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topjm [15]
3 years ago
8

Which sets of side lengths represent Pythagorean triples? Check all that apply.

Mathematics
1 answer:
marshall27 [118]3 years ago
8 0

Answer:

08, 15, 17

Step-by-step explanation:

Use the Pythagorean theorem.

Where the sum of both legs squared is equal to the hypotenuse squared.

8² + 15² = 17²

64 + 225 = 17²

289 = 17²

289 = 289

The side lengths that represent Pythagorean triplets are 08, 15, 17.

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I have a roof that measures 175 ft by 45 ft. During the hurricane 11 inches of rain fell.
Nikolay [14]

Answer:

  4.1 billion

Step-by-step explanation:

1 ft = 30.48 cm

1 in = 2.54 cm

The volume of rain that fell on the roof is given by ...

  V = LWH

  V = (175 ft × 30.48 cm/ft)(45 ft × 30.48 cm/ft)(11 in × 2.54 cm/in)

  = 175×45×11×30.48²×2.54 cm³ = 204,412,236.336 cm³

At 20 drops per cm³, this will be ...

  20×204,412,236.336 ≈ 4,088,244,727 . . . . raindrops

About 4.1 billion raindrops fell on your roof.

5 0
2 years ago
Please help need this for today
Cloud [144]
All will apply except for (A, C, and E) because those lengths are less than 4/6. So (B, D and F) are the correct answers.
6 0
3 years ago
ASAP ASAP ASAP SHOW WORK TOO !!!!!!!!!!!!!!! thanks soo much
k0ka [10]
A.
(64x<span>² + 96x + 36) / (16x + 12)
= 4(16x</span><span>² + 24x + 9) / 4(4x + 3)
= 4*(4x + 3)(4x + 3) / 4(4x + 3)
= 4x + 3

b.
1.79 x 10^5 = 1.79 * 100,000 = 179,000

c.
(5.9736 x 10^24) + (4.8685 x 10^24)
= (5.9736 + 4.8685) x 10^24
= 10.8421 x 10^24
= 1.08421 x 10^25</span>
4 0
3 years ago
Petes boat can travel 48 miles upstream in 4 hours. The return trip takes 3 hours. Find the speed of the boat in still water and
tankabanditka [31]
So... hmm bear in mind, when the boat goes upstream, it goes against the stream, so, if the boat has speed rate of say "b", and the stream has a rate of "r", then the speed going up is b - r, the boat's rate minus the streams, because the stream is subtracting speed as it goes up

going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r

notice, the distance is the same, upstream as well as downstream
thus   \bf \begin{cases}&#10;b=\textit{rate of the boat}\\&#10;r=\textit{rate of the river}&#10;\end{cases}\qquad thus&#10;\\\\\\&#10;&#10;\begin{array}{lccclll}&#10;&distance&rate&time(hrs)\\&#10;&----&----&----\\&#10;upstream&48&b-r&4\\&#10;downstream&48&b+4&3&#10;\end{array}&#10;\\\\\\&#10;&#10;\begin{cases}&#10;48=(b-r)(4)\to 48=4b-4r\\\\&#10;\frac{48-4b}{-4}=r\\&#10;--------------\\&#10;48=(b+r)(3)\\&#10;-----------------------------\\\\&#10;thus\\\\&#10;48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3)&#10;\end{cases}

solve for "r", to see what the stream's rate is

what about the boat's? well, just plug the value for "r" on either equation and solve for "b"
5 0
3 years ago
When you subtract one negative integer from another, will your answer be greater than or less than the integer you started with?
SVEN [57.7K]
So say for example u have -7 - (-5), think of subtracting integers as adding the opposite, so ur adding the opposite of -5, the opposite of -5 is 5, so ur adding -7 and 5= -2
another one: -15 - (-18)
again, adding the opposite. -15 plus positive 18= 3

8 0
3 years ago
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