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3 years ago

What is the value of x for the point with a y-value of 6?

Mathematics

2 answers:

alina1380 [7]3 years ago

4 0

4 is the correct answer or x=4 ✌️

Nikolay [14]3 years ago

3 0

The answer would be 4.
If you look at the graph and look for the number six along the y-axis (y), you can find that the point is above the number 4 on the x-axis (x).

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Pie

Answer:

Step-by-step explanation:

:o who is it?!?!?!

3 0

3 years ago

Read 2 more answers

Could anyone explain to me about adding and subtracting exponents. I don’t understand these equations with their exponents!

meriva

Answer:

whenever you're multiplying terms that have exponents you multiply the coefficients and add their exponents:

Step-by-step explanation:

I think the answer to your first question should be:

30a²b - 30ab²

Your second answer is good

whenever you're multiplying terms that have exponents you multiply the coefficients and add their exponents:

for example, 6a²b³c x 2abc = 6(2)(a²)(a)(b³)(b)(c)(c)

= 12a³b⁴c²

8 0

2 years ago

There are 4 students and 1/2 sh butcher paper how much paper will each student get

inna [77]

Answer:

Step-by-step explanation:

what is sh? sheets? if so try dividing 4 by 1/2

4 0

2 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

2 years ago

HELP PLS ITS OVER DUE PLEASE MATCH THESE! please

Radda [10]

I’ve attached a photo with the definition and my answers
Hope this helps!

6 0

2 years ago

Read 2 more answers