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Alex
3 years ago
15

21/24 reduced fraction

Mathematics
2 answers:
svetoff [14.1K]3 years ago
6 0
21/24
Divide 21 by 3. You get 7.
Divide 24 by 3. You get 8.
7/8 is your answer
daser333 [38]3 years ago
5 0
Find the Greatest Common Factor (GCF) of <span>21,24

</span>Method 1: By Listing FactorsList out all factors of each number, and find the first common one.

Factors of 21

 : 1, 3, 7, 21

Factors of 24

 : 1, 2, 3, 4, 6, 8, 12, 24

Therefore, the GCF is <span>3. 

</span><span>Divide both the numerator and the denominator by the GCF 
</span>
<span><span>21÷3</span><span>​​</span>
                            = 7/8    = Solution 
</span>24÷<span>3 </span>
<span>

</span>
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450

Step-by-step explanation:

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4 years ago
How can a triangular prism have a greater volume than a rectangular prism?
son4ous [18]

Answer:

Step-by-step explanation:

Assume that the two prisms have bases of equal area.

Then the volume of the rectangular prism is V = (base area)(height).

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We could compare the two volumes by creating the ratio inequality

  (1/3) (base area)(t-height)

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      (base area)(r-height)

The triangular prism will have the greater volume for (1/3)((t-height) > r-height.

3 0
3 years ago
What is the value the 0 in the number 601,099
Gekata [30.6K]
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6 0
3 years ago
Determine whether the sets have a subset relationship. R = {right triangles} O = {obtuse triangles} a) R and O are disjoint. b)
DerKrebs [107]

Answer:

f

Step-by-step explanation:

A right angled triangle has one angle equal to 90°

an obtuse angle triangle has one angle greater tan 90°

a) An o and right-angled triangle btuse angled triangle can have one angle equal.

b) R and O are not equivalent by definition

c)  subset of R and Oare not all triangles as R and O are categories of all triangles. R and O are subset of all triangles

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5 0
3 years ago
One side of a rectangle is 20 cm larger than the other side. If you make the smaller side two times larger and the larger side t
grandymaker [24]
Let L be the length
Let w be the width
Let p be the perimeter
L+w+L+w=p
L=w+20
3L+2w+3L+2w=240
Sub the first equation in for L in the second equation and solve for w
3(w+20)+2w+3(w+20)+2w=240
3w+60+2w+3w+60+2w=240
10w+120=240
10w=240-120
10w=120
W=120/10
W=12
Sub w into the first equation and solve for L
L=w+20
L=12+20
L=32
Hope this helps!
3 0
4 years ago
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