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3 years ago

Solve for a a^2+17=42?

Mathematics

1 answer:

grandymaker [24]3 years ago

6 0

We are given the following equation:

$a^2+17=42$

Subtract both sides by 17

$a^2=25$

Take the positive/negative square root of both sides

$a=\pm5$

This should be your answer. Let me know if you have any questions, thanks!

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5x+2y=7 as a ordered pair

Elenna [48]

(1, 1), 5 + 2 = 7, not that hard!

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3 years ago

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Which equation is true?

ch4aika [34]

Answer : $9x^2 - 25 = (3x - 5)(3x + 5)$

(1) $9x^2 - 25 = (3x - 5)(3x - 5)$

Use FOIL method to mulitply (3x – 5)(3x – 5)

$(3x - 5)(3x - 5) = 9x^2 - 30x + 25$

(2) $9x^2 - 25 = (3x - 5)(3x + 5)$

Use FOIL method to mulitply (3x - 5)(3x + 5)

$(3x - 5)(3x - 5) = 9x^2 - 15x + 15x - 25 = 9x^2 - 25$

(3) $9x^2 - 25 = -(3x + 5)(3x + 5)$

Use FOIL method to mulitply -(3x + 5)(3x + 5)

$-(3x + 5)(3x + 5) = -(9x^2 + 30x + 25) = -9x^2 -30x -25$

(4) $9x^2 - 25 = -(3x + 5)(3x - 5)$

Use FOIL method to mulitply -(3x + 5)(3x - 5)

$-(3x + 5)(3x - 5) = -(9x^2 - 25) = -9x^2 + 25$

So equation 2 is true

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3 years ago

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The cost to enter the zoo is $20 for adults and $15 for children. In one day, 45 adults and 600 children visited the zoo. How mu

jenyasd209 [6]

Answer:

the answer is 9,900

Step-by-step explanation:

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2 years ago

The school office ordered 24 boxes of pens. Their total cost was $19152. What was the cost of 1 box of pens?

aleksley [76]

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3 years ago

An educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to make an est

nirvana33 [79]

Answer:

The 80% confidence interval for the mean number of words a third grader can read per minute is between 40.4 wpm and 40.8 wpm.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.8}{2} = 0.1$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.1 = 0.99$ , so $z = 1.28$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.28*\frac{4.1}{\sqrt{840}} = 0.2$

The lower end of the interval is the sample mean subtracted by M. So it is 40.6 - 0.2 = 40.4 words per minute.

The upper end of the interval is the sample mean added to M. So it is 40.6 + 0.2 = 40.8 words per minute.

The 80% confidence interval for the mean number of words a third grader can read per minute is between 40.4 wpm and 40.8 wpm.

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3 years ago