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3 years ago

Which equation is equivalent to √x^2+81 = x+10

Mathematics

2 answers:

Yuki888 [10]3 years ago

6 0

Answer:

D on edg

Step-by-step explanation:

jeka57 [31]3 years ago

4 0

Keywords:

<em>equation, operations, equivalent, binomial, square root </em>

For this case we have an equation in which we must apply operations to rewrite it in an equivalent way. We must start by raising both sides of the equation to the square. Thus, we eliminate the square root of the left side of equality and finally solve the binomial of the right side of equality.

So we have:

$\sqrt {x ^ 2 + 81} = x + 10\\(\sqrt {x ^ 2 + 81}) ^ 2 = (x + 10) ^ 2\\x ^ 2 + 81 = (x + 10) ^ 2$

By definition: $(a + b) ^ 2 = a ^ 2 + 2ab + b ^ 2$

$x ^ 2 + 81 = x ^ 2 + 2 (x) (10) + (10) ^ 2\\x ^ 2 + 81 = x ^ 2 + 20x +100$

Thus, $x ^ 2 + 81 = x ^ 2 + 20x +100$ is equivalent to $\sqrt {x ^ 2 + 81} = x + 10$

Answer:

$x ^ 2 + 81 = x ^ 2 + 20x +100$

Option D

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What’s is the answer

kvasek [131]

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For example:

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When you have something like this:

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Your answer is the second option

3 0

3 years ago

Plz help me i dotn know this

miv72 [106K]

Answer:

it's D. I think 105 cubic feet

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3 years ago

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Uma caixa contém 40 bolinhas de jogo de tênis, sendo 10 verde, 14 azuis e 16 vermelhas. Retirando uma bola ao acaso da caixa, qu

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Step-by-step explanation:

Since the total number of marbles is 40, that is our size of the sample. there are 16 red ones on that sample, so the probability of extracting a red marble from the box is given by the ratio of successes divided the total number of marbles:

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An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equ

kompoz [17]

Answer:

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Normal probability distribution.

Problems of normally distributed samples can be solved using the z-score formula.

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$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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