All categories

2 years ago

Mr. Williams is driving on a highway at an average speed of 50 miles per hour. His destination is 100 miles away. The

Mathematics

1 answer:

svetlana [45]2 years ago

4 0

Answer:

t = -d/50 + 2

0.5 hour

Step-by-step explanation:

Given the equation:

d = 50 - 100t

The inverse function:

A.) solving for t

d = 100 - 50t

d - 100 = - 50t

Divide both sides by - 50

d/-50 - (100/-50) = - 50t/-50

-d/50 - (-2) = t

t = -d/50 + 2

B) using the inverse function:

t = -d/50 + 2

Miles driven (d) = 75, find time (t)

t = - 75/50 + 2

t = - 1.5 + 2

t = 0.5

t is in hours, therefore time left to travel is 0.5 hours or 30 minutes

You might be interested in

If one side of a cube is 15m in length,find its volume

sergeinik [125]

Answer:

$V = 3375~m^{3}$

Step-by-step explanation:

The formula of the volume of a cube is $V = a^{3}$ where "a" would correspond to the length of an edge/side/height/etc.

If we insert our value into the formula, we get the following result:

$V = a^{3}$

$V = 15m^{3}$

$V = 3375~m^{3}~(cubic~meters)$

Hope it helped,

BiologiaMagister

5 0

3 years ago

Read 2 more answers

Urgent! Need help solving

OLga [1]

Answer:

d 64

Step-by-step explanation:

f(-3)=1/2 ^-6=2^6=64

5 0

3 years ago

Which algebraic expression is equivalent to the expression below?

spayn [35]

<span>8(3x + 6) + 13---expand using distributive property
=24x + 48+13 ...combine like terms
=24x +61....simplify

answer is C.24x + 61

hope it helps</span>

5 0

3 years ago

Suppose a man offered to work for 30 days at the following salary: one cent for the first day two cents for the second day four

Ivahew [28]

It would be to little because it would only be 30 cents

8 0

3 years ago

Write a definite integral that represents the area of the region. (Do not evaluate the integral.)

joja [24]

Answer:

$\int_{-1}^{1}7(x^{3}-x)\:dx$

Step-by-step explanation:

1) The other curve is $y=0$ then the common points of both curves are x-intercepts, the roots of $y=7(x^{3}-x)$

$y=7(x^{3}-x)\Rightarrow 7(x^3-x)=0 \Rightarrow 7(x^{3}-x)=7x(x-1)(x-1)\Rightarrow \\S=\left ( 0,0 \right ),\left ( -1,0 \right ),\left ( 1,0 \right )$

2). Then those intersection points are the upper and the lower limits. Plugging in to this formula for they belong to the interval [-1,1]:

$\int_{a}^{b}|f(x)-g(x)dx$

$\int_{a}^{b}|f(x)-g(x)dx \Rightarrow \int_{-1}^{1}|7(x^{3}-x)-0|dx \Rightarrow \int_{-1}^{1}7(x^{3}-x)\:dx$

8 0

3 years ago