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lutik1710 [3]
3 years ago
11

One week, Kerry travels 125 miles and uses 5 gallons of gasoline. The next week, she travels 175 miles and uses 7 gallons of gas

oline. Which best describes the function that can be used to represent m, the number of miles traveled, and g, the number of gallons used?
Mathematics
2 answers:
Arlecino [84]3 years ago
7 0

Answer:

m=25g  direct variation

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form y/x=k or y=kx

In this problem

Let

m------> the number of miles traveled

g------> the number of gallons used

we have that

\frac{125}{5}\frac{miles}{gallons}=25\frac{miles}{gallons}

\frac{175}{7}\frac{miles}{gallons}=25\frac{miles}{gallons}

The linear direct variation that represent the situation is equal to the equation

\frac{m}{g}=k  or m=kg

where

k is the constant of proportionality

In a direct variation the constant k is equal to the slope m of the line, and the line passes through the origin

k=25\frac{miles}{gallons}

substitute

m=kg -------> m=25g



Leya [2.2K]3 years ago
6 0
First find out how much in that one week, the amount of miles per gallon

125/5 = 25, or 25 mpg

Find out how much for the next week

175/7 = 25, or 25 mpg

Note: 
g = gallons
m = miles

you're equation will be:
m = 25(g)

For every gallon (g), you multiply it by 25 to get miles (m).


hope this helps
You might be interested in
For a 95% confidence interval what would the sample size need to be as it's most conservative to have a 2% margin of error
aleksley [76]

Answer:

<em>The size of the sample 'n' = 2401</em>

Step-by-step explanation:

<u><em>Step:1</em></u>

Given that the margin of error = 2 % = 0.02

The margin of error is determined by

              M.E = \frac{Z_{0.05} \sqrt{p(1-p)} }{\sqrt{n} }

we know that the proportion

               \sqrt{p(1-p)} < \frac{1}{2}

<u><em>Step:2</em></u>

<em>The margin of error is determined by </em>

<em>               </em>0.02 = \frac{1.96( \frac{1}{2} ) }{\sqrt{n} }<em></em>

              \sqrt{n} =  \frac{1.96}{2 X 0.02}

              √n  = 49

Squaring on both sides, we get

                n = 49 × 49

                n = 2401

<u>Final answer:-</u>

<em>The size of the sample 'n' = 2401 </em>

8 0
3 years ago
There is a 0.05 chance that we will have an earthquake this year. What is the probability we won't have an earthquake this year
viktelen [127]

Answer:

0.95‬ chance or 95% chance.

Step-by-step explanation:

If there is a 0.05 chance of there being and earthquake subtract 0.05 from 1 to find out what the chance of not having an earthquake it.  Since probability is a fraction out of a whole percentages can be used to find the answer.  

3 0
3 years ago
Rewrite the expression (x+3)(3x+8)-3X(x+3)
inysia [295]
Multiply (x+3)(3x+8)      x times 3x =3x^2 then multiply x times 8 = 8x
                                 keep multiplying 3 times 3x =9x  then 3 times 8=24 
                          then multiply the -3x(x+3)=  -3x^2-9x
your hole expression will look like this 3x^2+8x+9x+ 24-3x^2-9x

if you need to simplify you can do by adding or subtracting them.
for example you can add 8x+9x= 17x

      
6 0
4 years ago
21. x² +9x - 22<br> Factors of cSum of Factors<br> 7<br> 7<br> ?<br> ?<br> 7<br> 7<br> 9<br> 7
Eduardwww [97]

Answer:

\large \begin{array}{| c | c |}\cline{1-2} \sf Factors\:of\:c & \sf Sum\:of\:Factors\\\cline{1-2} 1, -22 & -21 \\\cline{1-2} -1, 22 & 21\\\cline{1-2} 2, -11 & -9 \\\cline{1-2} -2, 11 & 9 \\\cline{1-2}\end{array}

Step-by-step explanation:

Quadratic equation: ax² + bx + c = 0, where a ≠ 0

x² + 9x - 22

  • a = 1
  • b = 9
  • c = -22

Factors of c (-22):

1 × -22     or     -1 × 22

2 × -11      or     -2 × 11

Sum of factors:

1 + (-22) = -21          or          -1 + 22 = 21

2 + (-11) = -9            or           -2 + 11 = 9

Table:

\large \begin{array}{| c | c |}\cline{1-2} \sf Factors\:of\:c & \sf Sum\:of\:Factors\\\cline{1-2} 1, -22 & -21 \\\cline{1-2} -1, 22 & 21\\\cline{1-2} 2, -11 & -9 \\\cline{1-2} -2, 11 & 9 \\\cline{1-2}\end{array}

Hope this helps!

6 0
3 years ago
For the function defined by f(t)=2-t, 0≤t&lt;1, sketch 3 periods and find:
Oksi-84 [34.3K]
The half-range sine series is the expansion for f(t) with the assumption that f(t) is considered to be an odd function over its full range, -1. So for (a), you're essentially finding the full range expansion of the function

f(t)=\begin{cases}2-t&\text{for }0\le t

with period 2 so that f(t)=f(t+2n) for |t| and integers n.

Now, since f(t) is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L

where

b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt

In this case, L=1, so

b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt
b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}
b_n=\dfrac{4-2(-1)^n}{n\pi}

The half-range sine series expansion for f(t) is then

f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t

which can be further simplified by considering the even/odd cases of n, but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming f(t) is even/symmetric across its full range. In other words, you are finding the full range series expansion for

f(t)=\begin{cases}2-t&\text{for }0\le t

Now the sine series expansion vanishes, leaving you with

f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L

where

a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt

for n\ge0. Again, L=1. You should find that

a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3

a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt
a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}
a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}

Here, splitting into even/odd cases actually reduces this further. Notice that when n is even, the expression above simplifies to

a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0

while for odd n, you have

a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}

So the half-range cosine series expansion would be

f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t

Attached are plots of the first few terms of each series overlaid onto plots of f(t). In the half-range sine series (right), I use n=10 terms, and in the half-range cosine series (left), I use k=2 or n=2(2)-1=3 terms. (It's a bit more difficult to distinguish f(t) from the latter because the cosine series converges so much faster.)

5 0
4 years ago
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