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ZanzabumX [31]
3 years ago
8

Which fractions are equivalent? Explain how you know.

Mathematics
1 answer:
podryga [215]3 years ago
8 0

For this case we have the following fractions:

A. \frac {20} {5} = 4

B .- \frac {20} {5} = - 4

C. \frac {20} {- 5} = - 4

D. \frac {-20} {5} = - 4

E .- \frac {20} {5} = - 4

As noted, the last four fractions result in -4. So, they are equivalent fractions.

Answer:

Option B, C, D, E. They are equivalent fractions.

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Gre4nikov [31]
73 I believe I’m not sure
7 0
2 years ago
Ok so I need help on this ,asap plz :)
soldier1979 [14.2K]
Okay so to do the sqrt(75) you see if there are hidden squares. Split it to be the sqrt(25) and sqrt(3) then simplify. In the end, you get 5sqrt(3). Hope this helps!
4 0
3 years ago
The Natchez Trace Bridge in Franklin, Tennessee, is 1,500 feet long. Suppose you build a model of the bridge using a scale of 1
noname [10]
Set this up as a ratio putting inches on top and feet on bottom. Then your first ratio would be 1/25 since 1 inch is the same as 25 feet.  In between them goes an equals sign and the the other ratio has an x on top (because inches is what we are looking for) and the 1500 on the bottom. Cross multiply to solve for x:
1(1500) = x(25).  Divide both sides by 25 to get 60
5 0
3 years ago
Help Meeeeeeee!!!!!!!!
vesna_86 [32]

Answer: 1st one

Step-by-step explanation:

8 0
3 years ago
Point G is the centroid of the right △ABC with m∠C=90° and m∠B=30°. Find AG if CG=4 ft.
o-na [289]

Answer: \text{Length of AG=}\frac{2\sqrt{63}}{3}

Explanation:  

Please follow the diagram in attachment.  

As we know median from vertex C to hypotenuse is CM  

\therefore CM=\frac{1}{2}AB

We are given length of CG=4  

Median divide by centroid 2:1  

CG:GM=2:1  

Where, CG=4

\therefore GM=2 ft

Length of CM=4+2= 6 ft  

\therefore CM=\frac{1}{2}AB\Rightarrow AB=12

In \triangle ABC, \angle C=90^0

Using trigonometry ratio identities  

AC=AB\sin 30^0\Rightarrow AC=6 ft

BC=AB\cos 30^0\Rightarrow BC=6\sqrt{3} ft  

CN=\frac{1}{2}BC\Rightarrow CN=3\sqrt{3} ft

In \triangle CAN, \angle C=90^0  

Using pythagoreous theorem  

AN=\sqrt{6^2+(3\sqrt{3})^2\Rightarrow \sqrt{63}

Length of AG=2/3 AN

\text{Length of AG=}\frac{2\sqrt{63}}{3} ft


5 0
3 years ago
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