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erma4kov [3.2K]
2 years ago
12

PLZ HELP i will give brainliest

Mathematics
1 answer:
alexandr1967 [171]2 years ago
6 0

Answer: okay wow I dunno the answer sorry bud

Step-by-step explanation:

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Perform the indicated operation. 22 6/11 ∙ 1 1/2
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3. 33 9/11

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3 years ago
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4 0
3 years ago
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Solve the equatuon<br><br> x/2-(2/(x+1))=1<br><br> x=?
Snezhnost [94]

\frac{x}{2}-\frac{2}{x+1}=1

We have 2 denominators that we need to get rid of. Whenever there are the denominators, all we have to do is multiply all whole equation with the denominators.

Our denominators are both 2 and x+1. Therefore, we multiply the whole equation by 2(x+1)

\frac{x}{2}[2(x+1)]-\frac{2}{x+1}[2(x+1)] = 1[2(x+1)]

Then shorten the fractions.

\frac{x}{2}[2(x+1)]-\frac{2}{x+1}[2(x+1)] = 1[2(x+1)]\\x(x+1)-2(2)=1(2x+2)

Distribute in all.

x^2+x-4=2x+2

We should get like this. Because the polynomial is 2-degree, I'd suggest you to move all terms to one place. Therefore, moving 2x+2 to another side and subtract.

x^2+x-4-2x-2=0\\x^2-x-6=0\\

We are almost there. All we have to do is, solving for x by factoring. (Although there are more than just factoring but factoring this polynomial is faster.)

(x-3)(x+2)=0\\x=3,-2

Thus, the answer is x = 3, -2

7 0
2 years ago
For the love of God help me !! I'm desperate for it tomorrow
USPshnik [31]
D:5x-2>0 \wedge x>0 \wedge x-1>0\\D:5x>2 \wedge x>0 \wedge x>1\\D: x>\frac{2}{5} \wedge x>1\\D:x>1\\\log_2(5x-2)-\log_2x-\log_2(x-1)=2\\\log_2\frac{5x-2}{x(x-1)}=\log_24\\\frac{5x-2}{x(x-1)}=4\\4x(x-1)=5x-2\\4x^2-4x=5x-2\\4x^2-9x+2=0\\4x^2-x-8x+2=0\\x(4x-1)-2(4x-1)=0\\(x-2)(4x-1)=0\\x=2 \vee x=\frac{1}{4}\\\frac{1}{4}\not \in D \Rightarrow \boxed{x=2}
8 0
3 years ago
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How do you simplify this equation
dezoksy [38]

\bf \cfrac{\quad2-\frac{4}{x+2} \quad }{5+\frac{10}{x+2}}\implies \cfrac{\stackrel{LCD~of~x+2}{\frac{2(x+2)-4}{x+2}}}{\stackrel{LCD~of~x+2}{\frac{5(x+2)+10}{x+2}}}\implies \cfrac{\stackrel{distributing}{\frac{2x+4-4}{x+2}}}{\stackrel{distributing}{\frac{5x+10+10}{x+2}}}\implies \cfrac{\quad\frac{2x}{x+2} \quad }{\frac{5x+20}{x+2}} \\\\\\ \cfrac{2x}{\underline{x+2}}\cdot \cfrac{\underline{x+2}}{5x+20}\implies \cfrac{2x}{5x+20}\implies \cfrac{2x}{5(x+4)}

4 0
3 years ago
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