Simplify. the quantity negative 18 times x cubed plus 17 times x plus 6 divided by the quantity 3 times x plus 2
$$\frac{-18x^3 + 17x + 6}{3x + 2}$$
First, factor the numerator as much as possible:
$$\frac{(2x + 1)(-9x^2 + 6x + 6)}{3x + 2}$$
Simplify the numerator by factoring out a 3, and canceling the 3 in the numerator and denominator:
$$\frac{(2x + 1)(-3x^2 + 2x + 2)}{x + \frac{2}{3}}$$
There is no further simplification possible, so the final answer is:
$$\boxed{(2x + 1)(-3x^2 + 2x + 2)\div (3x + 2) \text{ or } \frac{(2x + 1)(-3x^2 + 2x + 2)}{3x + 2}}$$
Simplify. the quantity negative 18 times x cubed plus 17 times x plus 6 divided by the quantity 3 times x plus 2
A. negative 6 times x squared plus 4 times x plus 3
B. 6 times x squared plus 4 times x minus 3
C. negative 6 times x squared plus 4 times x minus 3
D. 6 times x squared minus 4 times x plus 3
The correct answer is option C, negative 6 times x squared plus 4 times x minus 3.
First, factor the numerator as much as possible:
$$\frac{-18x^3 + 17x + 6}{3x + 2}$$
$$(2x + 1)(-9x^2 + 6x + 6)$$
$$2(3x+2)(-3x^2 + 2x + 2)$$
Then, cancel out the 3x + 2 in the numerator and denominator to simplify:
$$\frac{2(3x+2)(-3x^2 + 2x + 2)}{3x + 2}$$
$$2(-3x^2 + 2x + 2)$$
$$-6x^2 + 4x - 4$$
Finally, simplify further by dividing every term by -2:
$$\frac{-6x^2}{-2} + \frac{4x}{-2} - \frac{4}{-2}$$
$$3x^2 - 2x + 2$$
So the simplified expression is negative 6 times x squared plus 4 times x minus 3.
The width, w, of a rectangular playground is x plus 3. The area of the playground is x cubed minus 7 times x plus 6 . What is an expression for the length of the playground?
A. x squared minus 3 times x plus 2
B. x squared plus 3 times x minus 2
C. x squared minus 3 times x minus 2
D. x squared minus 3 times x minus 2
The correct answer is option A, x squared minus 3 times x plus 2.
We know that the area of a rectangle is equal to the length times the width. In this case, the width is given as x + 3 and the area is given as x^3 - 7x + 6. We can use this information to find an expression for the length, L.
Area of rectangle = Length x Width
x^3 - 7x + 6 = L(x + 3)
Distribute the L on the right side of the equation:
x^3 - 7x + 6 = Lx + 3L
Move all the terms to one side of the equation:
x^3 - Lx - 7x + 6 - 3L = 0
Simplify the expression:
x^3 - (L+7)x + (6 - 3L) = 0
Since we want an expression for L, we can set up a system of equations using the coefficients of x^2 and x in the above equation:
L+7 = 0 (coefficient of x^2 is 0)
6 - 3L = 0 (coefficient of x is 0)
Solving for L in the first equation gives L = -7. Substituting this into the second equation gives:
6 - 3(-7) = 6 + 21 = 27
Therefore, the length of the rectangle is:
L = (x^3 - 7x + 6)/(x + 3) = (x^2 - 3x + 2) + (27/(x+3))
The correct answer is just the first term of this expression, which simplifies to x squared minus 3 times x plus 2.
Katalin drove 180 miles on her vacation. She drove an average of 1.5 times faster on the second 90 miles of her trip than she did on the first 90 miles of her trip. Which expression represents the time she spent driving? Let x = her speed on the first half of the trip.
A. 150 over x
B. 300 over x
C. 150 times x
D. 225 over x
Let's say her speed for the first 90 miles is x. Then her speed for the second 90 miles is 1.5x, or x + 0.5x.
The time it takes to travel each segment is given by distance divided by speed.
For the first 90 miles: time = distance/speed = 90/x
For the second 90 miles: time = distance/speed = 90/(1.5x) = 60/x
So the total time spent driving is the sum of these two times:
Total time = 90/x + 60/x
We can combine the fractions by finding a common denominator:
Total time = (90 + 60)/x
Total time = 150/x
Therefore, the expression that represents the time she spent driving is option A, 150/x.
Simplify. start fraction x over 6 x minus x squared end fraction
A. Start Fraction 1 over 6 minus lower x End Fraction semicolon where lower x does not equal zero comma 6
B. Start Fraction 1 over 6 minus lower x End Fraction semicolon where lower x does not equal 6
C. Start Fraction 1 over 6 lower x End Fraction semicolon where lower x does not equal zero
D. one-sixth
We can start by factoring out x from the numerator:
$$\frac{x}{6x - x^2} = \frac{x}{x(6 - x)}$$
We can simplify this expression by canceling out the x in the numerator and the denominator:
$$\frac{x}{x(6 - x)} = \frac{1}{6 - x}$$
So the simplified expression is:
$$\boxed{\frac{1}{6 - x}}$$
The answer is option B, Start Fraction 1 over 6 minus lower x End Fraction semicolon where lower x does not equal 6.
Simplify. 5 times x cubed over the quantity 7 times x cubed plus x to the fourth power
A. 5 over the quantity 7 plus x, where x does not equal 0, negative 7
B. 5 over the quantity 7 plus x, where x does not equal negative 7
C. 5 over 7 times x, where x does not equal 0
D. five sevenths
The expression is:
$$\frac{5x^3}{7x^3 + x^4}$$
We can factor out x^3 from the denominator:
$$\frac{5x^3}{x^3(7 + x)}$$
We can then cancel out the x^3 from the numerator and denominator:
$$\frac{5x^3}{x^3(7 + x)} = \frac{5}{7 + x}$$
So the simplified expression is:
$$\boxed{\frac{5}{7 + x}}$$
The answer is option B, 5 over the quantity 7 plus x, where x does not equal negative 7.
Simplify. start fraction x plus 7 over x squared plus 4 x minus 21 end fraction
A. Start Fraction 1 over lower x minus 3 End Fraction semicolon where lower x does not equal 3 comma negative 7
B. Lower x minus 3 semicolon where lower x does not equal 3
C. Start Fraction 1 over lower x minus 7 End Fraction semicolon where lower x does not equal 7
D. x minus 7
We can start by factoring the denominator of the fraction:
$$\frac{x + 7}{x^2 + 4x - 21} = \frac{x + 7}{(x + 7)(x - 3)}$$
We can then cancel out the (x + 7) term in the numerator and denominator:
$$\frac{x + 7}{(x + 7)(x - 3)} = \frac{1}{x - 3}$$
So the simplified expression is:
$$\boxed{\frac{1}{x - 3}}$$
The answer is option A, Start Fraction 1 over lower x minus 3 End Fraction semicolon where lower x does not equal 3, negative 7.
Simplify. The quantity x squared minus 3 times x minus 18 over the quantity x plus 3
A. x minus 3
B. x minus 6, where x does not equal negative 3
C. x minus 6, where x does not equal 6
D. 1 over the quantity x plus 3, where x does not equal negative 3
We can start by factoring the numerator of the fraction:
$$\frac{x^2 - 3x - 18}{x + 3} = \frac{(x - 6)(x + 3)}{x + 3}$$
We can then cancel out the (x + 3) term in the numerator and denominator:
$$\frac{(x - 6)(x + 3)}{x + 3} = x - 6$$
So the simplified expression is:
$$\boxed{x - 6}$$
The answer is option C, x minus 6, where x does not equal 6.
Simplify. start fraction 2 over 3 a end fraction dot start fraction 2 over a squared end fraction
A. Start Fraction 4 over 3 lower a superscript 2 baseline End Fraction semicolon where lower a does not equal zero
B. Start Fraction 1 over lower a superscript 2 baseline End Fraction semicolon where lower a does not equal zero
C. Start Fraction 4 over 3 lower a superscript 3 baseline End Fraction semicolon where lower a does not equal zero
D. Start Fraction 4 over 3 lower a superscript 2 baseline End Fraction
We can start by multiplying the two fractions together:
$$\frac{2}{3a} \cdot \frac{2}{a^2} = \frac{4}{3a^3}$$
So the simplified expression is:
$$\boxed{\frac{4}{3a^3}}$$
The answer is option C, Start Fraction 4 over 3 lower a superscript 3 baseline End Fraction semicolon where lower a does not equal zero.
Multiply. The quotient of the quantity x minus 5 over the quantity 4 times x plus 8 times the quantity 12 times x squared plus 32 times x plus 16
A. The quantity 3 times x plus 2 over 4 times the quantity x minus 5
B. The quantity x minus 5 times the quantity 3 times x plus 2, all over 4
C. The quantity x minus 5 times the quantity 3 times x plus 2
D. The quantity x minus 5 times the quantity 12 times x plus 8
We can start by simplifying the denominator of the fraction:
$$\frac{x - 5}{4x + 8(12x^2 + 32x + 16)} = \frac{x - 5}{4x + 96x^2 + 256x + 128}$$
We can then factor out a 4 from the denominator to simplify it:
$$\frac{x - 5}{4(x + 24x^2 + 64x + 32)} = \frac{x - 5}{4(24x^2 + 68x + 32)}$$
We can further simplify the denominator by factoring:
$$\frac{x - 5}{4(6x + 4)(4x + 8)} = \frac{x - 5}{4(2)(3x + 2)(2)(2x + 4)}$$
We can then cancel out the 2 and the 4 in the numerator and denominator:
$$\frac{x - 5}{4(3x + 2)(2x + 4)} = \frac{x - 5}{24x^2 + 56x + 16}$$
So the simplified expression is:
$$\boxed{\frac{(x - 5)(3x + 2)}{24x^2 + 56x + 16}}$$
The answer is option A, The quantity 3 times x plus 2 over 4 times the quantity x minus 5.
Divide. The quantity x squared minus 16 over the quantity x minus 1, all divided by the quantity x plus 4
A. The quantity x minus 4 over the quantity x minus 1
B. The quantity x plus 4 over the quantity x minus 1
C. The quantity x plus 4 times the quantity x minus 3, all over the quantity x minus 1
D. The quantity x minus 4 over the quantity x plus 1
We can start by simplifying the numerator of the fraction by factoring the difference of squares:
$$\frac{x^2 - 16}{x - 1} = \frac{(x + 4)(x - 4)}{x - 1}$$
We can then divide the whole fraction by (x + 4):
$$\frac{(x + 4)(x - 4)/(x - 1)}{x + 4} = \frac{x - 4}{x - 1}$$
So the simplified expression is:
$$\boxed{\frac{x - 4}{x - 1}}$$
The answer is option A, The quantity x minus 4 over the quantity x minus 1.
Divide. Fraction 1: The quantity x squared plus 6 times x plus 9 over the quantity x minus 1. Fraction 2: the quantity x squared minus 9 over the quantity x squared minus 2 times x plus 1. Find Fraction 1 over Fraction 2.
A. The quantity x plus 3 times the quantity x minus 1 divided by the quantity x minus 3
B. The quantity x minus 3 times the quantity x plus 1 divided by the quantity x plus 3
C. The quantity x plus 3 times the quantity x plus 1, all over the quantity x minus 3
D. The quantity x minus 3 times the quantity x minus 1, all over the quantity x plus 3
We can start by simplifying the numerators and denominators separately:
Fraction 1:
$$\frac{x^2 + 6x + 9}{x - 1} = \frac{(x + 3)^2}{x - 1}$$
Fraction 2:
$$\frac{x^2 - 9}{x^2 - 2x + 1} = \frac{(x + 3)(x - 3)}{(x - 1)^2}$$
To divide two fractions, we can multiply the first fraction by the reciprocal of the second fraction:
$$\frac{(x + 3)^2}{x - 1} \cdot \frac{(x - 1)^2}{(x + 3)(x - 3)} = \frac{(x + 3)(x + 3)(x - 1)(x - 1)}{(x - 1)(x + 3)(x - 3)}$$
We can then cancel out the common factors in the numerator and denominator:
$$\frac{(x + 3)(x - 1)}{x - 3}$$
So the simplified expression is:
$$\boxed{\frac{(x + 3)(x - 1)}{x - 3}}$$
The answer is option A, The quantity x plus 3 times the quantity x minus 1 divided by the quantity x minus 3.
Divide. The quantity 12 times t to the sixth power minus 8 times t to the eighth power, all divided by 2 times t cubed
A. 4 times m to the seventh power minus 5 times m to the fifth power
B. 4 times m to the fourth power minus 10 times m to the fifth power
C. 8 times m to the seventh power minus 10 times m squared
D. 4 times m to the fourth power minus 5 times m squared
We can start by factoring out 4t^6 from the numerator:
$$\frac{12t^6 - 8t^8}{2t^3} = \frac{4t^6(3 - 2t^2)}{2t^3}$$
We can simplify this expression by canceling out the 2, and the t^3 in the numerator and denominator:
$$\frac{4t^6(3 - 2t^2)}{2t^3} = \boxed{2t^3(3 - 2t^2)}$$
The answer is option D, 4 times m to the fourth power minus 5 times m squared.
Divide. The quantity 10 times d to the ninth power minus 6 times d to the eighth power plus 4 times d to the fifth power, all divided by 2 times d cubed
A. 5 times d to the sixth power minus 3 times d to the sixth power plus 2 times d squared
B. 5 times d to the sixth power minus 6 times d to the eighth power plus 4 times d to the fifth power
C. 10 times d to the ninth power minus 6 times d to the eighth power plus 2 times d squared
D. 5 times d to the ninth power minus 3 times d to the ninth power plus 2 times d to the fifth power