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3 years ago

PLEASE HELP. i need to show my work for the correct answer. please help

Mathematics

1 answer:

e-lub [12.9K]3 years ago

7 0

Answer:

I THINK IM WRONG BUT

Step-by-step explanation:

X3 NUZZELS POINCES ON U

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Solve the system of linear equations using elimination. <br><br> −2x − y = 3<br> −9x − y = 17

Maurinko [17]

-2×-y=3-9×-y=17

=2×+9×=1y+1y=17+3

11×=2y=20

11=y=20÷2

11y=10

y=10/11

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3 years ago

Marneshia walked three over eight of a mile in three over five of an hour. What equation can be used to calculate her unit rate

lana [24]

The answer is a (three eights over three fifths= five eights)

3 0

3 years ago

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Of the 8760 hours in a year, most televisions are on for 2190 hours. What percent is this

Dimas [21]

Answer:

Hello!! Your answer is 25% I think!

Hope this helps!!

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2 years ago

Prove that if r is any rational number, then 3r2 − 2r + 4 is rational. The following properties may be used in your proof. Prope

ziro4ka [17]

Given:

Expression is

$3r^2-2r+4$

To prove:

If r is any rational number, then $3r^2-2r+4$ is rational.

Step-by-step explanation:

Property 1: Every integer is a rational number. It is Theorem 4.3.1.

Property 2: The sum of any two rational numbers is rational. It is Theorem 4.3.2.

Property 3: The product of any two rational numbers is rational. It is Exercise 15 in Section 4.3.

Let r be any rational number.

We have,

$3r^2-2r+4$

It can be written as

$3(r\times r)-2r+4$

Now,

3, -2 and 4 are rational numbers by property 1.

$r^2=r\times r$ is rational by Property 3.

$3r^2\text{ and }-2r$ are rational by Property 3.

$3r^2+(-2r)+4$ is rational by property 2.

So, $3r^2-2r+4$ is rational.

Hence proved.

6 0

2 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.

$\displaystyle\lim_{x\to8^+}\frac1{\sqrt{x-8}}=+\infty$
$\displaystyle\lim_{x\to\infty}\frac1{\sqrt{x-8}}=0$

Since $\dfrac1{\sqrt{x-8}}$ is positive everywhere, the range is $(0,\infty)$

3 0

2 years ago