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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?
Mathematics
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
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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
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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:

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3 years ago
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