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4 years ago

Can someone hurry, I have like 5 mins.

Mathematics

1 answer:

3241004551 [841]4 years ago

7 0

The answer is letter a

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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

PLS HELP MY HWS DUE IN 20MINS PLS SHOW WORK TOO TY <br><br> solve for x

Tems11 [23]

Answer:

13x+2+5x-7+3x-4=180

13x+5x+3x=180-2+7+4

21x/21=189/21

X=9

Please put the degrees sign there

Step-by-step explanation:

5 0

3 years ago

The domain and range in this graph will be:

Nataliya [291]

Answer:

Step-by-step explanation:

this is a line going from -infinity to +infinity in x and in y direction.

for b or c to be right the function would have to be a single point, as nothing else is valid as value.

8 0

2 years ago

How can you study with out getting distracted

jarptica [38.1K]

If you play music then you won't get distracted by the outside world or if you study in a quiet room

6 0

3 years ago

Read 2 more answers

How do I find the restrictions on x if there are any? <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bx%20-%201%7D%20%20%3

Natali5045456 [20]

We have the expression:

$\frac{1}{x - 1}=\frac{5}{x - 10}$

When we have rational functions, where the denominator is a function of x, we have a restriction for the domain for any value of x that makes the denominator equal to 0.

That is because if the denominator is 0, then we have a function f(x) that is a division by zero and is undefined.

If we have a value that makes f(x) to be undefined, then this value of x does not belong to the domain of f(x).

Expression:

$\begin{gathered} \frac{1}{x-1}=\frac{5}{x-10} \\ \frac{x-1}{1}=\frac{x-10}{5} \\ x-1=\frac{x}{5}-\frac{10}{5} \\ x-1=\frac{1}{5}x-2 \\ x-\frac{1}{5}x=-2+1 \\ \frac{4}{5}x=-1 \\ x=-1\cdot\frac{5}{4} \\ x=-\frac{5}{4} \end{gathered}$

Answer: There is no restriction for x in the expression.

6 0

1 year ago