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2 years ago

7

Math plz help it's hard (8th grade math)

2 answers:

Mrac [35]2 years ago

8 0

This is what I got I dont know if It should be an equation but that's all i can get to, gl

atroni [7]2 years ago

7 0

Answer: yes i got the same thing as the other person

Step-by-step explanation:

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Help 7th grade math need fast please

Vika [28.1K]

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7 1/4 boxes weigh total of 66 7/8 what does one box weigh

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Step-by-step explanation:

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2 years ago

12e^5 divide by 3e^3

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2 years ago

What are the solutions to the following system?

bija089 [108]

The solutions are $(\sqrt{2},-1) \text { and }(-\sqrt{2},-1)$ .

Solution:

Given equation:

$-2 x^{2}+y=-5$

Add $2 x^{2}$ on both sides.

$-2 x^{2}+y+2 x^{2}=-5+2 x^{2}$

$y=-5+2 x^{2}$ -------- (1)

$y=-3 x^{2}+5$ (given) -------- (2)

Equate (1) and (2).

$-5+2 x^{2}=-3 x^{2}+5$

Add $3 x^{2}$ on both sides.

$-5+2 x^{2}+3 x^{2}=-3 x^{2}+5 +3 x^{2}$

$-5+5x^{2}=5$

Add 5 on both sides.

$-5+5x^{2}+5=5+5$

$5x^{2}=10$

Divide by 5 on both sides, we get

$x^{2}=2$

Taking square root on both sides, we get

$x=\sqrt{2}, x=-\sqrt{2}$

Substitute $x=\sqrt{2}$ in (1).

$y=-5+2(\sqrt{2})^2$

$y=-5+4$

$y=-1$

Substitute $x=-\sqrt{2}$ in (1).

$y=-5+2(-\sqrt{2})^2$

$y=-5+4$

$y=-1$

Therefore the solutions are $(\sqrt{2},-1) \text { and }(-\sqrt{2},-1)$ .

Option C is the correct answer.

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3 years ago

What is the pattern here?

tigry1 [53]

To get the number at the back, you would need to multiply the number in front by 3 and minus 1.
The answer would be A

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2 years ago