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Naily [24]
3 years ago
9

Please help with these problems! Will mark answer as brainliest and give 30 points. You have to show your work for each problem.

Mathematics
1 answer:
gulaghasi [49]3 years ago
4 0

Answer:

1) (2x + 2)(x + 2)

2)  2(x + 1)(x - 1)

3) (x^2 + 3)(x^2 - 3)

4) 2x(x + 5)(x - 2)

5) (5x + 2y)(5x - 2y)

Step-by-step explanation:

1)

2x^2 + 6x + 4

= 2x^2 + 4x + 2x + 4

= 2x(x + 2) + 2(x + 2)

= (2x + 2)(x + 2)

2)

2x^2 - 2

= 2(x^ 2 - 1) (The difference of two squares)

= 2(x + 1)(x - 1)

3)

x^4 - 9 (The difference of two squares)

=(x^2 + 3)(x^2 - 3)

4)

2x^3 + 6x^2 - 20x

= 2x(x^2 + 3x - 10)

= 2x(x + 5)(x - 2)

5)

25x^2 - 4y^2         (The difference of two squares)

= (5x + 2y)(5x - 2y)

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-16 + 12 how this works is that you have to add -16 + 12 and you will get your answer
Rasek [7]

Answer: -4

Step-by-step explanation:

The outcome is negative since -16 is bigger than 12.

If you do the math then you will get -4

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3 years ago
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A video game store has a 20% sale on all video games. If the game you wanted cost $32.00, what would the discount be?
Charra [1.4K]

Answer:

$6.4

Step-by-step explanation:

You just do 32 × .2 to give you the discount of 6.4

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7 0
3 years ago
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What is an equation of the line that passes through the point (3,-1) and (8, 9)?
Evgesh-ka [11]

Answer:

4) y = 2x - 7

Step-by-step explanation:

First we have to find the slope by using the following formula....

let (3, - 1) be (x_{1}, y_{1} ), let (8, 9) be (x_{2} ,y_{2} ), and m be the slope, then....

m = \frac{y_{2} -y_{1} }{x_{2} - x_{1} } = \frac{9 + 1}{8 - 3} = \frac{10}{5} = 2

Now that we know the slope we can find the equation of the function by using point-slope for of the function which look like this.....

y -y_{1} = m(x - x_{1} )\\y + 1 = 2(x - 3)\\y + 1 = 2x - 6\\y = 2x - 7

3 0
3 years ago
The value x = 3 is a solution to each of the following except which?
ololo11 [35]

Answer:

x \: is \: not \: a \: solution \: in \to \: 5x + 4 < 19. \\

Step-by-step explanation:

\: if \: x = 3 \: then \\  4x-1=2x +5  \: is \: same \: as  \to \:  \\ 4(3) - 1  = 2(3) + 5\\ 12 - 1 = 6 + 5 \\ 11 =11 \to \: \\  \boxed{ x \: is \:a \:  solution \: here}  \\  \\ if \: x = 3 \\ then \to \\ 5x + 4    <   19\: is \: same \:as \to \\ 5(3) + 4 < 19 \\ 15 + 4 < 19 \\ 19 \: is \: equal \: to \: 19 : \\ 19 \:  can \: not \: be \: less \: than \: 19 \\ \boxed{ henc e\: x \: is \: not \: a \: solution \: here}

♨Rage♨

6 0
3 years ago
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