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marshall27 [118]
3 years ago
7

What is the number of solid 3-inch by 2-inch by 2-inch rectangular prisms that can be

Mathematics
1 answer:
Lerok [7]3 years ago
3 0

Answer:

It will be about 12 inches

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1/2 of Ila’s workspace is covered in paper. 1/3 of the paper is covered in yellow sticky notes. What fraction of Ila’s workspace
Gennadij [26K]

Answer:

\frac{1}{6}          

Step-by-step explanation:

We have been given that 1/2 of Ila’s work-space is covered in paper. 1/3 of the paper is covered in yellow sticky notes.

Ila’s work-space is covered in yellow sticky notes would be 1/3 of 1/2.

\text{Ila’s workspace is covered in yellow sticky notes}=\frac{1}{3}\times \frac{1}{2}  

\text{Ila’s workspace is covered in yellow sticky notes}=\frac{1\times1}{3\times 2}  

\text{Ila’s workspace is covered in yellow sticky notes}=\frac{1}{6}

Therefore, \frac{1}{6} of Ila’s workspace is covered in yellow sticky notes.

7 0
3 years ago
Simplify 8 2(10 – r). a. 7r – 18 b. –2r – 18 c. –2r – 9 d. 28 – 2r
fiasKO [112]
8 + 2(10 - r) = 8 + 20 - 2r = 28 - 2r
8 0
4 years ago
If nine times a certain number is subtracted from 7 the result is 52. Find the number.
mezya [45]

Answer:

Let the number to find be x

7 - 9x = 52

-9x = 52-7

- 9x = -45

Divide by - 9

x = 5

The number is 5

Hope this helps.

6 0
4 years ago
Read 2 more answers
PLEASE HELP 30 POINTS <br> Prove that AM is perpendicular to BD
ch4aika [34]

Answer:

Step-by-step explanation:

AM is perpendicular to BD because they from a right angle and the definition of perpendicular is when two lines meet they form a right angle. AM goes straight down and BD form the horizontal line. These two lines form a right angle so AM is perpendicular to BD.  

3 0
3 years ago
3
Ierofanga [76]

Answer:

A

Step-by-step explanation:

Sum the product of the components in the first row of A with the corresponding components of the first column in B

Repeat this with the components in the second row of A with the corresponding components of the second column in B, that is

AB

= \left[\begin{array}{ccc}2&1\\3&4\\\end{array}\right] \left[\begin{array}{ccc}3&1\\5&2\\\end{array}\right]

= \left[\begin{array}{ccc}2(3)+1(5)&2(1)+1(2)\\3(3)+4(5)&3(1)+4(2)\\\end{array}\right]

= \left[\begin{array}{ccc}6+5&2+2\\9+20&3+8\\\end{array}\right]

= \left[\begin{array}{ccc}11&4\\29&11\\\end{array}\right]  → A

6 0
3 years ago
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