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3 years ago

6

For every 60 miles, the car uses 2 gallons of gas. gas costs $2.79 a gallon. what will be the total cost for a one way trip of 9

0 miles?

1 answer:

dybincka [34]3 years ago

7 0

$8.37. Since 2 gallons of gas are used every 60 miles, we can calculate that 1 gallon of gas is used every 30 miles. 90mi/30mi = 3 gallons. 3 gallons x $2.79 = $8.37.

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Use substitution to solve the<br> following system of equations.<br> 9x + 5y = 32 AND y = -3x + 4

Darya [45]

Answer:

x = -2

y = 10

Step-by-step explanation:

Since they already gave y as -3x +4, you will then have to substitute that amount into the formula

9x + 5y = 32

By doing that you get

9x +5(-3x+4) = 32

9x -15x +20 = 32

Then, you ill take 20 to the other side to subtract 32, which will get you

-6x = 12

Then, you will get x = -2

Then substitute that for the equation

y = -3x + 4

You will get

y = -3(-2) + 4,

You will get y as 10

SO then your final answer is x = -2, y = 10

4 0

3 years ago

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

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3 years ago

Geometry Question Help if you can please

docker41 [41]

Answer:

≈ 78.55 cm

Step-by-step explanation:

Using the ratio of arc (AB) to circumference (C) is equal to the ratio of angle subtended at centre by arc AB to 360°, that is

$\frac{AB}{C}$ = $\frac{110}{360}$ , that is

$\frac{24}{C}$ = $\frac{110}{360}$ ( cross- multiply )

110C = 8640 ( divide both sides by 110 )

C = $\frac{8640}{110}$ ≈ 78.55 cm ( to 2 dec. places )

5 0

3 years ago

Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If X is the number among 15 randomly sel

Vlad [161]

Answer:

The required probability is 0.94

Step-by-step explanation:

Consider the provided information.

There are 400 refrigerators, of which 40 have defective compressors.

Therefore N = 400 and X = 40

The probability of defective compressors is:

$\frac{40}{400}=0.10$

It is given that If X is the number among 15 randomly selected refrigerators that have defective compressors,

That means n=15

Apply the probability density function.

$P(X=x)=^nC_xp^x(1-p)^{n-x}$

We need to find P(X ≤ 3)

$P(X\leq3) =P(X=0)+P(X=1)+P(X=2)+P(X=3)\\P(X\leq3) =\frac{15!}{15!}(0.1)^0(1-0.1)^{15}+\frac{15!}{14!}(0.1)^1(1-0.1)^{14}+\frac{15!}{13!2!}(0.1)^2(1-0.1)^{13}+\frac{15!}{12!3!}(0.1)^3(1-0.1)^{12}\\$

$P(X\leq3) =0.944444369992\approx 0.94$

Hence, the required probability is 0.94

4 0

3 years ago

A parking meter charges $2 per hour. as a driver parks her car, she notices that the parking meter has 15 minutes remaining on i

Ne4ueva [31]

<span>The parking meter charges $2 per hour, means that each half hour costs $1.

Thus, 1$ pays for half an hour, so d $ pay for d half hours, that is d/2 hours, which is 0.5d hours.

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Answer: B</span>

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4 years ago

Read 2 more answers