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den301095 [7]
3 years ago
13

Answer this question to chick you knowledge

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
7 0
I don’t know bro but I’m just trying to help you but I don’t know so skip that question for.
You might be interested in
If correct then I’ll mark brainlist
Leto [7]

Answer:

(-11, -12)

Step-by-step explanation:

we need one variable to x out

on the note of x, lets cancel out the xs

6x+3y=-102

3x+y=-45 < same as 6x+2y-90

6x+3y=-102

-

6x+2y=-90

y=-12

substitute y in

3x-12=-45

3x=-33

x=-11

(-11, -12)

Please give brainliest :)

5 0
3 years ago
What do you add to get 24 and multiply to get 144
valina [46]
That would be 12.

12 + 12 = 24

12 x 12 = 144


7 0
3 years ago
Read 2 more answers
PLEASE HELP I HAVE AN HOUR LEFT!!
Yuki888 [10]

The statement that correctly describes the horizontal asymptote of g(x) is:

Limit of g (x) as x approaches plus-or-minus infinity = 6, so g(x) has an asymptote at y = 6.

<h3>What are the asymptotes of a function f(x)?</h3>

  • The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.
  • The horizontal asymptote is the limit of f(x) as x goes to infinity, as long as this value is different of infinity.

In this problem, the function is:

g(x) = \frac{42x^3 - 15}{7x^3 - 4x^2 - 3}

The horizontal asymptote is given as follows:

y = \lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} \frac{42x^3 - 15}{7x^3 - 4x^2 - 3} = \lim_{x \rightarrow \infty} \frac{42x^3}{7x^3} = \lim_{x \rightarrow \infty} 6 = 6

Hence the correct statement is:

Limit of g (x) as x approaches plus-or-minus infinity = 6, so g(x) has an asymptote at y = 6.

More can be learned about asymptotes at brainly.com/question/16948935

#SPJ1

5 0
2 years ago
Please help!!
Inessa [10]

Answer:   \text{x}\sqrt{3\text{x}}   which is Choice A

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

Work Shown:

Let y = \sqrt{3\text{x}}

So,

4\text{x}\sqrt{3\text{x}}-\text{x}\sqrt{3\text{x}}-2\text{x}\sqrt{3\text{x}}\\\\4\text{x}y-\text{x}y-2\text{x}y\\\\(4\text{x}-\text{x}-2\text{x})y\\\\\text{x}y\\\\\text{x}\sqrt{3\text{x}}\\\\

This therefore means:

4\text{x}\sqrt{3\text{x}}-\text{x}\sqrt{3\text{x}}-2\text{x}\sqrt{3\text{x}}=\text{x}\sqrt{3\text{x}}\\\\

as long is x is nonnegative.

6 0
1 year ago
Solve the inequality and enter your solution as an inequality comparing the
sammy [17]

Answer:x>1

Step-by-step explanation:

Subract 11 from both sides

8 0
2 years ago
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