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3 years ago

Can 7, 8 and 11 be the sides of a right triangle? Justify your answer

Mathematics

1 answer:

Margaret [11]3 years ago

5 0

It would make an obtuse scalene triangle. The area would be 27.928 and the perimeter would be 26.

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W(x)=x+4; find w(xt4)

shusha [124]

Answer: W(x + 4) = x + 8

Step-by-step explanation:

I am assuming the "t" is supposed to be an addition sign "+"

w(x + 4) means to replace all of the x's in the given function with x + 4

W(x) = x + 4

W(x + 4) = (x + 4) + 4

= x + 8

7 0

3 years ago

Hi I need help please I’ll give brainliesst to show ever helps and explains

saw5 [17]

Answer: 233.19

Step-by-step explanation:

equation to find volume: width x length x height

the left box:

(14-7) x 3.25 x 3.25= 73.9375 (round after adding)

the right box:

7 x 3.25 x 7= 159.25

73.9375 + 159.25= 233.1875

Round to nearest hundredth: 233.19

8 0

2 years ago

What fraction represents 0.64 ? With bar notation 64/9999 64/999 64/99 64/9

Bad White [126]

Answer:

16/25

This is because you have to make it into a percent which is 64% then you have to do 64/100 and simplify it.

4 0

3 years ago

PLEASE HELP ME THIS IS DUE ASAP!!! I will mark brainliest if right!!! Explain!

Fed [463]

Answer:

Step-by-step explanation:

6 0

3 years ago

How to prove this???

swat32

$\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 
$

go to right side now

$4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)$

use $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ and $x^3 + y^3 = x^2 - xy + y^2$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)$

so $\sin^2 A + \cos^2 A = 1$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side$

4 0

3 years ago