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3 years ago

Help please. 7th grade, area question.

Mathematics

1 answer:

andrezito [222]3 years ago

6 0

Answer:

The area of the warning table is approximately 372.09 cm^2

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The ceiling of Katie’s living room is a square that is 20 ft long on each side. To decorate for a party, she plans to hang crepe

tatyana61 [14]

The room is a square, meaning all four sides are 20ft long.

Because we know all the side lengths, we can find the perimeter of the square by adding the side lengths together.

(20 + 20 + 20 + 20) = 80

Now that we know the perimeter, we have to match this number to the feet of crepe paper.

We can find this by dividing the total length needed by the length a single roll provides.

80 / 8 = 10

Therefore, we know that Katie will have to buy a minimum of 10 rolls if she wants to cover the entire perimeter of the room.

Hope this helps! :)

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3 years ago

4x+7=15x-1-3x<br> Solve for x plss

ANEK [815]

Answer: x=1

Step-by-step explanation:

4x-12x=-1-7

-8x=-8

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3 years ago

Linda earns $600 each week working her full time job. Her employer has a 16% tax deduction on all monies earned each week

shutvik [7]

Answer:

$504 a week

Step-by-step explanation:

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3 years ago

4% of what amount of money is $19?

IRINA_888 [86]

$47.5... you divide 0.4 from 19

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4 years ago

4pi/3 find the exact trigonometric ratios for the angle whose radian measue is given

bulgar [2K]

Answer:

All trigonometric ratios.

Step-by-step explanation:

We have to find all the trigonometric ratios.

Given:

$\theta = \dfrac{4\pi}{3} = \pi + \dfrac{\pi}{3}$

Formula:

$\sin(\pi +x) = -\sin(x)\\\cos(\pi +x) = -\cos(x)$

$\sin(\pi + \dfrac{\pi}{3}) = -\sin(\dfrac{\pi}{3}) = -\dfrac{\sqrt{3}}{2}$

$\cos ((\pi + \dfrac{\pi}{3})) = -\cos(\dfrac{\pi}{3}) = -\dfrac{1}{2}$

Formula:

$\tan\theta =\dfrac{\sin\theta}{\cos\theta}\\\cot\theta = \dfrac{1}{\tan\theta}\\\sec\theta = \dfrac{1}{\cos\theta}\\\csc\theta = \dfrac{1}{\sin\theta}$

$\tan(\pi + \dfrac{\pi}{3})= \dfrac{\sin(\pi + \dfrac{\pi}{3})}{\cos(\pi + \dfrac{\pi}{3})}\\\\\tan(\pi + \dfrac{\pi}{3}) = \dfrac{\frac{-\sqrt{3}}{2}}{\frac{-1}{2}} = \sqrt{3}$

$\cot(\pi + \dfrac{\pi}{3}) = \dfrac{1}{\tan(\pi + \dfrac{\pi}{3})} = \dfrac{1}{\sqrt{3}}$

$\sec(\pi + \dfrac{\pi}{3}) = \dfrac{1}{\cos(\pi + \dfrac{\pi}{3})} = -2$

$\csc(\pi + \dfrac{\pi}{3}) = \dfrac{1}{\sin(\pi + \dfrac{\pi}{3})} = -\dfrac{2}{\sqrt{3}}$

8 0

3 years ago

Help please. 7th grade, area question. ​

Help please. 7th grade, area question.