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Oksanka [162]
3 years ago
6

What is the volume of the ones that have it and how do I find the expression

Mathematics
1 answer:
kherson [118]3 years ago
7 0

Answer:

For the shape you must do 3*3*3 + 2*2*2 + 1 unit cubed, or 36 units^3

For Jessy's jewelry box, you must multiply the side lengths, 3*10*3, or 90 inches^3

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Enter an expression which would represent the simplified form of 4x + 8 subtracted from 2x - 10?
Ksju [112]

Answer:

2x - 2

Step-by-step explanation:

Move the numbers to be near their mates

4x - 2x + 8 - 10

Then solve as much as you can

4x - 2 = 2x, 8 - 10 = -2

<u>2x - 2</u>

8 0
2 years ago
Suppose your friends have the following ice cream flavor preferences: 70% of your friends like Chocolate (C). The remaining do n
Marianna [84]

Answer:

  B.  0.18

Step-by-step explanation:

I find the question oddly worded. Apparently, 25% of 70% of your friends like Chocolate and also like Sprinkles. You need to find the product of these numbers to determine the proportion of friends who like Chocolate that also like Sprinkles. That product is ...

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2 years ago
When adding two numbers, such as 123 and 423, care is taken to first line them up and then add like digits. How does expanding t
Ivanshal [37]
Since a polynomial is where we have like terms such as (1 x 10²) and (4 x 10²), we can add these up using the distributive property to get (5 x 10²) but still keep the 10². For example, it's similar to if we had 2x²+3x²=5x². The x² is still there, but we add up the 2 and 3. Similarly, we can add these up for 10^1 and 10^0
4 0
3 years ago
1/6+1/12=<br><img src="https://tex.z-dn.net/?f=1%20%5Cdiv%206%20%2B%201%20%5Cdiv%2012" id="TexFormula1" title="1 \div 6 + 1 \div
erica [24]

\frac{1}{6} + \frac{1}{12}


We know that:


\frac{a}{b} + \frac{c}{d} = \frac{a.d + b.c}{bd}


So,


\frac{1}{6} + \frac{1}{12} = \frac{1.12+6.1}{6.12}

\frac{12+6}{72}

\frac{18}{72}

\frac{9}{36}

\frac{3}{12}

\frac{1}{4}

8 0
3 years ago
Help please 6x^2-18
prohojiy [21]

Answer:

6(x^2-3)

Step-by-step explanation:

5 0
3 years ago
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