All categories

2 years ago

Jennifer is 20 miles north of her house, and she is driving north on the highway at a rate of 55 miles per hour whats the slpoe

Mathematics

1 answer:

LenaWriter [7]2 years ago

3 0

Answer:

The slope is 55.

Step-by-step explanation:

Slope determines the direction and the rate of increase or decrease between the two points on a surface or line.

In the question, it is given initially she is 20 miles north of her house

⇒ When x = 0 the y-coordinate is 20

So, the y-intercept is 20

Now, it is given that the rate of change is 55 miles per hour

Hence, the slope is same as 55 because it is the actual rate of change between the positions of Jennifer with time.

Hence, The slope is 55.

You might be interested in

If (k-6)(k+2)-(k²-20)=2k+6, whatis tge value of k?

xxMikexx [17]

K=1/3

simplify each bracket, put all numbers on the left hand side, equal it to 0 and solve for k

7 0

1 year ago

I need the answer for c

Temka [501]

The girth because it makes the whole tree bigger and the branches and everything would get bigger too

3 0

2 years ago

The volume of a rectangular prism is more than 156 cubic inches. The prism has a length of 8 inches and a width of 3 inches. Sel

Anon25 [30]

Multiply All Of The Numbers Thats What I Think You Should Do!

3 0

3 years ago

2.3 The diameter of a circle is 8 m long. Calculate its: radius

slavikrds [6]

Answer:

Step-by-step explanation:

Diameter is equal to 2r

so,

8÷2=4

4 0

2 years ago

Read 2 more answers

What is the value of \dfrac{d}{dx}\left(\dfrac{2x+3}{3x^2-4}\right) dx d ( 3x 2 −4 2x+3 )start fraction, d, divided by, d, x

stellarik [79]

Answer:

Step-by-step explanation:

We are asked to find the value of expression $\frac{d}{dx}(\frac{2x+3}{3x^2-4})$ at $x=-1$ .

First of all, we will find the derivative of the given expression using "Quotient Rule of Derivatives" as shown below:

$(\frac{f(x)}{g(x)})'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^2}$

$\frac{d}{dx}(\frac{2x+3}{3x^2-4})$

$\frac{\frac{d}{dx}(2x+3)*(3x^2-4)-(2x+3)*\frac{d}{dx}(3x^2-4)}{(3x^2-4)^2}$

$\frac{2*(3x^2-4)-(2x+3)*(6x)}{(3x^2-4)^2}$

$\frac{6x^2-8-12x^2-18x}{(3x^2-4)^2}$

$\frac{-6x^2-18x-8}{(3x^2-4)^2}$

Therefore, our required derivative is $\frac{-6x^2-18x-8}{(3x^2-4)^2}$ .

Now, we will substitute $x=-1$ in our derivative to find the required value as:

$\frac{-6(-1)^2-18(-1)-8}{(3(-1)^2-4)^2}$

$\frac{-6(1)+18-8}{(3(1)-4)^2}$

$\frac{-6+18-8}{(3-4)^2}$

$\frac{4}{(-1)^2}$

$\frac{4}{1}$

$4$

Therefore, the value of expression $\frac{d}{dx}(\frac{2x+3}{3x^2-4})$ at $x=-1$ is 4.

6 0

2 years ago