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3 years ago

10

Tina and her friends want to go bowling. Family Bowling Alley charges $1 for shoes and $4 dollars per game. Knight Owl Bowling c

harges $2 for shoes and $3.75 per game. How many games will it take for the cost of bowling at each alley be the same?

1 answer:

Serhud [2]3 years ago

6 0

Answer:

Number of games = 4

Step-by-step explanation:

Let the number of games = x

Family Bowling Alley charges $1 for shoes and $4 dollars per game

If x games is played

Cost= 1+4(x)

Knight Owl Bowling charges $2 for shoes and $3.75 per game.

If x games is played

Cost = 2 + 3.75(x)

When the both cost are equal

1+4(x)= 2+3.75(x)

4x-3.75x = 2-1

0.25x = 1

X= 1/0.25

X= 4

Number of games = 4

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velikii [3]

we have that

*-------------------------*--------------------------------*

E F G

EF=2x-12

FG= 3x-15

EG=23

we know that

EF + FG = EG

so

[2x - 12] + [3x - 15] = 23 simplify

5x - 27 = 23 add 27 to both sides

5x = 50 divide both sides by 5

x = 10

EF=2x-12-------> EF=2*10-12-------> EF=8

FG= 3x-15------> FG=3*10-15------> FG=15

therefore

the answer part a) is

the value of x is 10

the answer part b) is

the value of EF is 8

the answer part c) is

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2 years ago

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Verdich [7]

Answer:

1. 600 watts

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Step-by-step explanation:

1. Energy = 36 000 J , t = 1 minute = 60 seconds

Power = $\frac{energy}{time}$

= $\frac{36000}{60}$

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2 years ago

Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ

atroni [7]

Answer: $y=Ce^(^3^t^{^9}^)$

Step-by-step explanation:

Beginning with the first differential equation:

$\frac{dy}{dt} =27t^8y$

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

$\frac{1}{y} \frac{dy}{dt} =27t^8$

Multiply both sides by 'dt' to get:

$\frac{1}{y}dy =27t^8dt$

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

$\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt$

$ln(y)=27(\frac{1}{9} t^9)+C$

$ln(y)=3t^9+C$

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

$e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)$

$y=e^(^3^t^{^9} ^+^C^)$

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

$y=e^(^3^t^{^9}^)e^C$

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

$y=Ce^(^3^t^{^9}^)$

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

$\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8$

Now check if the derivative equals the right side of the original differential equation:

$(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)$

$Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)$

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

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3 years ago

a 18 ft tall statue standing next to a globe casts a 12 ft shadow. Of the globe casts a shadow that is 2 ft ling, then how tall

Igoryamba

<h3>Answer:</h3>

3 ft

<h3>Step-by-step explanation:</h3>

The statue's height is 1.5 times the length of its shadow, so we expect the same relationship for the globe.

... 1.5 × 2 ft = 3 ft

_____

<em>Comment on the problem</em>

As a practical matter, with the sun high enough in the sky to cast a shadow shorter than the object's height, it will be quite difficult to measure the length of the shadow of the point at the top of the globe. The shadow of other parts of the globe will interfere.

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