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3 years ago

PLEASE HELP!!!!!!!!!! Solve. −8y ≥ 56 y ≤ 7 y ≥ –7 y ≤ −7 y ≥ 7

Mathematics

1 answer:

iris [78.8K]3 years ago

6 0

Answer:

Step-by-step explanation:

the answer is

-8y>=56

-y>=7

y<=7

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how far would a 4.0 cm tall object be located from a reflecting surface with a focal lenght that is equal to 5 cm if it is to cr

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Answer:

HERE IS YOUR ANSWER

Step-by-step explanation:

Use the mirror equation:

1/di + 1/do = 1/f

where di = -10 cm and f = +15 cm. (Note that di is negative if the image is virtual.)

Substitute and solve for do.

1/do + 1/(-10 cm) = 1/(15 cm)

1/do = 1/(15 cm) - 1/(-10 cm) = 5/(30 cm)

do = 6 cm

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In ΔFGH, the measure of ∠H=90°, the measure of ∠F=60°, and HF = 60 feet. Find the length of GH to the nearest foot.

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Step-by-step explanation:

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3 years ago

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pochemuha

Answer: A(-4,3) B(9,3) C(7,0) D(-6,-4)

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3 years ago

Solve and explain 4 20 ___ = _____ x - 5 x - 7

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X = 9/2

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3 years ago

Send it quick guys...

Paladinen [302]

Answer:

194

Step-by-step explanation:

Recall 2 rules to solve this easily:

1. $(a+b)^2=a^2+2ab+b^2$

2. $(a-b)^2=a^2-2ab+b^2$

Now, if we let a = 7 and $b=4\sqrt{3}$ , we can say the problem is basically of the form:

$(a+b)^2 + (a-b)^2$

This can be simplified using the rules:

$(a+b)^2 + (a-b)^2\\a^2+2ab+b^2+a^2-2ab+b^2\\2a^2+2b^2$

Now we can plug the values of a and b we initially thought of (remember though $\sqrt{x} \sqrt{x} =x$ ):

$2a^2+2b^2\\2(7)^2+2(4\sqrt{3})^2\\ 2(49)+2(4\sqrt{3})(4\sqrt{3})\\98+2(4)(4)(\sqrt{3} )(\sqrt{3} )\\98+2(4)(4)(3)\\98+96 \\194$

194 is the final answer.

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4 years ago