Can someone please help me and write out the answer

Which of the following numbers could be the sides of a right triangle. A)8,15,17. B)6,8,12 C)9,11,21 D)4,13,16

ArbitrLikvidat [17]

Answer:

A) 8, 15, 17.

Step-by-step explanation:

Right triangle obey the Pythagorean theorem. Thus, we choose the two smaller numbers (being the cathetus) and if after applying the P. Theorem we get the biggest of each option (the hypotenuse) that means that those numbers could be the sides of a right triangle.

The Pythagorean theorem states that: $a^2+b^2=c^2$

Thus:

$\sqrt{a^2+b^2}=c$

Option A:

$\sqrt{8^2+15^2}=17$ → 17 = 17 OK!

Option B:

$\sqrt{6^2+8^2} = 10$ → 10 ≠ 12 NO

Option C:

$\sqrt{9^2+11^2} = [tex]\sqrt{202}$ [/tex] → $\sqrt{202}$ ≠ 21 NO

Option D:

$\sqrt{4^2+13^2} = \sqrt{185}$ → $\sqrt{185}$ ≠ 16 NO

3 0

3 years ago

50 ÷ (-5) = Please simplify

mafiozo [28]

Answer:

-10

Step-by-step explanation:

50 ÷ (-5)

First ignore the brackets and the negative sign.

Then divide 50 by 5 to get 10

Then return the negative sign so the answer is; -10

3 0

4 years ago

20 % of 2 is equal to A. 20 B. 4 C. 0.4 D. 0.04

ExtremeBDS [4]

I think the answer is 0.4

4 0

3 years ago

1. if csc β = 7/3 and cot β = - 2√10 / 3, Find sec β

slava [35]

Step-by-step explanation:

1.

$\tan \beta = \frac{1}{ \cot \beta } = - \frac{3}{2 \sqrt{10} } = - \frac{3 \sqrt{10} }{20}$

$\csc \beta \tan \beta = \frac{1}{ \cos \beta } = \sec \beta$

Therefore,

$\sec \beta = ( \frac{7}{3} )( - \frac{3 \sqrt{10} }{20} ) = - \frac{7 \sqrt{10} }{20}$

2.

$\csc y = \frac{1}{ \sin y} = - \frac{ \sqrt{6} }{2}$

$= > \sin y = - \frac{ \sqrt{6} }{3}$

Use the identity

$\cos y = \sqrt{1 - \sin ^{2} y} \: \: \: \: \: \: \: \: \: \: \: \\ = \sqrt{1 - {( - \frac{ \sqrt{6} }{3}) }^{2} } = - \frac{ \sqrt{3} }{3}$

We chose the negative value of the cosine because of the condition where cot y > 0. Otherwise, choosing the positive root will yield a negative cotangent value. Now that we know the sine and cosine of y, we can now solve for the tangent:

$\tan \beta = \frac{ \sin y}{ \cos y} =( - \frac{ \sqrt{6} }{3} )( - \frac{3}{ \sqrt{3} } ) = \sqrt{2}$

3. Recall that sec x = 1/cos x, therefore cos x = 5/6. Solving for sin x,

$\sin x = \sqrt{1 - \cos ^{2} x} = \sqrt{ \frac{11}{6} }$

Solving for tan x:

$\tan x = \frac{ \sin x}{ \cos x} = (\frac{ \sqrt{11} }{ \sqrt{6} } )( \frac{6}{5} ) = \frac{ \sqrt{66} }{5}$

5 0

3 years ago

What two-dimensional figure will result from slicing this rectangular pyramid parallel to the base?

andreev551 [17]

Answer: The answer is D. Trapezoid.

Step-by-step explanation: As shown in the attached figure, a rectangular pyramid ABCDE is drawn. We are slicing this rectangular pyramid parallel to the base BCDE at the points F, G, H and I.

We can clearly see from the figure that upper half of the sliced figure will be similar to the pyramid BCDE and the lower sliced figure will be a trapezoid. These are the three-dimensional figures.

Also, the sliced two-dimensional figure FGHI will be a rectangle, because

the pyramid is a rectangular one and so, FI=GH, FG=HI and all the angles are right angles.

Thus, the resulting two-dimensional figure will be a rectagle.

7 0

3 years ago

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Can someone please help me and write out the answer​

Can someone please help me and write out the answer