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3 years ago

7

(8.4 108) ÷ (2.2 106)

1 answer:

puteri [66]3 years ago

8 0

Answer:

3.804

Step-by-step explanation:

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PLEASE HELP!!! WILL MARK AS BRAINLISTS!!!

Veseljchak [2.6K]

Answer:

vertical asymptote is: x=4

horizontal asymptote is: y=1

pls mark as brainliest :)

6 0

3 years ago

3(5j+2)=2(3j-6) how do i do this

ohaa [14]

3(5j+2)=2(3j-6) do distributed to remove parenthesis 15j+6=6j-3 . Then move all j to left side and the second term to right side 15j-6j=-3-6. Now solve both sides 9j= -9. Now divide both sides by 9 to get j alone to j = -1

5 0

3 years ago

Read 2 more answers

((7, 6), (0,-1), (-7, -1), (-3, 2), (-5, 3) } Domain. Range

Olin [163]

The domain is {-7,-5,-3,0,7}.
The range is {-1,-1,2,3,6}.
:) hope you get a good grade.

4 0

3 years ago

Solve using the distributive property of whole numbers. <br> 27,345 x 113 x-27,345 x 13

rewona [7]

Answer:

2,734,500

Step-by-step explanation:

27,345 × 113 - 27,345 × 13 ← factor out 27,345 from each term

= 27,345(113 - 13)

= 27,345 × 100

= 2,734,500

8 0

2 years ago

Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$

Viefleur [7K]

QUESTION 1

Given that:

$f(x)=2x^2+3x-9$ ,

$g(x)=5x+11$ ,

and

$h(x)=-3x^2+1$

Then;

$f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)$

$f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1$

Group similar terms;

$f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1$

Simplify;

$f(x)-g(x)+h(x)=-x^2-2x-19$

QUESTION 2

Given that;

$f(x)=4x-7$ .

$g(x)=(x+1)^2$

and

$s(x)=f(x)+g(x)$

Substitute the functions;

$s(x)=4x-7+(x+1)^2$

Substitute x=3

$s(3)=4(3)-7+(3+1)^2$

$s(3)=12-7+(4)^2$

$s(3)=5+16$

$s(3)=21$

QUESTION 3

Given:

$f(x)=3x+2$

$g(x)=x^2-5x-1$

$f(g(x))=f(x^2-5x-1)$

This implies that;

$f(g(x))=3(x^2-5x-1)+2$

Expand the parenthesis;

$f(g(x))=3x^2-15x-3+2$

$f(g(x))=3x^2-15x-1$

QUESTION 4

The given function is;

$f(x)=3(x-6)^2+1$

Let

$y=3(x-6)^2+1$

$\Rightarrow y-1=3(x-6)^2$

$\Rightarrow \frac{y-1}{3}=(x-6)^2$

$\Rightarrow \sqrt{\frac{y-1}{3}}=x-6$

$\Rightarrow x=6+\sqrt{\frac{y-1}{3}}$

The range is:

$\frac{y-1}{3}\ge0$

$y-1\ge0$

$y\ge1$

The interval notation is;

$[1,+\infty)$

6 0

4 years ago