3 years ago

If a pair of parallel lines are rotated

Mathematics

1 answer:

Sever21 [200]3 years ago

4 0

They are still parrallel lines

You might be interested in

The half-life of radium is 1690 years. If 20 grams are present now, how much will be present in 250 years?

Aneli [31]

Answer:

18.05 g

Step-by-step explanation:

The multiplier of the initial quantity is ...

(1/2)^(t/1690)

so, for t=250, this becomes ...

(1/2)^(250/1690) ≈ 0.902545

Then the quantity remaining of the initial 20 g is ...

(20 g)(0.902545) ≈ 18.05 g

3 0

3 years ago

Nancy is buying a new pair of boots. The store is having a sale and all boots are 20% off. Which expression represents the sale

leva [86]

There are two expressions that can work for this: 0.8x and x-0.2x Hope this helps :)

3 0

3 years ago

A recipe calls for 1 2/3 cups of sugar. If you

Kruka [31]

Answer: 5

Step-by-step explanation: You have to multiply 1 2/3 and 3.

I hope this helps you out!

7 0

3 years ago

Read 2 more answers

Liquid measurements charts

grin007 [14]

Whats the questions???

4 0

3 years ago

Let f be a differentiable function such that f(3)=15, f(6)=3, f ' (3)= -8, and f ' (6)= -2. The function g is differentiable and

Ray Of Light [21]

Answer:

A. $g'(3)=-\frac{1}{2}$

Step-by-step explanation:

We have been given that f be a differentiable function such that $f(3)=15$ , $f(6)=3$ , $f'(3)= -8$ , and $f'(6)= -2$ . The function g is differentiable and $g(x)= f^{-1} (x)$ for all x.

We know that when one function is inverse of other function, so:

$g(f(x))=x$

Upon taking derivative of both sides of our equation, we will get:

$g'(f(x))*f'(x) =1$

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

Plugging $x=6$ into our equation, we will get:

$g'(f(6))=\frac{1}{f'(6)}$

Since $g(x)= f^{-1} (x)$ , then $g'(f(6))=g'(3)$ .

$g'(3)=\frac{1}{f'(6)}$

Since we have been given that $f'(6)= -2$ , so we will get:

$g'(3)=\frac{1}{-2}$

$g'(3)=-\frac{1}{2}$

Therefore, $g'(3)=-\frac{1}{2}$ and option A is the correct choice.

8 0

3 years ago