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Monica [59]
3 years ago
7

PLS HELP ASAP! HURRY, GIVING BRAINLY!

Mathematics
1 answer:
Kazeer [188]3 years ago
7 0
The second answer I believe to be correct :)
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Which ordered pair is a solution to the inequality?<br><br> Help please!!
Murrr4er [49]

Choice A is the answer which is the point (1,-1)

=========================================

How I got this answer:

Plug each point into the inequality. If you get a true statement after simplifying, then that point is in the solution set and therefore a solution. Otherwise, it's not a solution.

-------------

checking choice A

plug in (x,y) = (1,-1)

y - 2x \le -3

-1 - 2(1) \le -3

-3 \le -3

This is true because -3 is equal to itself. So this is the answer.

-------------

checking choice B

plug in (x,y) = (2,4)

y - 2x \le -3

4 - 2(2) \le -3

0 \le -3

This is false because 0 is not to the left of -3, nor is 0 equal to -3. We can cross this off the list.

-------------

checking choice C

plug in (x,y) = (-2,3)

y - 2x \le -3

3 - 2(-2) \le -3

7 \le -3

This is false because 7 is not to the left of -3, nor is 7 equal to -3. We can cross this off the list.

-------------

checking choice D

plug in (x,y) = (3,4)

y - 2x \le -3

4 - 2(3) \le -3

-2 \le -3

This is false because -2 is not to the left of -3, nor is -2 equal to -3. We can cross this off the list.


4 0
3 years ago
3. Approximate square root v15
baherus [9]

Answer:

<h2>4</h2>

Step-by-step explanation:

\sqrt{15}\\=3.87298\dots \\\\= 4

4 0
4 years ago
Identify the expression with nonnegative limit values. More info on the pic. PLEASE HELP.
marshall27 [118]

Answer:

\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\  \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right

Step-by-step explanation:

a) \lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}

b) \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}

c) \lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0

d) \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}

e) \lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2

f) \lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}

4 0
3 years ago
How many whole numbers between 99 and 999 contain the digit '1'?
Sedbober [7]

Answer:

162

Step-by-step explanation:

The number with "1" digit starts with 101 then 110 and so on.

6 0
3 years ago
Read 2 more answers
What is 22.3 rounded to the nearest hundred
WITCHER [35]

Answer:

100

Step-by-step explanation:

8 0
4 years ago
Read 2 more answers
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