Christy drove 300 miles on her vacation. She drove an average of 1.25 times faster on the second 150 miles of her trip than she did on the first 150 miles of her trip. Which expression represents the time she spent driving? Let x = her speed on the first half of the trip.
A. 375 over x
B. 337.5 over x
C. 270 times x
D. 270 over x
Let's call Christy's speed on the first 150 miles "x". Then her speed on the second 150 miles would be 1.25x (since she drove 1.25 times faster).
We know that distance = rate x time, so we can set up two equations:
150 = x * t1 (where t1 is the time it took Christy to drive the first 150 miles)
150 = 1.25x * t2 (where t2 is the time it took Christy to drive the second 150 miles)
We can solve for t1 and t2 by dividing both sides of each equation by their respective rates:
t1 = 150/x
t2 = 120/x
(Note that we used 1.25x for the rate on the second leg of the trip, since she was driving 1.25 times faster than x).
Now we can add the two times together to get the total time Christy spent driving:
total time = t1 + t2
total time = 150/x + 120/x
total time = 270/x
Therefore, the expression that represents the time Christy spent driving is D. 270/x.
Simplify. start fraction negative 14 x cubed over x cubed minus 5 x superscript 4 baseline end fraction
A. Start Fraction negative 14 over 5 lower x minus 1 End Fraction semicolon where lower x does not equal Start Fraction 1 over 5 End Fraction comma zero
B. Start Fraction negative 14 lower x over 1 minus 5 lower x End Fraction semicolon where lower x does not equal Start Fraction 1 over 5 End Fraction
C. Start Fraction 1 minus 5 lower x over negative 14 lower x End Fraction semicolon where lower x does not equal zero
D. Start Fraction negative 14 over 1 minus 5 lower x End Fraction semicolon where lower x does not equal Start Fraction 1 over 5 End Fraction comma zero
We can factor out the greatest common factor of the numerator, which is -14x³.
start fraction negative 14 x cubed over x cubed minus 5 x superscript 4 baseline end fraction = start fraction negative 14x³ (1 - 5x) over x³ end fraction
Now we can simplify by canceling out the x³ in the numerator and denominator:
start fraction negative 14x³ (1 - 5x) over x³ end fraction = start fraction negative 14 (1 - 5x) over 1 end fraction = -14(1-5x)
Therefore, the simplified expression is A. Start Fraction negative 14 over 5 lower x minus 1 End Fraction, where lower x does not equal Start Fraction 1 over 5 End Fraction, 0.
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
We can use long division to simplify the expression:
```
-6x² + 5x - 4
-------------------
3x + 2 | -18x³ + 17x + 6
18x³ + 12x²
-------------
5x² + 17x
5x² + 3x
--------
14x + 6
```
Therefore,
the quantity negative 18 times x cubed plus 17 times x plus 6 divided by the quantity 3 times x plus 2
= (-6x² + 5x - 4) + (14x + 6) / (3x + 2)
= -6x² + 19x + 2 / (3x + 2)
Therefore, the simplified expression is option B. 6 times x squared plus 4 times x minus 3.
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 factor out x in the denominator:
start fraction x over 6 x - x squared end fraction = start fraction x(1 - x) over x(6 - x) end fraction
Now we can cancel out the x in the numerator and denominator:
start fraction x(1 - x) over x(6 - x) end fraction = start fraction 1 - x over 6 - x end fraction
Finally, we can rearrange the terms in the numerator to get:
start fraction 1 - x over 6 - x end fraction = start fraction -x + 1 over -x + 6 end fraction
Therefore, the simplified expression is A. Start Fraction 1 over 6 minus lower x End Fraction, where lower x does not equal zero, 6.
Do the data in the table represent a direct variation or an inverse variation? Write an equation to model the data in the table.
x 1 3 5 10
y 4 12 20 40
A. direct variation; y equals 4 times x
B. direct variation; y equals one-fourth x
C. inverse variation; x times y equals 4
D. inverse variation; x times y equals one fourth
Based on the values in the table, we can see that as x increases, y also increases. However, the relationship between x and y is not simply multiplication or division by a constant factor. Instead, the product of x and y remains the same:
1 * 4 = 4
3 * 12 = 36
5 * 20 = 100
10 * 40 = 400
This implies an inverse variation between x and y.
The equation to model the data can be written as:
xy = 4
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 factor the denominator:
x² + 4x - 21 = (x + 7)(x - 3)
Then, we can rewrite the expression as:
start fraction x + 7 over x squared + 4x - 21 end fraction = start fraction x + 7 over (x + 7)(x - 3) end fraction
Now, we can simplify by canceling out the common factor of x+7 in the numerator and the denominator:
start fraction x + 7 over (x + 7)(x - 3) end fraction = start fraction 1 over x - 3 end fraction
Therefore, the simplified expression is A. Start Fraction 1 over lower x minus 3 End Fraction, where lower x does not equal 3, -7.
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
To divide fractions, we can multiply by the reciprocal of the second fraction:
start fraction start fraction x squared minus 16 over x minus 1 end fraction divided by x + 4 end fraction = start fraction x squared - 16 over x - 1 end fraction times start fraction 1 over x + 4 end fraction
Factor the numerator:
start fraction x squared - 16 over x - 1 end fraction times start fraction 1 over x + 4 end fraction = start fraction (x + 4)(x - 4) over x - 1 end fraction times start fraction 1 over x + 4 end fraction
Cancel out the common factor of x+4:
start fraction (x + 4)(x - 4) over x - 1 end fraction times start fraction 1 over x + 4 end fraction = start fraction x - 4 over x - 1 end fraction
Therefore, the answer is A. The quantity x minus 4 over the quantity x minus 1.
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
We know that the area of a rectangle is equal to its length times its width. We are given that the width of the playground is x+3, so we can set up the equation:
Area = Length times Width
x³ - 7x + 6 = Length(x+3)
We can simplify the right side by distributing the Length:
x³ - 7x + 6 = Length*x + 3*Length
Now we can solve for Length by isolating it on one side:
x³ - 7x + 6 = Length*x + 3*Length
x³ - 7x + 6 = x*Length + 3*Length
x³ - 7x + 6 = (x+3)*Length
Divide both sides by x+3:
start fraction x cubed - 7x + 6 over x + 3 end fraction = Length
We can simplify the expression on the left side by long division:
```
x^2 - 3x + 2
---------------
x + 3 | x^3 + 0x^2 - 7x + 6
x^3 + 3x^2
-----------
-3x^2 - 7x
-3x^2 - 9x
---------
2x + 6
```
Therefore,
start fraction x cubed - 7x + 6 over x + 3 end fraction = x² - 3x + 2
Therefore, the expression for the length of the playground is A. x squared minus 3 times x plus 2.
Divide. The quantity 8 times m to the seventh power minus 10 times m to the fifth power, all divided by 2 times m 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 simplify the expression:
start fraction 8m^7 - 10m^5 over 2m^3 end fraction = 4m^(7-3) - 5m^(5-3)
= 4m^4 - 5m^2
Therefore, the answer is option D. 4 times m to the fourth power minus 5 times m squared.
Simplify into one fraction.
start fraction 9 over x minus 1 end fraction plus start fraction 5 over x plus 4 end fraction
A. start fraction 14 over 2 x plus 3 end fraction
B. start fraction 14 x plus 31 over left parenthesis x minus 1 right parenthesis left parenthesis x plus 4 right parenthesis end fraction
C. start fraction 14 x plus 41 over left parenthesis x minus 1 right parenthesis left parenthesis x plus 4 right parenthesis end fraction
D. start fraction 14 over left parenthesis x minus 1 right parenthesis left parenthesis x plus 4 right parenthesis end fraction
To add two fractions, we need to find a common denominator. In this case, the common denominator is (x-1)(x+4).
We can rewrite the expressions with the common denominator:
start fraction 9 over x - 1 end fraction + start fraction 5 over x + 4 end fraction = start fraction 9(x+4) over (x-1)(x+4) end fraction + start fraction 5(x-1) over (x-1)(x+4) end fraction
= start fraction 9x + 36 + 5x - 5 over (x-1)(x+4) end fraction
= start fraction 14x + 31 over (x-1)(x+4) end fraction
Therefore, the answer is option B. start fraction 14 x plus 31 over left parenthesis x minus 1 right parenthesis left parenthesis x plus 4 right parenthesis.
Simplify. The quotient 7 over 2 times a times the quotient 5 over a squared
A. 35 over 2 times a squared, where a does not equal 0
B. 35 over 2 times a squared
C. 12 over 2 times a cubed, where a does not equal 0
D. 35 over 2 times a cubed, where a does not equal 0
We can simplify the expression by multiplying the two fractions:
start fraction 7 over 2a end fraction times start fraction 5 over a squared end fraction = start fraction 35 over 2a cubed end fraction
Therefore, the simplified expression is option D. start fraction 35 over 2 times a cubed, where a does not equal 0.
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