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3 years ago

12

Roy drives a sedan that gets mileage of 30 miles per gallon. If fuel consumption is represented by g and mileage by m, which fun

ction represents the relationship between m and g?

1 answer:

BARSIC [14]3 years ago

6 0

30m/g

There isn't much to it. What are the options?

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HELP! WILL GIVE BRAINLIEST!

romanna [79]

Answer:

I Know it is B trust me

Step-by-step explanation:

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3 years ago

Read 2 more answers

If you can help please do! Thank you!

Aleks [24]

Answer:

see explanation

Step-by-step explanation:

The diagonals of a parallelogram bisect each other , then

CO = OA = b

DO = OB = a

(a)

CA = CO + OA = b + b = 2b

(b)

DC = DO + OC = a - b

(c)

CB = CO + OB = b + a = a + b

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2 years ago

8) Sumeya bought 2 packs of scrunchies for $31.75. How much

frutty [35]

Divide $31.75 by 2 to find out the cost for 1 pack. Then multiply that total times 3 to get the total for 3 packs.

31.75/2 = 15.875

15.875 x 3 = 47.625, rounded to $47.63

5 0

3 years ago

Lowest common multiple of 20 and 54

Ivahew [28]

The LCM is 540 because 54X10=540 which 20 goes into

8 0

3 years ago

Prove the divisibility:<br><br>45^45·15^15 by 75^30

garri49 [273]

Answer:

$3^{75}$ .

Step-by-step explanation:

We have been an division problem: $\frac{45^{45}*15^{15}}{75^{30}}$ .

We will simplify our division problem using rules of exponents.

Using product rule of exponents $(a*b)^n=a^n*b^n$ we can write:

$45^{45}=(9*5)^{45}=9^{45}*5^{45}$

$15^{15}=(3*5)^{15}=3^{15}*5^{15}$

$75^{30}=(15*5)^{30}=15^{30}*5^{30}$

Substituting these values in our division problem we will get,

$\frac{9^{45}*5^{45}*3^{15}*5^{15}}{15^{30}*5^{30}}$

Using power rule of exponents $a^n*a^m=a^{n+m}$ we will get,

$\frac{9^{45}*5^{(45+15)}*3^{15}}{15^{30}*5^{30}}$

$\frac{9^{45}*5^{60}*3^{15}}{15^{30}*5^{30}}$

Using product rule of exponents $(a*b)^n=a^n*b^n$ we will get,

$\frac{(3*3)^{45}*5^{60}*3^{15}}{(3*5)^{30}*5^{30}}$

$\frac{3^{45}*3^{45}*5^{60}*3^{15}}{3^{30}*5^{30}*5^{30}}$

Using power rule of exponents $a^n*a^m=a^{n+m}$ we will get,

$\frac{3^{(45+45+15)}*5^{60}}{3^{30}*5^{(30+30)}}$

$\frac{3^{105}*5^{60}}{3^{30}*5^{60}}$

$\frac{3^{105}}{3^{30}}$

Using quotient rule of exponent $\frac{a^m}{a^n}=a^{m-n}$ we will get,

$\frac{3^{105}}{3^{30}}=3^{105-30}$

$3^{105-30}=3^{75}$

Therefore, our resulting quotient will be $3^{75}$ .

7 0

3 years ago