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3 years ago

I've been stuck on this question for a while now, I need help!

Mathematics

1 answer:

MaRussiya [10]3 years ago

6 0

In order to solve it you must make each component negative. Multiply them by negative one!

M < 2

Now which numbers would satisfy it? M is less than two... so, m could be one, or any negative number, yeah?

Multiply by negative one. - M and -2 are what you have. If it were one... negative one is less than negative two. Turns out M is less than two.

M < 6

M is less than six. It could be 1, 2, 3, 4, or 5.

-M and -6

Is negative 3 less than -6? All of the numbers are! So, m is actually less than 6

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nikdorinn [45]

Answer:

should be c

Step-by-step explanation:

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2 years ago

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Solve by Elimination/Addition<br><br> {−6x−2y=−1618x+6y=48

tia_tia [17]

Answer:

201.5x-y=-6

Step-by-step explanation:

−6x−2y=−1618x+6y=48

Collect like terms

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3 years ago

They numbers 2,5 or ___ are numbers that would make 34,? 21 divisible by 3.

Nikitich [7]

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The numbers 2, 5 or<u> 27</u>

Step-by-step explanation:

The sum of those three numbers is 34. 2+5+27=34

Oh and 21 divided by 3 is 7

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3 years ago

Help me please!!!!!!!!

Alexxx [7]

Answer:

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Step-by-step explanation:

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3 years ago

How do find the domain and range of <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7Bx-8%7D%7D" id="TexFormula1" tit

Kisachek [45]

$\sqrt{x-8}$ is undefined if the argument $x-8$ is negative, so you first need to require that

$x-8\ge0\implies x\ge8$

We're not done yet, though, because $\dfrac1{\sqrt{x-8}}$ still doesn't exist when $x=8$ , so we remove this from the domain and we're left with $x>8$ , or in interval notation, $(8,\infty)$

To find the range, consider the limits of the function as you approach either endpoint of the domain.

$\displaystyle\lim_{x\to8^+}\frac1{\sqrt{x-8}}=+\infty$
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Since $\dfrac1{\sqrt{x-8}}$ is positive everywhere, the range is $(0,\infty)$

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3 years ago