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3 years ago

TRUE OF FALSE : -4 <-2

Mathematics

1 answer:

Licemer1 [7]3 years ago

6 0

Answer: false but not sure

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PLEASE HELP MEEEEeeee

skelet666 [1.2K]

Answer:

D. 2^7

Step-by-step explanation:

2^3 x 2^4 would equal 128. So, if you do the math for all of them:

A isn't correct because it equals 16.

B isn't correct because it equals 4096.

C isn't correct because it equals 16384.

D is correct because 2^7 is 128.

Hope this helped!

7 0

2 years ago

Read 2 more answers

Solve for g. <br> -3 + 5 + 6g = 11 - 3g

Veronika [31]

The answer would be I believe
$g = 1$

8 0

3 years ago

write 25/8 as whole or mixed number in simplest form and write 32/7 as whole or mixed number in simplest form 31/4 as whole or m

ioda

6 1/4 for 25/8

4 4/7 for 32/7

7 3/4 for 31/4

7 0

3 years ago

Read 2 more answers

How would I solve this step by step?

hoa [83]

Answer:

Step-by-step explanation:

you would add 60 + 90 then subtract it from 180

4 0

2 years ago

Read 2 more answers

Given f: ℝ → ℝ and : ℝ → ℝ , for the following, find f(g(x)) and g(f(x)), and state the domain.

horrorfan [7]

Answer:

Remember, $(f\circ g)(x)=f(g(x))$ and the range of g must be in the domain of f.

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

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

The domain of f(g(x)) and g(f(x)) is the set of reals.

$f(g(x))=f(\sqrt{x}-2)=(\sqrt{x}-2)^2-x=\sqrt{x}^2-2*2*\sqrt{x}+2^2-x=-4\sqrt{x}+4$

$g(f(x))=g(x^2-x)=\sqrt{x^2-x}-2$

The domain of f(g(x)) is the set of nonnegative reals and the domain of g(f(x)) is the set of number such that $x^2-x\geq 0$

$f(g(x))=f(\frac{1}{x-1})=(\frac{1}{x-1})^2=\frac{1}{(x-1)^2}$

$g(f(x))=g(x^2)=\frac{1}{x^2-1}$

The domain of f(g(x)) is the set of reals except the 1 and the domain of g(f(x)) is the set of reals except the 1 and -1

$f(g(x))=f(\frac{1}{x-1})=\frac{1}{(\frac{1}{x-1}-1)}=\frac{1}{\frac{2-x}{x-1}}=\frac{x-1}{2-x}$

$g(f(x))=g(\frac{1}{x+2})=\frac{1}{(\frac{1}{x+2}-1)}=\frac{1}{\frac{-x-1}{x+2}}=\frac{x+2}{-x-1}$

The domain of f(g(x)) is the set of reals except 2, and the domain of g(f(x)) is the set of reals except -1.

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

$g(f(x))=g(x-1)=log(2(x-1)+6)=log(2x+4)$

The domain of f(g(x)) is the set of nonnegative reals except -3. The domain of g(f(x)) is the set of nonnegative reals except -2.

3 0

3 years ago