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Simora [160]
3 years ago
11

5 is 90% of what number

Mathematics
2 answers:
avanturin [10]3 years ago
6 0
5 is 90% of what number

let the number be x.

5 = 90% of x

5 = (90/100)x

5 = 0.90x

0.90x = 5

x = 5/0.9

x ≈ 5.556

The number is ≈ 5.556
Olin [163]3 years ago
4 0
5.5 with a recurring decimal (5.55555555555 etc.)
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Verify the identity (tan x + 1)^2 + (tan x-1)^2= 2 sec^2 x
Elina [12.6K]

(\tan x+1)^2+(\tan x-1)^2=2\sec^2x\\\\\text{use}\ \tan x=\dfrac{\sin x}{\cos x}\\\\L_s=\left(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\cos x}\right)^2\\\\=\left(\dfrac{\sin x+\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x-\cos x}{\cos x}\right)^2\\\\=\dfrac{(\sin x+\cos x)^2}{\cos^2x}+\dfrac{(\sin x-\cos x)^2}{\cos^2x}\\\\\text{use}\ (a\pm b)^2=a^2\pm2ab+b^2

=\dfrac{\sin^2x+2\sin x\cos x+\cos^2}{\cos^2x}+\dfrac{\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{\sin^2x+2\sin x\cos x+\cos^2+\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{2\sin^2x+2\cos^2x}{\cos^2x}=\dfrac{2(\sin^2x+\cos^2x)}{\cos^2x}\\\\\text{use}\ \sin^2x+\cos^2x=1\\\\=\dfrac{2(1)}{\cos^2x}=2\cdot\dfrac{1}{\cos^2x}=2\left(\dfrac{1}{\cos x}\right)^2\\\\\text{use}\ \sec x=\dfrac{1}{\cos x}\\\\=2(\sec^2x)=2\sec^2x=R_s\\\\L_s=R_s\Rightarrow The\ identity

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3 years ago
A coin will be tossed 100 times. You get to pick 11 numbers. If the number of heads turns out to equal one of your 11 numbers, y
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bezimeni [28]
First we have to get x by itself, so we multiply two on each side, 
 
then we get 5x=660/4, I like to simplify here, so I would change it to 5x=165, then we divide each side by 5

x=33 

:)
5 0
3 years ago
Help me i'll help you back
Dmitriy789 [7]

Answer:

A, C, B, respectively

Step-by-step explanation:

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