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3 years ago

Someone help me with this algebra question please100% correct only

Mathematics

2 answers:

likoan [24]3 years ago

6 0

The Distributive Property is an instance such as this:

a(x+y), where you multiply <em /><em>a and x, and a and y, </em>and add those two values together.

For example:
a(x+y) = ax + ay, since we multiplied a and x together and a and y together and added the remaining values.

Another example:
2(4+x) = (2 x 4) + 2x = 8 + 2x, since we multiplied 2 and 4 together and 2 and x together and added the remaining values.

Here we have 6+4(3x-2)+5.

There is only one set of parentheses where we can use the distributive property.

Distribute 4(3x-2).

We multiply 4 and 3x together, and multiply 4 and -2 together, and add up the remaining two values.

4(3x-2)=12x-8

Substitute 12x-8 for 4(3x-2).

6+12x-8+5
12x+6-8+5
12x+11-8
12x+3

Your answer is Choice B, 12x+3.

Archy [21]3 years ago

4 0

I would say ..........B

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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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