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4 years ago

An avogadro's number of any element is equivalent to

Mathematics

1 answer:

Galina-37 [17]4 years ago

8 0

6.02 x 10^25 ..........................................................

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Using the common denominator of 56, what is an equivalent fraction for 3/8 and 3/7?

anyanavicka [17]

Answer: $\frac{3}{8}$ = $\frac{21}{56}$ $\frac{3}{7}$ = $\frac{24}{56}$

Step-by-step explanation:

In order to get from 8 to 56, we have to multiply by 7.

We always do the same to the top! 3 multiplied by 7 = 21.

In order to get from 7 to 56, we have to multiply by 8.

We always do the same to the top! 3 multiplied by 8 = 24.

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3 years ago

PJ's Pizza Place bases its pizza prices on the size of the pizza and the number of toppings. The price of a medium pizza, in dol

Dovator [93]

Answer:

The price of a medium pizza with 3 toppings is $15.

Step-by-step explanation:

If "t" represents the number of toppings, you will use the expression "9+2t" and plug in 3 for t (since there are 3 toppings)

So it will look like:

9+2(3)

Then you're gonna use PEMDAS, multiply 2*3

9+6

And then you get an answer of $15

Hope this helps :))

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2 years ago

Which expression is equivalent to (a^-3 b/a^-5 b3)^-3? Assume a=0, b=0.

prohojiy [21]

$Use:\\\\\dfrac{a^n}{a^m}=a^{n-m}\\\\(a^n)^m=a^{nm}\\\\a^{-n}=\dfrac{1}{a^n}\\\\(ab)^n=a^nb^n$

$\left(\dfrac{a^{-3}b}{a^{-5}b^3}\right)^{-3}=\left(a^{-3-(-5)}b^{1-3}\right)^{-3}=\left(a^{-3+5}b^{-2}\right)^{-3}=\left(a^2b^{-2}\right)^{-3}\\\\=(a^2)^{-3}(b^{-2})^{-3}=a^{2(-3)}b^{-2(-3)}=a^{-6}b^6=\dfrac{b^6}{a^6}$

7 0

3 years ago

To find the amount of interest Jesse will pay to borrow money, you can use the formula I = PRT. I = interest; P = principal; R =

Cerrena [4.2K]

Answer: A $2000

Step-by-step explanation:

Interest = prt / 100

= 5000 x .08 x 5

= 2000

5 0

3 years ago

One week, Kerry travels 125 miles and uses 5 gallons of gasoline. The next week, she travels 175 miles and uses 7 gallons of gas

bazaltina [42]

This is good information, but the question part is missing. Put the question in the comments and I'll try my best to help!

6 0

3 years ago