The time, t, required to drive a certain distance varies inversely with the speed, r. If it takes 7 hours to drive the distance at 55 miles per hour, how long will it take to drive the same distance at 65 miles per hour? (1 point)

5.92 hours. You get this by multiplying 7 and 55 to get 385, then divide 385 by 65 to get 5.92.

time/7=55/65

solve for time

385.00 hours?

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

Can someone help me with this please

Let's say that Christy's speed on the first 150 miles of her trip is x. Then her speed on the second 150 miles of her trip is 1.25x (1.25 times faster than x).

We can use the formula time = distance/speed to find the time Christy spent driving each half of her trip:

Time for first 150 miles = 150/x
Time for second 150 miles = 150/(1.25x) = 120/x

Total time = (150/x) + (120/x) = (270/x)

Therefore, the expression that represents the time she spent driving is D. 270/x.

Solve.

The quotient 1 over the quantity 2 times x minus 8 minus the quotient 3 over the quantity x minus 4 equals negative 5
A. x = 3
B. x equals nine-halves
C. x = 6
D. x equals negative 24 over 5

We can start by simplifying the left-hand side of the equation using the common denominator of (2x-8)(x-4):

1/(2x-8) - 3/(x-4) = (x-4)/(2x-8)(x-4) - 6/(2x-8)(x-4)
= (x-10)/(2x-8)(x-4)

Now we can set up the equation:

(x-10)/(2x-8)(x-4) = -5

Multiplying both sides by the denominator gives:

x - 10 = -5(2x-8)(x-4)
x - 10 = -10x^2 + 46x - 160
10x^2 - 45x + 150 = 0
2x^2 - 9x + 30 = 0

Using the quadratic formula, we get:

x = (9 ± sqrt(9^2 - 4(2)(30))) / (2(2))
x = (9 ± sqrt(9))/4

x = 3 or x = 5/2

Checking both solutions, we find that only x = 5/2 makes the denominators in the original equation non-zero, so the answer is B. x = 5/2.

Solve.

start fraction 8 over x end fraction plus start fraction 15 over x squared end fraction equals negative 1
A. x equals 5 comma negative 3
B. x equals negative 5 comma negative 3
C. x equals negative 5 comma 3
D. x = 5, 3

We can start by getting a common denominator for the two fractions on the left-hand side:

8/x + 15/x^2 = (8x + 15) / x^2

Now we can set up the equation:

(8x + 15) / x^2 = -1

Multiplying both sides by x^2 gives:

8x + 15 = -x^2
x^2 + 8x + 15 = 0

This quadratic factors as:

(x + 3)(x + 5) = 0

So the solutions are x = -3 and x = -5.

However, we need to check these solutions in the original equation, since we multiplied by x^2 which could be zero and make the equation undefined.

Checking x = -3:
8/(-3) + 15/(-3)^2 = -1
-8 + 5 = -1
This is not true, so x = -3 is not a solution.

Checking x = -5:
8/(-5) + 15/(-5)^2 = -1
-8/5 + 3/5 = -1
This is true, so x = -5 is the solution.

Therefore, the answer is (C) x = -5, 3.

What are the excluded values of the function? y equals 5 over the quantity 6 times x plus 72

A. x = 0
B. x = 12
C. x = 72
D. x = 11