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pishuonlain [190]

3 years ago

11

You are told that the average rate of change of a particular function is always negative. What can you conclude about the graph

of that function and why?

1 answer:

ira [324]3 years ago

6 0

A negative average rate of change means that as x increases, y decreases. That means that the graph goes down to the right.

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the value of a car that cost $16,000 decreased to $12,000 after 2 years what is the percent of decrease

choli [55]

Answer: 25%

Step-by-step explanation:

16 000 - 12 000 = 4000

16 000 / 4000 = 4

100 / 4 = 25

7 0

3 years ago

Read 2 more answers

3(x-2)+2(x+1)=-14<br> plz show work

Rom4ik [11]

Hello!

First, let's write the problem.
$3\left(x-2\right)+2\left(x+1\right)=-14$
Apply the distributive property on the left side of the equation.
$=3x-6+2x+2$
Add like terms.
$=5x-4$
Let's plug that in into the original equation.
$5x-4=-14$
Add 4 to both sides.
$5x-4+4=-14+4$
$5x=-10$
Divide both sides by 5.
$\frac{5x}{5}=\frac{-10}{5}$

Our final answer would be,
$x=-2$

You can feel free to let me know if you have any questions regarding this!
Thanks!

- TetraFish

5 0

3 years ago

Read 2 more answers

Help Asap 20 points!! And Brainlist!!!

Alenkasestr [34]

Answer: 2(5c - 1x)

Step-by-step explanation: It's asking you to take out the greatest common factor (GCF).

5 0

3 years ago

A road perpendicular to a highway leads to a farmhouse located d miles away. An automobile traveling on this highway passes thro

pshichka [43]

Answer:

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}$

Step-by-step explanation:

A road is perpendicular to a highway leading to a farmhouse d miles away.

An automobile passes through the point of intersection with a constant speed $\frac{dx}{dt}$ = r mph

Let x be the distance of automobile from the point of intersection and distance between the automobile and farmhouse is 'h' miles.

Then by Pythagoras theorem,

h² = d² + x²

By taking derivative on both the sides of the equation,

$(2h)\frac{dh}{dt}=(2x)\frac{dx}{dt}$

$(h)\frac{dh}{dt}=(x)\frac{dx}{dt}$

$(h)\frac{dh}{dt}=rx$

$\frac{dh}{dt}=\frac{rx}{h}$

When automobile is 30 miles past the intersection,

For x = 30

$\frac{dh}{dt}=\frac{30r}{h}$

Since $h=\sqrt{d^{2}+(30)^{2}}$

Therefore,

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+(30)^{2}}}$

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}$

3 0

3 years ago

Help me pls can someone do all 15 questions of my homework I need help badly

Tema [17]

Answer:

bye

Step-by-step explanation:

bye

6 0

3 years ago