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sammy [17]
2 years ago
6

Please show x < 80 on a number line

Mathematics
1 answer:
nevsk [136]2 years ago
7 0

Answer:

on the number line make an open circle at 80 (am open circle is this: ○) and connect that to an arrow that points back to 0 and the negative numbers. It doesn't matter how far back as long as you make it an arrow

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B. reflected across tge x axis

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WIIL GIVE BRAINLIEST IF YOU HELP ME
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What is the question?

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Estimate it as a whole number
denis-greek [22]

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2

Step-by-step explanation:

First, you do the subtraction 120-21.4 which is 98.7. So since you know its x^7, the number isn't going to be big. By guessing and checking, we start with 2. 2^7 is 128. We know that 1^7 would just be 1, so the closest whole number would be two.

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Can yall help me out
Ainat [17]

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Option D, Yes, Devon should have evaluated f(3)

Step-by-step explanation:

f(x) = x^3 + x^2 − 10x + 8

<u>Step 1:  Evaluate using F(3) NOT f(-3)</u>

f(3) = 3^3 + 3^2 - 10(3) + 8

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Answer:  Option D, Yes, Devon should have evaluated f(3)

6 0
3 years ago
What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve. mc017-1.jpg and x = 1 mc017-2.jpg and x = –1 mc
Wittaler [7]

We are given equation :x^6+6x^3+5=0

Use\:the\:rational\:root\:theorem

\mathrm{Therefore,\:we\:need\:to\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:5}{1}

-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1

\mathrm{Compute\:}\frac{x^6+6x^3+5}{x+1}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }x^5-x^4+x^3+5x^2-5x+5

Therefore, final factored form it

x^6\:+\:6x^3\:+\:5=\left(x+1\right)\left(x^5-x^4+x^3+5x^2-5x+5\right)

We can't factor it more.

Therefore,

x+1=0.

x=-1.

Therefore, the real solution of the equation would be -1.


5 0
3 years ago
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