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3 years ago

Find the missing right triangle side lengths in the models of the Pythagorean Theorem below

Mathematics

2 answers:

Sladkaya [172]3 years ago

8 0

Answer:

The picture on the left doesn’t have a right triangle.

maw [93]3 years ago

5 0

Answer: 24, 10, 26 and 4.6, 5, 6.7

Step-by-step explanation: for the first triangle the square with the area 576 has side lengths of 24 because 24 is the square root of 576, the square with the area 676 has a side length of 26 because 26 is the square root of 576, and we already know the side lengths of the third triangle are 10 for the second triange we know the side length of the square with the area 21 is about 4.6 because √21 ≈ 4.6 and the square with the area 25 the side length is 5 because √25 = 5 and the last square the side length is 6.7 because √46 ≈ 6.7

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Rewrite the expression in terms of the given function 1/1-sinx - sinx/1+sinx

Feliz [49]

Your question seems a bit incomplete, but for starters you can write

$\dfrac1{1-\sin x}-\dfrac{\sin x}{1+\sin x}=\dfrac{1+\sin x}{(1-\sin x)(1+\sin x)}-\dfrac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)}=\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}$

Expanding where necessary, recalling that $(1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x$ , you have

$\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\dfrac{1+\sin x-\sin x+\sin^2x}{\cos^2x}=\dfrac{1+\sin^2x}{\cos^2x}$

and you can stop there, or continue to rewrite in terms of the reciprocal functions,

$\dfrac{1+\sin^2x}{\cos^2x}=\sec^2x+\tan^2x$

Now, since $1+\tan^2x=\sec^2x$ , the final form could also take

$\sec^2x+\tan^2x=\sec^2x+(\sec^2x-1)=2\sec^2x-1$

or

$\sec^2x+\tan^2x=(1+\tan^2x)+\tan^2x=1+2\tan^2x$

7 0

3 years ago

What shape will you get if you take a cross section of a square pyramid parallel to its base? ANSWERS: A rectangle A triangle A

BigorU [14]

Answer:

A square

Step-by-step explanation:

Since the base is a square, any plane of the pyramid parallel to the base is shaped like a square (except the very top one which is a point).

5 0

2 years ago

A gas occupies 1.00 L at standard temperature. What is the volume at 606 K?

tino4ka555 [31]

Answer:

the volume at 606k is 2.22....the third one

Step-by-step explanation:

v1=1.00 t1=273 v2=? t2=606

v1/t1=v2/t2

1.00/273=v2/606

1.00×606=273v2

606=273v2

606/273=v2

2.21 L=v2

3 0

2 years ago

Solve (x - 3)^2+7=97, where X is a real number. Round your answer to the nearest hundredth.

Eddi Din [679]

Answer:

x ≈ - 6.49, x ≈ 12.49

Step-by-step explanation:

Given

(x - 3)² + 7 = 97 ( subtract 7 from both sides )

(x - 3)² = 90 ( take the square root of both sides )

x - 3 = ± $\sqrt{90}$ ( add 3 to both sides )

x = 3 ± $\sqrt{90}$

Then

x = 3 - $\sqrt{90}$ ≈ - 6.49 ( to the nearest hundredth )

x = 3 + $\sqrt{90}$ ≈ 12.49 ( to the nearest hundredth )

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2 years ago

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kirill115 [55]

Answer:

Step-by-step explanation:

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3 years ago