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Tanya [424]
3 years ago
7

I WILL NAME YOU BRAINLIEST AND GIVE YOU ALL MY POINTS IF YOU SOLVE THIS!!!!! In the diagram, each shape has a value and the sums

of the shapes in each of the first three rows is given. Find the sum of the shapes in the fourth row.
Mathematics
2 answers:
DedPeter [7]3 years ago
8 0

Answer:

Can you put up the diagram so I can solve it?

Step-by-step explanation:

cluponka [151]3 years ago
7 0

Answer:

...

Step-by-step explanation:

...

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Write the improper fraction 31/20 as a mixed number?
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Verify the identity (tan x + 1)^2 + (tan x-1)^2= 2 sec^2 x
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(\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
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goblinko [34]

Answer:

the answer for 323.34  is 10,982

Step-by-step explanation:

I tried very hard to answer this question.

you have to multiply 323.34  to get  10,982.

hope this hlep you.

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If α and β are the zeroes of the polynomial ax^2 + bx + c, find the value of α^2 + β^2
Pie

Answer:

Step-by-step explanation:

α + β = -b/a

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α² + β² = (α + β)² - 2αβ

           = (\frac{-b}{a})^{2}-2\frac{c}{a}\\\\= \frac{b^{2}}{a^{2}}-\frac{2c}{a}\\\\=\frac{b^{2}}{a^{2}}-\frac{2c*a}{a*a}\\\\=\frac{b^{2}-2ca}{a^{2}}

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