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Ksivusya [100]
3 years ago
6

I need help with this please!!!

Mathematics
1 answer:
Andru [333]3 years ago
5 0

Answer:

U = \frac{TE}{3}

Step-by-step explanation:

Given

T = \frac{3U}{E}

Multiply both sides by E to clear the fraction

TE = 3U ( divide both sides by 3 )

\frac{TE}{3} = U

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The rides at an amusement park have restrictions based on height. For the Swan Ride, the minimum height is 2 feet and the maximu
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11 feet

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Which two tables represent the same function?
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Answer:

The 1st and the 5th tables represent the same function

Step-by-step explanation:

* Lets explain how to solve the problem

- There are five tables of functions, two of them are equal

- To find the two equal function lets find their equations

- The form of the equation of a line whose endpoints are (x1 , y1) and

  (x2 , y2) is \frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

* Lets make the equation of each table

# (x1 , y1) = (4 , 8) and (x2 , y2) = (6 , 7)

∵ x1 = 4 , x2 = 6 and y1 = 8 , y2 = 7

∴ \frac{y-8}{x-4}=\frac{7-8}{6-4}

∴ \frac{y-8}{x-4}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 8) = -1(x - 4) ⇒ simplify

∴ 2y - 16 = -x + 4 ⇒ add x and 16 for two sides

∴ x + 2y = 20 ⇒ (1)

# (x1 , y1) = (4 , 5) and (x2 , y2) = (6 , 4)

∵ x1 = 4 , x2 = 6 and y1 = 5 , y2 = 4

∴ \frac{y-5}{x-4}=\frac{4-5}{6-4}

∴ \frac{y-5}{x-4}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 5) = -1(x - 4) ⇒ simplify

∴ 2y - 10 = -x + 4 ⇒ add x and 10 for two sides

∴ x + 2y = 14 ⇒ (2)

# (x1 , y1) = (2 , 8) and (x2 , y2) = (8 , 5)

∵ x1 = 2 , x2 = 8 and y1 = 8 , y2 = 5

∴ \frac{y-8}{x-2}=\frac{5-8}{8-2}

∴ \frac{y-8}{x-2}=\frac{-3}{6}=====\frac{y-8}{x-2}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 8) = -1(x - 2) ⇒ simplify

∴ 2y - 16 = -x + 2 ⇒ add x and 16 for two sides

∴ x + 2y = 18 ⇒ (3)

# (x1 , y1) = (2 , 10) and (x2 , y2) = (6 , 14)

∵ x1 = 2 , x2 = 6 and y1 = 10 , y2 = 14

∴ \frac{y-10}{x-2}=\frac{14-10}{6-2}

∴ \frac{y-10}{x-2}=\frac{4}{4}======\frac{y-10}{x-2}=1

- By using cross multiplication

∴ (y - 10) = (x - 2)

∴ y - 10 = x - 2 ⇒ add 2 and subtract y in the two sides

∴ -8 = x - y ⇒ switch the two sides

∴ x - y = -8 ⇒ (4)

# (x1 , y1) = (2 , 9) and (x2 , y2) = (8 , 6)

∵ x1 = 2 , x2 = 8 and y1 = 9 , y2 = 6

∴ \frac{y-9}{x-2}=\frac{6-9}{8-2}

∴ \frac{y-9}{x-2}=\frac{-3}{6}======\frac{y-9}{x-2}=\frac{-1}{2}

- By using cross multiplication

∴ 2(y - 9) = -1(x - 2) ⇒ simplify

∴ 2y - 18 = -x + 2 ⇒ add x and 18 for two sides

∴ x + 2y = 20 ⇒ (5)

- Equations (1) and (5) are the same

∴ The 1st and the 5th tables represent the same function

3 0
3 years ago
Could someone pkease help. i am not sure if i have the correct answer, thanks
Mama L [17]
I believe you are correct keep it up
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2 years ago
Write an expression for the cost of 1 uniform, add $13, $11 and $8 .
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Answer:

Hey there!

Let the cost of one uniform be c, so the expression becomes

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