Write an expression to represent:

Kazeer [188]

Lets say that the length of a rectangle is 12 cm and the width is 5 cm, you would add 12+12+5+5 (as in the length on both length sides and the width on both sides) so the perimeter of this rectangle is: 34 cm

8 0

3 years ago

Read 2 more answers

What is the value of f=(x)=16x when x= 1/2

zalisa [80]

For the second question, your first option is correct, you would rewrite 8 in power-base form 8=2^3 as the first step to solving the equation, power base form is exponential form.

4 0

3 years ago

I need to get the left side to equal the right side. Keeping the right side alone and not changing it.

likoan [24]

<u>To prove the trigonometric equation:</u>

$\sin ^{3} x+\cos ^{3} x=(\sin x+\cos x)(1-\sin x \cos x)$

$RHS=(\sin x+\cos x)(1-\sin x \cos x)$

We know that $\sin^2x +\cos^2x=1$ , substitute this in place of 1.

$=(\sin x+\cos x)(\sin^2x +\cos^2x-\sin x \cos x)$

Multiply each term of the first term with each term of the 2nd term.

$=\sin^3x + \sin x \cos^2x-\sin^2 x \cos x+\cos x \sin^2 x + \cos^3 x-\sin x\cos^2 x$

Group like terms together.

$=\sin^3x +( \sin x \cos^2x-\sin x\cos^2 x)+(\cos x \sin^2 x-\sin^2 x \cos x) + \cos^3 x$

$=\sin^3x +( 0)+(0) + \cos^3 x$

$=\sin^3x + \cos^3 x$

= LHS

RHS = LHS

$\sin ^{3} x+\cos ^{3} x=(\sin x+\cos x)(1-\sin x \cos x)$

Hence proved.

7 0

4 years ago

Please help I am on a times test

Damm [24]

Answer:

3125

Step-by-step explanation:

12.5x25x10

312.5x10

3125cm

so your answer will be D

5 0

3 years ago

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.

a)

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

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

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

b)

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

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

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 $x^2-x\geq 0$

c)

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

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

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

d)

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

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

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.

e)

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

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

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