All categories

3 years ago

Helpppppp please asp

Mathematics

1 answer:

Ksju [112]3 years ago

7 0

I believe the answer is D: 6×9÷3.

The question is a word problem, and they are asking you to put it into a numerical problem, meaning with numbers. 6 times 9 divided by 3 put into a numerical equation is 6×9÷3.

If this is incorrect, please, don't refrain to tell me. Thank you.

You might be interested in

DULU

Blababa [14]

Step-by-step explanation:

Find the GCD of the two lengths give

8 0

2 years ago

How did I get the answer to these two problems

saveliy_v [14]

You already have the answer

6 0

3 years ago

What are the phase shift and period for the function y 3cos[4(0 + 459)]-4?

RUDIKE [14]

I dont know the answer

3 0

3 years ago

What is -8.64 -0.024? 360 - 36 -360 36

liq [111]

Answer:

(-8.64) / (-0.024) = 360

Step-by-step explanation:

You have to type like this in your question;

(-8.64) / (-0.024)

or $\frac{-0.024}{-8.64}$ .

Bye!

3 0

3 years ago

Read 2 more answers

The function g has an inverse. The function g − 1 determines the number of folds needed to give the folded paper a thickness of

Amanda [17]

Answer:

$\mathbf{g^{-1} (t) = log _2 \ 20 \ t}$

Step-by-step explanation:

Here is the full question

A standard piece of paper is 0.05 mm thick. Let's imagine taking a piece of paper and folding the paper in half multiple times. We'll assume we can make "perfect folds," where each fold makes the folded paper exactly twice as thick as before - and we can make as many folds as we want.

Write a function g that determines the thickness of the folded paper (in mm) in terms of the number folds made, n. (Notice that g(0) 0.05,)

$g(n )= (05)2^n$

The function g has an inverse. The function g⁻¹ determines the number of folds needed to give the folded paper a thickness of t mm. Write a function formula for g⁻¹).

SOLUTION:

If we represent g(n) with t;

Then

$\mathbf{t = (0.05)2^n} \\ \\ \mathbf{\dfrac{t}{0.05} = 2^n} \\ \\ \mathbf{\dfrac {100 \ t }{ 5} = 2^n} \\ \\ \mathbf{20 t = 2^n}$

Taking logarithm of both sides; we have :

$\mathbf{log (20 t) = n log 2} \ \ \ \mathbf{(since \ , log \ a^b = b \ log \ a) } \\ \\ \mathbf{n = \dfrac{log \ 20 t }{log \ 2 }} \\ \\ \mathbf{g^{-1} (t) = \dfrac{log \ 20 t }{log \ 2 }} \\ \\ \mathbf{g^{-1} (t) = log _2 \ 20 \ t}$

4 0

4 years ago