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alukav5142 [94]
2 years ago
13

Anne wants to take her friends to a roller rink. The employee she spoke to ahead of time said the total cost for all 9 people in

the group would be $81. This includes an entrance ticket and a $3 skate rental fee for each person.
If t represents the entrance ticket, explain what does this equation mean?
9(t + 3) = 81

How much is the entrance ticket?
Mathematics
2 answers:
Hoochie [10]2 years ago
7 0

Answer:

I think the answers should be 6

topjm [15]2 years ago
4 0

Answer:

the equation means the total cost of a ticket t and $3 skate rental for nine people is 81 dollars

entrance ticket costs $6

Step-by-step explanation:

solve the equation for t to get the ticket cost

9(t+3)=81

divide by 9 so t+3=9

minus 3 so t=6

that means the ticket costs 6 dollars

Hope this helps! :)

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Answer:

Area = 22.3 m^2

Perimeter 27.4 m^2

Step-by-step explanation:

Perimeter = __ m

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The area of the triangle is about 4.2,

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14.1 + 4.2 + 4 = 22.3

Area = 22.3 m^2

9.4 + 6 + 4 + 4 + 4 = 27.4

Perimeter = 27.4 m^2

Therefore the area of the shape is about 22.3 m^2 and the perimeter is about 27.4 m^2.

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2 years ago
Please help asap i just need the relationship and measure of x
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3 years ago
Read 2 more answers
Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).
lutik1710 [3]

Answer:

The arc length is \dfrac{21}{16}

Step-by-step explanation:

Given that,

The given curve between the specified points is

x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}

The points from (\dfrac{9}{16},1) to (\dfrac{9}{8},2)

We need to calculate the value of \dfrac{dx}{dy}

Using given equation

x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}

On differentiating w.r.to y

\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})

\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})

\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})

\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}

We need to calculate the arc length

Using formula of arc length

L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}

Put the value into the formula

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}

L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}

L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}

L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}

Put the limits

L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})

L=\dfrac{21}{16}

Hence, The arc length is \dfrac{21}{16}

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