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2 years ago

What is the solution of

Mathematics

1 answer:

Artemon [7]2 years ago

3 0

Yo sup??

we can simply try option verification for this problem.

when x=0 we get,

1 and 1≠5 therefore x=0 is not a solution.

when x=1 we get

√6-1≠5 therefore x=1 is not a solution.

when x=16 we get

9-4=5 therefore x=16 is a solution

Hence the correct answer is option C

Hope this helps.

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The surface area of the rectangular prism can be found using the expression 6x exponent 3 + 4x exponent 2 + 12x. if the value of

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Answer:

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Step-by-step explanation:

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3 years ago

If sinx=5/13 and x is in quadrant 1, then sin x/2=

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Which two tables represent the same function?

topjm [15]

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

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3 years ago