6x + 3x² – x – x² yeah im bad at math please help !

Mathematics

1 answer:

juin [17]2 years ago

5 0

Answer:

x • (2x + 5)

Step-by-step explanation:

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "x2" was replaced by "x^2". 1 more similar replacement(s).

STEP

Equation at the end of step 1

((6x + 3x2) - x) - x2

STEP

Pulling out like terms

3.1 Pull out like factors :

2x2 + 5x = x • (2x + 5)

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What sums are equal to 6/12?

nadya68 [22]

Hey there!

"What sums are equal to $\bold{\frac{6}{12}}$ ?"

$\bold{\frac{6}{12}=\frac{1}{2}=0.5 }$

It can't be $\bold{A. \frac{6}{12} + \frac{6}{12}+ \frac{6}{12}+\frac{6}{12}+\frac{6}{12}+\frac{6}{12}}$ because the answer would be $\bold{3}$
It can be $\bold{B.\frac{2}{12} + \frac{1}{12} + \frac{1}{12} +\frac{2}{12}}$
It can also be $\bold{C. \frac{5}{12} + \frac{1}{12}}$
It can also again be $\bold{\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{3}{12}}$
$\boxed{\boxed{\bold{Answers: B,C,D}}}}$

Good luck on your assignment and enjoy your day!

~ $\bold{LoveYourselfFirst:)}$

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2 years ago

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Verify the identity (tan x + 1)^2 + (tan x-1)^2= 2 sec^2 x

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

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