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aleksklad [387]
3 years ago
14

Solve the equation n2 − 29 = 260 by using square roots.

Mathematics
1 answer:
umka2103 [35]3 years ago
4 0

Answer:

Option A, n = ± 17

Step-by-step explanation:

<u>Step 1:  Add 29 to both sides</u>

n^2 - 29 = 260

n^2 - 29 + 29 = 260 + 29

n^2 = 289

<u>Step 2:  Square root both sides</u>

\sqrt{n^2} = \sqrt{289}

n = ± 17

Answer:  Option A, n = ± 17

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Choose ALL of the points that are solutions of the following system of linear inequalities shown below.
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Answer:

The answer to your question is A, D, F

Step-by-step explanation:

Process

1.- From the graph, determine which points are the solution to the inequalities.

The solution will be all those points where the three inequalities cross.

First line. Dotted blue line. We observe that the solution is all the points over this line.

Second line. Straight violet line. We observe that the solution is all the points below the inequality.

A. It is a solution

B. it is not a solution

C. It is not a solution

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Which of the following points lies on the circle whose center is at the origin and whose radius is 10?
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The answer is (6,-8)
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3 years ago
After a late night of studying, Ebony decides to grab a latte before class so she can stay awake through her morning lecture. Sh
Rzqust [24]

Answer:

P(Same\ Bill) = \frac{1}{3}

P(Second

P(Both\ Even) = \frac{1}{9}

Pr(One\ Odd) = \frac{4}{9}

P(Sum < 10) = \frac{1}{3}

Step-by-step explanation:

Given

Bills: \$1, \$5, \$10

Selection = 2\ bills

The sample space is as follows:

This implies that we construct possible outcome that Ebony selects a bill, returns the bill and then select another.

This means that there are possibilities that the same bill is selected twice.

So, the sample space is as follows:

S = \{(1,1), (1,5), (1,10), (5,1), (5,5), (5,10), (10,1), (10,5), (10,10)\}

n(S) = 9

Solving (a): P(Same\ Bill)

This means that the first and second bill selected are the same.

The outcome of this are:

Same = \{(1,1),(5,5),(10,10)\}

n(Same\ Bill) = 3

The probability is:

P(Same\ Bill) = \frac{n(Same\ Bill)}{n(S)}

P(Same\ Bill) = \frac{3}{9}

P(Same\ Bill) = \frac{1}{3}

Solving (a): P(Second  < First\ Bill)

This means that the second bill selected is less than the first.

The outcome of this are:

Second < First = \{(1,5), (1,10), (5,10)\}

n(Second < First) = 3

The probability is:

P(Second

P(Second

P(Second

Solving (c): P(Both\ Even)

This means that the first and the second bill are even

The outcome of this are:

Both\ Even = \{(10,10)\}

n(Both\ Even) = 1

The probability is:

P(Both\ Even) = \frac{n(Both\ Even)}{n(S)}

P(Both\ Even) = \frac{1}{9}

Solving (e): P(Sum < 10)

This question has missing details.

The correct question is to determine the probability that, the sum of both bills is less than 10

The outcome of this are:

One\ Odd = \{(1,10), (5,10), (10,1), (10,5)\}

n(One\ Odd) = 4

The probability is:

Pr(One\ Odd) = \frac{n(One\ Odd)}{n(S)}

Pr(One\ Odd) = \frac{4}{9}

 

Solving (d): P(One\ Odd)

This question has missing details.

The correct question is to determine the probability that, exactly one of the bills is 0dd

The outcome of this are:

Sum < 10 = \{(1,1), (1,5), (5,1)\}

n(Sum < 10) = 3

The probability is:

P(Sum < 10) = \frac{n(Sum < 10)}{n(S)}

P(Sum < 10) = \frac{3}{9}

P(Sum < 10) = \frac{1}{3}

 

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3 years ago
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Yo sup??

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koban [17]

Step-by-step explanation:

<em>→</em><em> </em><em>3</em><em>x</em><em> </em><em>+</em><em> </em><em>4</em><em> </em><em>=</em><em> </em><em>1</em><em>3</em>

<em>→</em><em> </em><em>3</em><em>x</em><em> </em><em>=</em><em> </em><em>1</em><em>3</em><em> </em><em>-</em><em> </em><em>4</em><em> </em><em>=</em><em> </em><em>9</em>

<em>→</em><em> </em><em>x </em><em>=</em><em> </em><em>9</em><em>/</em><em>3</em><em> </em><em>=</em><em> </em><em>3</em>

<em>→</em><em> </em><em>x </em><em>=</em><em> </em><em>3</em>

<em>option</em><em> </em><em>B </em><em>is </em><em>correct</em>

<em>hope </em><em>this</em><em> answer</em><em> helps</em><em> you</em><em> dear</em><em>!</em><em> </em><em>take </em><em>care</em>

6 0
3 years ago
Read 2 more answers
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