Afamily has two cars. the first car has a fuel efficiency of 40 miles per gallon of gas and the second has a fuel efficiency of 30 miles per gallon of gas. during one particular week, the two cars went a combined total of 1900 miles, for a total gas consumption of 55 gallons. how many gallons were consumed by each of the two cars that week?
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The first car consumed 25 gallons of fuel and the second car consumed 30 gallons of fuel.

Step-by-step explanation:

Let us assume that the first car consumed x gallons and the second car consumed y gallons that week.

So, x + y = 55 (1)

⇒ 30x + 30y = 1650 (2)

Now, the second car has a fuel efficiency of 30 miles per gallons and the first car has it of 40 miles per gallons.

So, if the two car combined went for 1900 miles then we can write

40x + 30y = 1900 (3)

Now, solving equations (2) and (3) we get,

40x - 30x = 1900 - 1650

⇒ 10x = 250

⇒ x = 25 gallons.

And from equation (1) we get,

y = 55 - x = 55 - 25 = 30 gallons.

Therefore, the first car consumed 25 gallons of fuel and the second car consumed 30 gallons of fuel. (Answer)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.

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