 66 satisfy the inequality.
The solution set to the compound inequality 2x574||2x - 5| - 7| \le 4 is the union of two intervals: [3,1][-3, 1] and [4,8][4, 8]. Among the options provided, 2-2 and 00 fall into the interval [3,1][-3, 1], and 66 falls into the interval [4,8][4, 8]. Therefore, these three values satisfy the original inequality.

Adım Adım Çözüm

1
Unfold the outer absolute value inequality.
42x574-4 \le |2x - 5| - 7 \le 4
For any expression UU and constant c0c \ge 0, Uc|U| \le c is equivalent to cUc-c \le U \le c.
2
Isolate the inner absolute value term by adding 77 across all parts.
32x5113 \le |2x - 5| \le 11
Adding a constant preserves inequality directions.
3
Break the compound inequality into two separate absolute value conditions.
Condition 1: 2x511|2x - 5| \le 11; Condition 2: 2x53|2x - 5| \ge 3
Both conditions must hold simultaneously for xx.
4
Solve Condition 1 (2x511|2x - 5| \le 11).
112x511    62x16    3x8-11 \le 2x - 5 \le 11 \implies -6 \le 2x \le 16 \implies -3 \le x \le 8
Expanding the bounded absolute value inequality and solving for xx.
5
Solve Condition 2 (2x53|2x - 5| \ge 3).
2x532x - 5 \ge 3 or 2x53    2x82x - 5 \le -3 \implies 2x \ge 8 or 2x2    x42x \le 2 \implies x \ge 4 or x1x \le 1
For Uc|U| \ge c, UcU \ge c or UcU \le -c.
6
Find the intersection of the two solution sets.
x[3,1][4,8]x \in [-3, 1] \cup [4, 8]
Intersecting [3,8][-3, 8] with (,1][4,)(-\infty, 1] \cup [4, \infty) yields [3,1][4,8][-3, 1] \cup [4, 8].
7
Test the provided choices against the solution set [3,1][4,8][-3, 1] \cup [4, 8].
2[3,1]-2 \in [-3, 1] (valid), 0[3,1]0 \in [-3, 1] (valid), 3[3,1][4,8]3 \notin [-3, 1] \cup [4, 8] (invalid), 6[4,8]6 \in [4, 8] (valid), 9[3,1][4,8]9 \notin [-3, 1] \cup [4, 8] (invalid).
Values lying within the solution intervals satisfy the original inequality.

Anahtar Kavram

Solving nested absolute value inequalities using multi-step compound interval intersections.
Soru 3Soru

A commercial coffee roaster creates a custom blend by mixing two existing bean blends: Blend X and Blend Y. Blend X consists of 60% Arabica beans and 40% Robusta beans by weight, whereas Blend Y consists of 30% Arabica beans and 70% Robusta beans by weight. The roaster mixes a quantity of Blend X with a quantity of Blend Y to produce a total of 50 pounds of a new mixture that is 42% Arabica beans by weight. How many pounds of Blend X are in the final mixture?

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Cevap: 20

Cevap

20 pounds
Let xx represent the number of pounds of Blend X. The remaining weight of the mixture, (50x)(50 - x) pounds, comes from Blend Y. Setting up the equation for the total weight of Arabica beans gives 0.60x+0.30(50x)=0.42(50)0.60x + 0.30(50 - x) = 0.42(50). Simplifying this expression yields 0.60x+150.30x=210.60x + 15 - 0.30x = 21, which reduces to 0.30x=60.30x = 6. Dividing by 0.300.30 gives x=20x = 20. Therefore, 20 pounds of Blend X were used.

Adım Adım Çözüm

1
Define variables for component weights
Let xx be the pounds of Blend X. The weight of Blend Y used is 50x50 - x pounds.
The total combined weight of the mixture is given as 50 pounds.
2
Formulate an equation for the total weight of Arabica beans
0.60x+0.30(50x)=0.42(50)0.60x + 0.30(50 - x) = 0.42(50), which simplifies to 0.60x+150.30x=210.60x + 15 - 0.30x = 21.
The sum of Arabica beans contributed by each blend must equal the total weight of Arabica beans in the combined mixture.
3
Solve the linear equation for xx
0.30x+15=21    0.30x=6    x=200.30x + 15 = 21 \implies 0.30x = 6 \implies x = 20.
Subtract 15 from both sides to isolate the variable term, then divide by 0.30.

Anahtar Kavram

Linear Modeling and Mixture Problems
Tahmini Süre:1m 30s
Soru 4Soru

If xx is a real number that satisfies the inequality 52x7|5 - 2x| \le 7, what is the minimum possible value of the expression 34x3 - 4x?

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Cevap: 21-21

Cevap

21-21
Solving 52x7|5 - 2x| \le 7 yields 752x7-7 \le 5 - 2x \le 7. Subtracting 55 gives 122x2-12 \le -2x \le 2. Dividing by 2-2 and flipping the inequality direction gives 1x6-1 \le x \le 6. To minimize 34x3 - 4x, we choose the maximum value of xx because the term 4x-4x decreases as xx increases. Substituting x=6x = 6 yields 34(6)=213 - 4(6) = -21.

Adım Adım Çözüm

1
Unpack the absolute value inequality into a compound inequality
752x7-7 \le 5 - 2x \le 7
For any real constant k0k \ge 0, uk|u| \le k is equivalent to kuk-k \le u \le k.
2
Isolate the variable term by subtracting 55 from all parts of the inequality
122x2-12 \le -2x \le 2
Subtracting a constant from all parts preserves inequality direction.
3
Divide all parts by 2-2 and reverse the inequality signs
6x16 \ge x \ge -1, which is equivalent to 1x6-1 \le x \le 6
Dividing an inequality by a negative number reverses the direction of the inequality signs.
4
Find the minimum value of 34x3 - 4x over the interval [1,6][-1, 6]
The minimum value occurs at x=6x = 6: 34(6)=324=213 - 4(6) = 3 - 24 = -21
Because the linear expression 34x3 - 4x has a negative coefficient for xx, it is a decreasing function; its minimum occurs at the largest allowed value of xx.

Anahtar Kavram

Linear Inequalities and Absolute Value Bounds
Tahmini Süre:1m 30s
Soru 5Soru

If x+y=8x + y = 8 and 2xy=72x - y = 7, what is the value of xx?

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Cevap: 5

Cevap

5
Adding the two equations yields (x+y)+(2xy)=8+7(x + y) + (2x - y) = 8 + 7, which simplifies to 3x=153x = 15. Dividing both sides by 3 gives the correct value, x=5x = 5.

Adım Adım Çözüm

1
Add the two equations together to eliminate the variable yy.
(x+y)+(2xy)=8+7    3x=15(x + y) + (2x - y) = 8 + 7 \implies 3x = 15
Since the coefficients of yy are +1+1 and 1-1, adding the equations eliminates yy directly.
2
Solve for xx by dividing both sides of the equation by 3.
x=153=5x = \frac{15}{3} = 5
Isolating xx gives the required solution.

Anahtar Kavram

Elimination Method in Systems of Linear Equations
Soru 6Soru

If 2x34x53=x+1512\frac{2x - 3}{4} - \frac{x - 5}{3} = \frac{x + 15}{12}, what is the value of 3x23x - 2?

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Cevap: 10

Cevap

The value of 3x23x - 2 is 1010.
Multiplying the given linear equation by the least common denominator 12 yields 3(2x3)4(x5)=x+153(2x - 3) - 4(x - 5) = x + 15. Expanding both sides gives 6x94x+20=x+156x - 9 - 4x + 20 = x + 15, which simplifies to 2x+11=x+152x + 11 = x + 15, leading to x=4x = 4. Substituting x=4x = 4 into 3x23x - 2 gives 3(4)2=103(4) - 2 = 10.

Adım Adım Çözüm

1
Clear the denominators by multiplying both sides of the equation by the least common multiple of 4, 3, and 12, which is 12.
12(2x34)12(x53)=12(x+1512)    3(2x3)4(x5)=x+1512 \cdot \left(\frac{2x - 3}{4}\right) - 12 \cdot \left(\frac{x - 5}{3}\right) = 12 \cdot \left(\frac{x + 15}{12}\right) \implies 3(2x - 3) - 4(x - 5) = x + 15
Eliminating fractions simplifies the algebraic manipulation.
2
Distribute the coefficients through the terms in parentheses, paying careful attention to the negative sign on the second term.
6x94x+20=x+156x - 9 - 4x + 20 = x + 15
Distributing 4-4 across (x5)(x - 5) yields 4x+20-4x + 20.
3
Combine like terms on the left side of the equation.
2x+11=x+152x + 11 = x + 15
Grouping variable terms (6x4x=2x)(6x - 4x = 2x) and constant terms (9+20=11)(-9 + 20 = 11).
4
Isolate the variable xx by subtracting xx and 1111 from both sides.
x=4x = 4
Solves the linear equation for xx.
5
Substitute x=4x = 4 into the target expression 3x23x - 2.
3(4)2=122=103(4) - 2 = 12 - 2 = 10
Calculates the final answer requested in the stem.

Anahtar Kavram

Linear Equations with Fractional Coefficients
Soru 7Soru
For all real numbers xx such that x3x \neq 3 and x5x \neq -5, the algebraic expression
(x29)24(x3)2(x3)(x+5)\frac{(x^2 - 9)^2 - 4(x - 3)^2}{(x - 3)(x + 5)}
can be simplified to the equivalent polynomial expression x2+ax+bx^2 + ax + b, where aa and bb are constants. What is the value of a+ba + b?
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Cevap: -5

Cevap

The value of a+ba + b is 5-5.
Factoring the numerator yields (x3)2(x+1)(x+5)(x - 3)^2(x + 1)(x + 5). Canceling the common factors (x3)(x - 3) and (x+5)(x + 5) with the denominator simplifies the expression to (x3)(x+1)=x22x3(x - 3)(x + 1) = x^2 - 2x - 3. Comparing this to x2+ax+bx^2 + ax + b identifies a=2a = -2 and b=3b = -3, giving a sum of a+b=5a + b = -5.

Adım Adım Çözüm

1
Factor the numerator by recognizing (x29)=(x3)(x+3)(x^2 - 9) = (x - 3)(x + 3)
(x29)24(x3)2=[(x3)(x+3)]24(x3)2=(x3)2(x+3)24(x3)2(x^2 - 9)^2 - 4(x - 3)^2 = [(x - 3)(x + 3)]^2 - 4(x - 3)^2 = (x - 3)^2 (x + 3)^2 - 4(x - 3)^2
Applying the difference of squares identity inside the squared term allows factoring out common factors.
2
Factor out (x3)2(x - 3)^2 from the numerator
(x3)2[(x+3)24](x - 3)^2 \left[ (x + 3)^2 - 4 \right]
Extracting the greatest common algebraic factor simplifies the remaining expression.
3
Apply difference of squares to (x+3)24(x + 3)^2 - 4
(x+3)222=((x+3)2)((x+3)+2)=(x+1)(x+5)(x + 3)^2 - 2^2 = ((x + 3) - 2)((x + 3) + 2) = (x + 1)(x + 5)
Recognizing (x+3)222(x + 3)^2 - 2^2 as A2B2A^2 - B^2 yields factored linear terms directly.
4
Substitute the fully factored numerator back into the rational expression and simplify
\frac{(x - 3)^2 (x + 1)(x + 5)}{(x - 3)(x + 5)} = (x - 3)(x + 1)
Canceling non-zero common factors (x3)(x - 3) and (x+5)(x + 5) simplifies the rational function.
5
Expand (x3)(x+1)(x - 3)(x + 1) and determine a+ba + b
(x3)(x+1)=x22x3(x - 3)(x + 1) = x^2 - 2x - 3, so a=2a = -2 and b=3b = -3. Therefore, a+b=2+(3)=5a + b = -2 + (-3) = -5.
Matching coefficients with x2+ax+bx^2 + ax + b gives a=2a = -2 and b=3b = -3.

Anahtar Kavram

Simplifying complex rational expressions through nested difference of squares factoring.
Soru 8Soru
Consider the system of linear equations in two variables xx and yy:
2x3y=a4x6y=b\begin{aligned} 2x - 3y &= a \\ 4x - 6y &= b \end{aligned}
where aa and bb are real constants. Which of the following statements MUST be true? Select all that apply.

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Cevap: If b=2ab = 2a, the system has infinitely many solutions.; If b2ab \neq 2a, the system has no solution.; If b=2a+1b = 2a + 1, the lines represented by the equations in the xyxy-plane are parallel and distinct.

Cevap

The correct statements are: 'If b=2ab = 2a, the system has infinitely many solutions.', 'If b2ab \neq 2a, the system has no solution.', and 'If b=2a+1b = 2a + 1, the lines represented by the equations in the xyxy-plane are parallel and distinct.'
Statements asserting that the system has infinitely many solutions when b=2ab = 2a, no solution when b2ab \neq 2a, and parallel distinct lines when b=2a+1b = 2a + 1 are all mathematically sound. Multiplying the first equation by 2 reveals that the left sides are identical (4x6y4x - 6y). Equality of the right sides (b=2ab = 2a) makes the lines identical, whereas inequality (b2ab \neq 2a) makes them parallel and distinct.

Adım Adım Çözüm

1
Analyze the coefficients of the system of equations.
The coefficients of xx and yy in the second equation (44 and 6-6) are exactly twice the coefficients of xx and yy in the first equation (22 and 3-3).
Comparing coefficient ratios determines line relationships (slopes).
2
Multiply the first equation by 2.
2(2x3y)=2(a)    4x6y=2a2(2x - 3y) = 2(a) \implies 4x - 6y = 2a.
This puts the left side of the first equation in exact alignment with the second equation (4x6y=b4x - 6y = b).
3
Evaluate the condition for infinitely many solutions.
If b=2ab = 2a, the two equations become 4x6y=2a4x - 6y = 2a and 4x6y=2a4x - 6y = 2a, which describe the exact same line, giving infinitely many solutions.
Coincident lines intersect at every point along the line.
4
Evaluate the condition for no solution.
If b2ab \neq 2a, subtracting the equations gives 0=b2a00 = b - 2a \neq 0, a contradiction. Hence, the lines are parallel and distinct, meaning no solution exists.
Distinct parallel lines never intersect.
5
Examine specific cases such as b=2a+1b = 2a + 1 and a=0,b=0a=0, b=0.
For b=2a+1b = 2a + 1, since 2a+12a2a + 1 \neq 2a, b2ab \neq 2a, confirming parallel distinct lines. For a=0,b=0a=0, b=0, b=2(0)=0b = 2(0) = 0, which yields infinitely many solutions rather than a unique solution.
Verifies specific option claims.

Anahtar Kavram

Systems of Linear Equations (Solvability and Geometric Interpretation)
Soru 9Soru

If the pair (x,y)(x, y) satisfies the system of linear equations below, what is the value of xyx - y?

3x5y=142x+7y=1\begin{aligned} 3x - 5y &= 14 \\ 2x + 7y &= -1 \end{aligned}
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Cevap: 4

Cevap

The value of xyx - y is 44.
Solving the system of linear equations by elimination gives x=3x = 3 and y=1y = -1. Substituting these values into the target expression xyx - y yields 3(1)=3+1=43 - (-1) = 3 + 1 = 4.

Adım Adım Çözüm

1
Eliminate one variable using the elimination method.
Multiply the first equation by 22 and the second equation by 33:
6x10y=286x - 10y = 28
6x+21y=36x + 21y = -3
Creating matching coefficients for xx allows elimination of xx by subtraction.
2
Subtract the first modified equation from the second modified equation to solve for yy.
(6x+21y)(6x10y)=328(6x + 21y) - (6x - 10y) = -3 - 28
31y=31    y=131y = -31 \implies y = -1
Subtracting eliminates xx, isolating yy.
3
Substitute y=1y = -1 back into one of the original equations to solve for xx.
3x5(1)=14    3x+5=14    3x=9    x=33x - 5(-1) = 14 \implies 3x + 5 = 14 \implies 3x = 9 \implies x = 3
Plugging in yy allows direct solution for xx.
4
Evaluate the requested expression xyx - y.
xy=3(1)=3+1=4x - y = 3 - (-1) = 3 + 1 = 4
Subtracting a negative value is equivalent to adding its positive value.

Anahtar Kavram

Solving systems of linear equations using elimination and evaluating algebraic expressions.
Tahmini Süre:1m 30s
Soru 10Soru

The quadratic equation x2kx+36=0x^2 - kx + 36 = 0, where kk is a positive constant, has two distinct real roots r1r_1 and r2r_2 such that r2r1=5r_2 - r_1 = 5. What is the value of kk?

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Cevap: 13

Cevap

The value of kk is 13.
According to Vieta's formulas, for the quadratic equation x2kx+36=0x^2 - kx + 36 = 0, the sum of the roots is r1+r2=kr_1 + r_2 = k and the product of the roots is r1r2=36r_1 r_2 = 36. Using the identity (r2r1)2=(r1+r2)24r1r2(r_2 - r_1)^2 = (r_1 + r_2)^2 - 4r_1 r_2, we substitute the known values r2r1=5r_2 - r_1 = 5, r1+r2=kr_1 + r_2 = k, and r1r2=36r_1 r_2 = 36. This gives 52=k24(36)5^2 = k^2 - 4(36), which simplifies to 25=k214425 = k^2 - 144. Solving for k2k^2 gives k2=169k^2 = 169. Since kk is specified as a positive constant, k=13k = 13.

Adım Adım Çözüm

1
Apply Vieta's formulas to the given quadratic equation
The sum of the roots is r1+r2=kr_1 + r_2 = k and the product of the roots is r1r2=36r_1 r_2 = 36.
For any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of roots is b/a-b/a and the product of roots is c/ac/a.
2
Relate the difference of the roots to their sum and product
(r2r1)2=(r1+r2)24r1r2(r_2 - r_1)^2 = (r_1 + r_2)^2 - 4r_1 r_2
Expanding both sides shows that r222r1r2+r12=r12+2r1r2+r224r1r2r_2^2 - 2r_1 r_2 + r_1^2 = r_1^2 + 2r_1 r_2 + r_2^2 - 4r_1 r_2, which is an algebraic identity.
3
Substitute the known values into the identity
52=k24(36)    25=k21445^2 = k^2 - 4(36) \implies 25 = k^2 - 144
We are given that r2r1=5r_2 - r_1 = 5, r1r2=36r_1 r_2 = 36, and r1+r2=kr_1 + r_2 = k.
4
Solve for the positive constant kk
k2=169    k=13k^2 = 169 \implies k = 13
Adding 144 to both sides gives k2=169k^2 = 169. Taking the positive square root because k>0k > 0 yields k=13k = 13.

Anahtar Kavram

Vieta's Formulas and Root Difference Identity
Soru 11Soru

If xx and yy are positive integers such that 2x+1+2x=3y+23y2^{x+1} + 2^x = 3^{y+2} - 3^y, what is the value of x+yx + y?

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Cevap: 4

Cevap

The value of x+yx + y is 44.
Factoring the left side gives 2x(2+1)=32x2^x(2 + 1) = 3 \cdot 2^x, while factoring the right side gives 3y(91)=83y=233y3^y(9 - 1) = 8 \cdot 3^y = 2^3 \cdot 3^y. Equating the two expressions gives 32x=233y3 \cdot 2^x = 2^3 \cdot 3^y. Rearranging terms to separate bases yields 2x3=3y12^{x-3} = 3^{y-1}. Because 2 and 3 share no common prime factors, this equality holds for integers if and only if both exponents are equal to 0. Solving x3=0x - 3 = 0 gives x=3x = 3, and solving y1=0y - 1 = 0 gives y=1y = 1. Both are positive integers. Thus, x+y=3+1=4x + y = 3 + 1 = 4.

Adım Adım Çözüm

1
Factor out common terms on both sides of the equation.
2x(2+1)=3y(91)    32x=83y2^x(2 + 1) = 3^y(9 - 1) \implies 3 \cdot 2^x = 8 \cdot 3^y
Factoring simplifies sums of powers with identical bases.
2
Rewrite integers using prime factorizations and re-group bases.
32x=233y    2x3=3y13 \cdot 2^x = 2^3 \cdot 3^y \implies 2^{x-3} = 3^{y-1}
Dividing both sides by 2332^3 \cdot 3 separates the base-2 and base-3 exponential terms.
3
Set each exponent to zero using prime independence.
x3=0    x=3x - 3 = 0 \implies x = 3 and y1=0    y=1y - 1 = 0 \implies y = 1
Powers of distinct prime numbers 2 and 3 can only be equal if both powers equal 11 (20=30=12^0 = 3^0 = 1).
4
Calculate the required sum x+yx + y.
3+1=43 + 1 = 4
Evaluates the requested combined value of the variables.

Anahtar Kavram

Solving exponential equations involving distinct prime bases through factoring and exponent properties.
Soru 12Soru
If xx is a real number satisfying the equation
(23x+2+23x)3(42x+142x)2=20009\frac{\left(2^{3x+2} + 2^{3x}\right)^3}{\left(4^{2x+1} - 4^{2x}\right)^2} = \frac{2000}{9}
what is the value of xx?
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Cevap: 4

Cevap

4
Factoring out 23x2^{3x} in the numerator gives 23x(22+1)=523x2^{3x}(2^2 + 1) = 5 \cdot 2^{3x}. Cubing this yields 12529x125 \cdot 2^{9x}. In the denominator, rewriting 42x4^{2x} as 24x2^{4x} and factoring gives 24x(41)=324x2^{4x}(4 - 1) = 3 \cdot 2^{4x}. Squaring this yields 928x9 \cdot 2^{8x}. Taking the quotient gives 12592x\frac{125}{9} \cdot 2^x. Setting this equal to 20009\frac{2000}{9} leads directly to 1252x=2000    2x=16    x=4125 \cdot 2^x = 2000 \implies 2^x = 16 \implies x = 4.

Adım Adım Çözüm

1
Factor out common exponential terms inside the parentheses
Numerator inside becomes 523x5 \cdot 2^{3x} and denominator inside becomes 324x3 \cdot 2^{4x}
Factoring out 23x2^{3x} from 23x+2+23x2^{3x+2} + 2^{3x} isolates the constant multiplier (4+1)(4+1), and expressing 42x4^{2x} as 24x2^{4x} allows base unification.
2
Raise the simplified terms to their respective outer powers
Numerator becomes 12529x125 \cdot 2^{9x} and denominator becomes 928x9 \cdot 2^{8x}
Using power rules (ab)n=anbn(a \cdot b)^n = a^n b^n and (am)n=amn(a^m)^n = a^{m n}.
3
Simplify the fraction by subtracting exponents of like bases
The left side simplifies to 12592x\frac{125}{9} \cdot 2^x
 me29x28x=29x8x=2x\ me{2^{9x}}{2^{8x}} = 2^{9x-8x} = 2^x using the quotient rule for exponents.
4
Solve the resulting single-variable exponential equation
2x=162^x = 16, which yields x=4x = 4
Multiplying both sides by 99 yields 1252x=2000125 \cdot 2^x = 2000, so 2x=16=242^x = 16 = 2^4.

Anahtar Kavram

Factoring exponential expressions and applying power of a power and quotient rules
Soru 13Soru

If xx is a positive real number, which of the following is equivalent to the expression (x3)2x4x8\frac{(x^3)^2 \cdot x^{-4}}{\sqrt{x^8}}?

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Cevap: x2x^{-2}

Cevap

The simplified expression is equivalent to x2x^{-2}.
Applying the rules of exponents systematically yields (x3)2=x6(x^3)^2 = x^6 in the numerator, which combines with x4x^{-4} to give x2x^2. The denominator x8\sqrt{x^8} simplifies to x8/2=x4x^{8/2} = x^4. Dividing x2x^2 by x4x^4 gives x24=x2x^{2-4} = x^{-2}.

Adım Adım Çözüm

1
Simplify the power raised to a power in the numerator.
(x3)2=x32=x6(x^3)^2 = x^{3 \cdot 2} = x^6
When raising a power to a power, multiply the exponents: (am)n=amn(a^m)^n = a^{m \cdot n}.
2
Multiply the terms in the numerator.
x6x4=x6+(4)=x2x^6 \cdot x^{-4} = x^{6 + (-4)} = x^2
When multiplying exponential expressions with the same base, add the exponents: aman=am+na^m \cdot a^n = a^{m+n}.
3
Simplify the radical in the denominator.
x8=(x8)12=x82=x4\sqrt{x^8} = (x^8)^{\frac{1}{2}} = x^{\frac{8}{2}} = x^4
Taking the square root of a non-negative term is equivalent to raising it to the power of 12\frac{1}{2}.
4
Divide the numerator by the denominator.
\frac{x^2}{x^4} = x^{2 - 4} = x^{-2}
When dividing exponential expressions with the same base, subtract the denominator exponent from the numerator exponent: \frac{a^m}{a^n} = a^{m-n}.

Anahtar Kavram

Laws of Exponents and Radical Simplification
Tahmini Süre:45s
Soru 14Soru

A manufacturing company produces two models of office chairs, Model P and Model Q. Producing each Model P chair requires 33 hours of assembly and 11 hour of finishing. Producing each Model Q chair requires 22 hours of assembly and 22 hours of finishing. On a given day, the assembly department was scheduled for 9696 total hours of work and the finishing department was scheduled for 4848 total hours of work. If both departments operated at full capacity and used all scheduled hours, how many Model P chairs were produced on that day?

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Cevap: 24

Cevap

24
Letting pp be the number of Model P chairs and qq be the number of Model Q chairs, the hours used by the assembly department give 3p+2q=963p + 2q = 96 and the hours used by the finishing department give p+2q=48p + 2q = 48. Subtracting the finishing equation from the assembly equation eliminates 2q2q, resulting in 2p=482p = 48, or p=24p = 24.

Adım Adım Çözüm

1
Define variables for the unknown quantities.
Let pp equal the number of Model P chairs produced and qq equal the number of Model Q chairs produced.
Assigning variables to the unknown quantities enables the construction of algebraic equations.
2
Formulate a system of linear equations representing total department hours.
Assembly department equation: 3p+2q=963p + 2q = 96; Finishing department equation: p+2q=48p + 2q = 48.
The sum of hours required for both models in each department must equal that department's total scheduled hours.
3
Solve the system of equations for pp using the elimination method.
Subtracting p+2q=48p + 2q = 48 from 3p+2q=963p + 2q = 96 yields 2p=482p = 48, which gives p=24p = 24.
Eliminating qq directly provides the value for pp, which corresponds to the target quantity requested in the problem.

Anahtar Kavram

Modeling real-world resource allocation using systems of linear equations
Soru 15Soru
For a real constant k1k \neq -1, consider the linear equation in one variable xx:
2(x3)k+1x+13=1\frac{2(x - 3)}{k + 1} - \frac{x + 1}{3} = 1
Which of the following values of kk result in a solution xx that is a positive integer? Select all such values.

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: -2; 2; 3

Cevap

The values of kk that yield a positive integer solution for xx are 2-2, 22, and 33.
Solving the linear equation for xx in terms of kk gives x=4k+225kx = \frac{4k + 22}{5 - k}. Substituting k=2k = -2 yields x=2x = 2, substituting k=2k = 2 yields x=10x = 10, and substituting k=3k = 3 yields x=17x = 17. All three resulting values of xx are positive integers.

Adım Adım Çözüm

1
Clear denominators by multiplying the entire equation by 3(k+1)3(k + 1).
6(x3)(x+1)(k+1)=3(k+1)6(x - 3) - (x + 1)(k + 1) = 3(k + 1)
Eliminating fractional terms simplifies isolation of the variable xx.
2
Expand all terms on both sides of the equation.
6x18(kx+x+k+1)=3k+3    5xkx19k=3k+36x - 18 - (kx + x + k + 1) = 3k + 3 \implies 5x - kx - 19 - k = 3k + 3
Carefully distribute negative signs and combine like terms.
3
Group terms containing xx on the left side and constant/k terms on the right side.
(5k)x=4k+22    x=4k+225k(5 - k)x = 4k + 22 \implies x = \frac{4k + 22}{5 - k}
Express xx explicitly as a rational function of the parameter kk.
4
Evaluate xx for each given option to determine which produce positive integers.
For k=2k = -2: x=14/7=2x = 14/7 = 2 (integer);
For k=0k = 0: x=22/5=4.4x = 22/5 = 4.4 (not integer);
For k=1k = 1: x=26/4=6.5x = 26/4 = 6.5 (not integer);
For k=2k = 2: x=30/3=10x = 30/3 = 10 (integer);
For k=3k = 3: x=34/2=17x = 34/2 = 17 (integer).
Test each candidate value of kk against the condition that xx must be a positive integer.

Anahtar Kavram

Solving parametric linear equations in one variable and evaluating integer solutions
Tahmini Süre:2m 30s
Soru 16Soru

Line LL is defined by the equation 2x+y=62x + y = 6 in the xyxy-plane. Which of the following statements about line LL must be true? Select all that apply.

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Cevap: The slope of line LL is 2-2.; The xx-intercept of line LL is (3,0)(3, 0).

Cevap

The slope of line LL is 2-2, and the xx-intercept of line LL is (3,0)(3, 0).
The given line equation 2x+y=62x + y = 6 can be rewritten in slope-intercept form y=2x+6y = -2x + 6. This directly shows that the slope is 2-2. Setting y=0y = 0 gives 2x=6    x=32x = 6 \implies x = 3, so the xx-intercept is (3,0)(3, 0). Therefore, both the statement that the slope is 2-2 and the statement that the xx-intercept is (3,0)(3, 0) are correct.

Adım Adım Çözüm

1
Convert the equation to slope-intercept form (y=mx+by = mx + b).
y=2x+6y = -2x + 6
This isolates yy to clearly reveal the slope m=2m = -2 and the yy-intercept (0,6)(0, 6).
2
Find the xx-intercept by setting y=0y = 0.
2x+0=6    x=32x + 0 = 6 \implies x = 3, giving coordinate (3,0)(3, 0)
The xx-intercept is the point where the line crosses the xx-axis.
3
Determine perpendicular slope rules.
Perpendicular slope =12=12= -\frac{1}{-2} = \frac{1}{2}
Perpendicular lines have negative reciprocal slopes, not identical slopes.

Anahtar Kavram

Linear equations, slope-intercept form, intercepts, and perpendicular slopes in coordinate geometry.
Tahmini Süre:1m 0s
Soru 17Soru

If (2x3)2=25(2x - 3)^2 = 25 and x<0x < 0, what is the value of xx?

Cevabı ve açıklamayı göster

Cevap: 1-1

Cevap

1-1
Taking the square root of both sides of (2x3)2=25(2x - 3)^2 = 25 yields two equations: 2x3=52x - 3 = 5 (which gives x=4x = 4) and 2x3=52x - 3 = -5 (which gives x=1x = -1). Because the problem specifies that x<0x < 0, the correct value must be 1-1.

Adım Adım Çözüm

1
Take the square root of both sides of the equation (2x3)2=25(2x - 3)^2 = 25
2x3=±52x - 3 = \pm 5
Applying the square root property to a squared binomial requires considering both positive and negative principal roots.
2
Set up two separate linear equations corresponding to the two cases
Case 1: 2x3=52x - 3 = 5 or Case 2: 2x3=52x - 3 = -5
To find all possible solutions for xx, evaluate both root possibilities.
3
Solve each linear equation for xx
Case 1 yields 2x=8    x=42x = 8 \implies x = 4. Case 2 yields 2x=2    x=12x = -2 \implies x = -1.
Isolate xx using standard algebraic operations.
4
Apply the given constraint x<0x < 0
x=1x = -1
Since 4>04 > 0, the positive root is eliminated, leaving x=1x = -1 as the sole valid solution.

Anahtar Kavram

Quadratic Equations and Factoring
Soru 18Soru

If x=5x = 5 is a solution to the quadratic equation x2(k+3)x+3k+1=0x^2 - (k + 3)x + 3k + 1 = 0, where kk is a constant, what is the value of the other solution?

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Cevap: 72\frac{7}{2}

Cevap

72\frac{7}{2}
Substituting x=5x = 5 into x2(k+3)x+3k+1=0x^2 - (k + 3)x + 3k + 1 = 0 yields 255(k+3)+3k+1=025 - 5(k + 3) + 3k + 1 = 0, which simplifies to 112k=011 - 2k = 0, so k=112k = \frac{11}{2}. Substituting k=112k = \frac{11}{2} back into the product of roots formula x1x2=3k+1x_1 \cdot x_2 = 3k + 1 gives 5x2=3(112)+1=3525 \cdot x_2 = 3\left(\frac{11}{2}\right) + 1 = \frac{35}{2}. Dividing by 55 yields the second root x2=72x_2 = \frac{7}{2}.

Adım Adım Çözüm

1
Substitute the known solution x=5x = 5 into the quadratic equation to solve for kk.
52(k+3)(5)+3k+1=0    255k15+3k+1=0    112k=0    k=1125^2 - (k + 3)(5) + 3k + 1 = 0 \implies 25 - 5k - 15 + 3k + 1 = 0 \implies 11 - 2k = 0 \implies k = \frac{11}{2}.
Since x=5x = 5 is a solution, it must satisfy the equation.
2
Use Vieta's formulas to find the other solution x2x_2.
Product of roots x1x2=3k+1=3(112)+1=352x_1 \cdot x_2 = 3k + 1 = 3\left(\frac{11}{2}\right) + 1 = \frac{35}{2}. Since x1=5x_1 = 5, 5x2=352    x2=725 \cdot x_2 = \frac{35}{2} \implies x_2 = \frac{7}{2}.
By Vieta's formulas for a standard quadratic x2+bx+c=0x^2 + bx + c = 0, the product of the roots equals cc.

Anahtar Kavram

Quadratic equations, Vieta's formulas, and parameter evaluation
Soru 19Soru

A hotel renovated its guest rooms and purchased a total of 100 lighting fixtures, consisting of wall sconces costing $45\$45 each and ceiling pendants costing $70\$70 each. If the average (arithmetic mean) cost per fixture for the entire purchase was $52\$52, how many wall sconces were purchased?

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Cevap: 72

Cevap

72
Setting up the weighted total cost equation 45x+70(100x)=520045x + 70(100 - x) = 5200 simplifies to 25x=1800-25x = -1800, giving x=72x = 72 wall sconces.

Adım Adım Çözüm

1
Define the variable for the unknown quantity
Let xx represent the number of wall sconces purchased. The number of ceiling pendants is 100x100 - x.
Defining the target quantity as xx allows the problem to be modeled using a single-variable linear equation.
2
Set up the algebraic equation for total cost
45x+70(100x)=52×10045x + 70(100 - x) = 52 \times 100
The total cost of all fixtures is the sum of the total sconce cost and total pendant cost, which equals the overall average cost multiplied by the total number of fixtures.
3
Expand and solve the linear equation
45x+700070x=5200    25x=1800    x=7245x + 7000 - 70x = 5200 \implies -25x = -1800 \implies x = 72
Simplifying algebraic terms isolates xx to find the exact number of wall sconces.

Anahtar Kavram

Linear Algebraic Modeling and Weighted Averages
Soru 20Soru

A chemistry laboratory prepares a 100 mL100\text{ mL} mixture using three solutions: Solution XX (10%10\% acid by volume), Solution YY (30%30\% acid by volume), and Solution ZZ (50%50\% acid by volume). The resulting mixture is 37%37\% acid by volume. If the volume of Solution YY used is 5 mL5\text{ mL} more than twice the volume of Solution XX, what is the volume, in mL\text{mL}, of Solution ZZ used in the mixture?

Cevabı ve açıklamayı göster

Cevap: 5050

Cevap

50 mL50\text{ mL}
Setting up the system of equations gives x+y+z=100x + y + z = 100, x+3y+5z=370x + 3y + 5z = 370, and y=2x+5y = 2x + 5. Substituting y=2x+5y = 2x + 5 into the first two equations yields 3x+z=953x + z = 95 and 7x+5z=3557x + 5z = 355. Solving for xx gives x=15x = 15, which leads to z=953(15)=50z = 95 - 3(15) = 50. Therefore, 50 mL50\text{ mL} of Solution Z was used.

Adım Adım Çözüm

1
Define variables and set up the system of linear equations
Let xx, yy, and zz be the volumes in mL\text{mL} of Solutions XX, YY, and ZZ, respectively.
1) Total volume: x+y+z=100x + y + z = 100
2) Total acid volume: 0.10x+0.30y+0.50z=0.37(100)    x+3y+5z=3700.10x + 0.30y + 0.50z = 0.37(100) \implies x + 3y + 5z = 370
3) Relationship between YY and XX: y=2x+5y = 2x + 5
Translate the word problem statements into mathematical equations.
2
Substitute y=2x+5y = 2x + 5 into equations (1) and (2) to reduce to a two-variable system
From equation (1):
x+(2x+5)+z=100    3x+z=95    z=953xx + (2x + 5) + z = 100 \implies 3x + z = 95 \implies z = 95 - 3x

From equation (2):
x+3(2x+5)+5z=370    7x+15+5z=370    7x+5z=355x + 3(2x + 5) + 5z = 370 \implies 7x + 15 + 5z = 370 \implies 7x + 5z = 355
Eliminating yy simplifies the system to two equations in xx and zz.
3
Substitute z=953xz = 95 - 3x into 7x+5z=3557x + 5z = 355 and solve for xx
7x+5(953x)=355    7x+47515x=355    8x=120    x=157x + 5(95 - 3x) = 355 \implies 7x + 475 - 15x = 355 \implies -8x = -120 \implies x = 15
Solves for the unknown volume of Solution X.
4
Calculate zz using z=953xz = 95 - 3x
z=953(15)=9545=50z = 95 - 3(15) = 95 - 45 = 50
Finds the requested volume of Solution Z.

Anahtar Kavram

Setting up and solving 3x3 systems of linear equations using substitution or elimination
Tahmini Süre:2m 0s
Sayfa 1 / 18Sonraki