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cricket20 [7]
3 years ago
9

Pat sewed 200 tote bags in 4 hours at what rate did pat sew the tote bags

Mathematics
1 answer:
Alik [6]3 years ago
6 0

Answer:

50 tote bags were sewed per hour

Step-by-step explanation:


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398.574986215 to the nearest ten-thousandth.
Julli [10]

Answer: 398.5750

Step-by-step explanation:

It is important to remember that the fourth digit after the decimal point is in the ten-thousandth place.

Then, given the  following number:

398.574986215

You can follow these steps in order ti round it to the nearest ten-thousandth:

1. You can identify that the digit in the ten-thousandth place is:

9

2. Identify the digit to the right of 9. This is:

8

3. Since:

 8>5

 You must round up. Increase the digit 9 by 1. (Notice that 9+1=10, then the digit to the left of  9 increases by 1 too). Then:

398.5750

3 0
4 years ago
Factorise x - 2xz - 2xy + 4y². Show all the steps.​
nirvana33 [79]

Answer:

1(x - 2xz - 2xy + 4y²)

Step-by-step explanation:

I notice that no term all of them have. which means itbwill be factorised via 1.

1(x - 2xz - 2xy + 4y²)

3 0
2 years ago
I don't really understand how to put anything into standard form. If anyone could help that would be great...thanks.
zalisa [80]
I believe you would first distribute within the parentheses and then make it so A and B are the only things on the left side and I and the other random characters are on the right.
5 0
3 years ago
Penny pays $14 a month for her book club membership. With the membership each book costs $5. write an algebraic expression for h
dsp73
14 + 5x, with x = no. of books
8 0
3 years ago
Read 2 more answers
ASAP someone please simply two exponents
zhannawk [14.2K]

Answer:

1.) m^{15}

2.) =\frac{1}{y^{15}}

Give me a comment if you want the explanation.

1.) \mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

\left(m^3\right)^5=m^{3\cdot \:5}

\mathrm{Multiply\:the\:numbers:}\:3\cdot \:5=15

=m^{15}

2.) \mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

\left(y^{-3}\right)^5=y^{-3\cdot \:5}

=y^{-3\cdot \:5}

=y^{-15}

\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

=\frac{1}{y^{15}}

4 0
3 years ago
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