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JulsSmile [24]
3 years ago
9

if c(m)=0.50m+30 represents the cost of renting a car, how many miles were driven if the cost is $130?

Mathematics
2 answers:
Lisa [10]3 years ago
4 0
M= the number of miles 
c= cost 
130(m)=0.50m+30 
subtract 30 on both sides 
100(m)=0.50m
divide .50 to isolate m
m=200
adell [148]3 years ago
4 0
200 miles were driven if the cost is $130.

To solve, identify what each part of the equation means, plug 130 into the right part of the equation, and solve for m.

C(M) ⇒ Cost for renting a car
0.50M ⇒ Cost per mile driven
M ⇒ Miles driven
30 ⇒ Initial cost

Replace C(M) with 130 and solve for M.
130 = 0.50M + 30
100 = 0.50M
200 = M

So, if the cost is $130, 200 miles were driven.
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Evaluate f (x) =2x-8 over the domain {0, 1, 3, 8} What is the range of f (x) ?
JulsSmile [24]

Answer:

{-8,-6.-2,8}

Step-by-step explanation:

we have

f(x)=2x-8

Evaluate f(x) for different values of x

For x=0 -----> f(0)=2(0)-8=-8

For x=1 -----> f(1)=2(1)-8=-6

For x=3 -----> f(3)=2(3)-8=-2

For x=8 -----> f(8)=2(8)-8=8

therefore

The range is equal to

{-8,-6.-2,8}

8 0
3 years ago
Solve this thing pls<br> <img src="https://tex.z-dn.net/?f=3%5Cleft%28q-7%5Cright%29%3D27" id="TexFormula1" title="3\left(q-7\ri
evablogger [386]

Answer:

  • Solution of equation ( q ) = <u>1</u><u>6</u>

Step-by-step explanation:

In this question we have given an equation that is <u>3 </u><u>(</u><u> </u><u>q </u><u>-</u><u> </u><u>7</u><u> </u><u>)</u><u> </u><u>=</u><u> </u><u>2</u><u>7</u><u> </u>and we have asked to solve this equation that means to find the value of <u> </u><u>q</u><u> </u><u>.</u>

<u>Solution : -</u>

\quad \: \longmapsto \:  3(q - 7) = 27

<u>Step </u><u>1</u><u> </u><u>:</u> Solving parenthesis :

\quad \: \longmapsto \:3q - 21 = 27

<u>Step </u><u>2</u><u> </u><u>:</u> Adding 21 on both sides :

\quad \: \longmapsto \:3q -  \cancel{ 21} +  \cancel{21} = 27  +  21

On further calculations we get :

\quad \: \longmapsto \:3q = 48

<u>Step </u><u>3 </u><u>:</u> Dividing by 3 from both sides :

\quad \: \longmapsto \: \frac{ \cancel{3}q}{ \cancel{3}}  =  \cancel {\frac{48}{3} }

On further calculations we get :

\quad \: \longmapsto \:   \pink{\boxed{\frak{q = 16}}}

  • <u>Therefore</u><u>,</u><u> </u><u>solution</u><u> </u><u>of </u><u>equation</u><u> </u><u>(</u><u> </u><u>q </u><u>)</u><u> </u><u>is </u><u>1</u><u>6</u><u> </u><u>.</u>

<u>Verifying</u><u> </u><u>:</u><u> </u><u>-</u>

Now we are very our answer by substituting value of q in the given equation . So ,

  • 3 ( q - 7 ) = 27

  • 3 ( 16 - 7 ) = 27

  • 3 ( 9 ) = 27

  • 27 = 27

  • L.H.S = R.H.S

  • Hence, Verified.

<u>Therefore</u><u>,</u><u> </u><u>our </u><u>solution</u><u> </u><u>is </u><u>correct</u><u> </u><u>.</u>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>
3 0
3 years ago
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Using GCF and the distributive property, write two equivalent expressions for 30+45?
yaroslaw [1]
5x15, 3x25 because 30+45=75 and both 5x15 and 3x25 equal 75
6 0
3 years ago
The tempature is -3 degress.The temperature is expected to decrease 2.5 degress each hour until the temperature reaches -18 degr
Andrews [41]

We can set this problem up as an algebraic equation, which we can then solve to find our answer.


We start at -3. So we can start our equation with that:

-3


However, this doesn't represent any change in temperature, just that we start at -3. To show change, we will use the variable x, where x stands for the number of hours. Each hour accounts for a change of -2.5 in degrees, meaning that we must multiply our x by -2.5. We can now add this term into our equation:

-3 - 2.5x


We are trying to find when we will reach -18 degrees, so we will set our entire equation equal to -18 and solve:

-3 - 2.5x = -18

-2.5x = -15

x = 6


It will take us 6 hours, or choice A.

6 0
3 years ago
Solve.<br><br> 3 + |t + 4| &lt; 10
vivado [14]

3 + |t + 4| < 10

Subtract 3

|t + 4| < 7

Because it is absolute value...

t + 4 < 7   AND  t + 4 > -7

t < 3   AND   t > -11

ANSWER:

t < 3   AND   t > -11

8 0
3 years ago
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