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3 years ago

I know its not B. Any other ideas?Topic: Radical Equations and Extraneous Roots

Mathematics

1 answer:

lbvjy [14]3 years ago

3 0

Hi! :)

In this case, we have to square both sides of the equation. When we square the left side, we will get x^2 + 81, because square root squared denies one another, and for the right side, we have to square the binomial, like this:

(x+10)^2 = x^2 + 20x + 100

So the correct answer is the last one : x^2 + 81 = x^2 + 20x + 100.

Hope this helps!

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Answer:

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the suspects weight in pounds is equal to the cost of Jacob Wilson's family meal. If the family ordered 6 steak dinners, at 15.7

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3 years ago

Which of the following firms could raise prices and expect an increase in revenues?

Katena32 [7]

Answer:

b. a firm whose product has an elasticity of 0.31

Step-by-step explanation:

A type of product that has elasticity less than 1 is considered as 'inelsatic goods. Inelastic goods would not have a decrease in demand even if they are increasing the price of their product.

This means that when the producer of this product increase the price, their profit will either stay stagnant or increase.

In most cases, products that are very crucial to fulfill our basic needs or day to day activity have this type of elasticity.

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3 years ago

Read 2 more answers

The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation St

saveliy_v [14]

Answer:

a)0.067

b)0.111

c)0.612

d)$687.28

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $385

Standard Deviation, σ = $110

We are given that the distribution of domestic airfares is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(domestic airfare is $550 or more)

P(x > 550)

$P( x > 550) = P( z > \displaystyle\frac{550 - 385}{110}) = P(z > 1.5)$

$= 1 - P(z \leq 1.5)$

Calculation the value from standard normal z table, we have,

$P(x > 550) = 1 - 0.933 = 0.067= 6.7\%$

b) P(domestic airfare is $250 or less)

$P(x \leq 250) = P(z \leq \displaystyle\frac{250-385}{110}) = P(z \leq -1.22)$

Calculating the value from the standard normal table we have,

$P( x \leq 250) = 0.111 = 11.1\%$

c))P(domestic airfare is between $300 and $500)

$P(300 \leq x \leq 500) = P(\displaystyle\frac{300 - 385}{110} \leq z \leq \displaystyle\frac{500-385}{110}) = P(-0.77 \leq z \leq 1.04)\\\\= P(z \leq 1.04) - P(z < -0.77)\\= 0.851 - 0.239 = 0.612 = 61.2\%$