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Svetradugi [14.3K]
3 years ago
5

ME AJUDEM PFV!!O valor da expressão (NA FOTO)simplificada é:

Mathematics
1 answer:
koban [17]3 years ago
7 0

D) 16.

Espero que esto ayude hermano.

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Me pueden ayudar a saber el resultado y la operacion de la ecuacion 8-5x=8+2x por favor me urge
Art [367]
Solve for x:
8 - 5 x = 2 x + 8
Subtract 2 x from both sides:
8 + (-5 x - 2 x) = (2 x - 2 x) + 8
-5 x - 2 x = -7 x:
-7 x + 8 = (2 x - 2 x) + 8
2 x - 2 x = 0:
8 - 7 x = 8
Subtract 8 from both sides:
(8 - 8) - 7 x = 8 - 8
8 - 8 = 0:
-7 x = 8 - 8
8 - 8 = 0:
-7 x = 0
Divide both sides of -7 x = 0 by -7:
(-7 x)/(-7) = 0/(-7)
(-7)/(-7) = 1:
x = 0/(-7)
0/(-7) = 0:
Answer: x = 0
6 0
3 years ago
I AWLAYS PROMISE BRAINLIEST ...... BRAINLEIST FOR THE CORRECT ASNWER
djyliett [7]

Answer:

<h3>its so hard im sorry i can't answer it</h3>
4 0
3 years ago
Read 2 more answers
Evaluate the expression when g=3 and h= 39.<br> h-69<br> L<br> B.<br> X6<br> ?
lawyer [7]
Answer:
39-18 is 21.
Explanation:
3 0
2 years ago
RESPOND QUICK
mrs_skeptik [129]

Answer:

A

Step-by-step explanation:

3/24=9/72        3*3=9    24*3=72      x=3

3/9=y/12         9/3=3        12/3=4      y=4

X=3 y=4

6 0
3 years ago
F(x)=X over x^3-2x^2+5x why will this have no zeros​
Mumz [18]

If you evaluate directly this function at x=0, you'll see that you have a zero denominator.

Nevertheless, the only way for a fraction to equal zero is to have a zero numerator, i.e.

\dfrac{x}{x^3-2x^2+5x}=0\iff x=0

So, this function can't have zeroes, because the only point that would annihilate the numerator would annihilate the denominator as well.

Moreover, we have

\displaystyle \lim_{x\to 0} \dfrac{x}{x^3-2x^2+5x} = \lim_{x\to 0} \dfrac{x}{x(x^2-2x+5)} = \lim_{x\to 0} \dfrac{1}{x^2-2x+5} = \dfrac{1}{5}

So, we can't even extend with continuity this function in such a way that f(0)=0

5 0
3 years ago
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