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vekshin1
2 years ago
13

Explain how you can determine whether or not the ratios 4/2 and 8/6 and form a proportion.

Mathematics
1 answer:
expeople1 [14]2 years ago
3 0

Answer:

Not proportional.

Step-by-step explanation:

Given ratios:

\frac{4}{2}, \frac{8}{6}

To tell whether the ratios are proportional or not.

Solution:

In order to tell whether the given ratios are proportional or not, we will reduce the fractions to the simplest numbers by dividing the numerator and denominator by their greatest common factor and then comparing them. If they are equal then the ratios are proportional else not proportional.

So, the simplest form of \frac{4}{2} can be calculated by dividing both by 2 as it is the G.C.F.

\frac{4}{2}=\frac{4\div2}{2\div 2}=\frac{2}{1}

the simplest form of \frac{8}{6} can be calculated by dividing both by 2 as it is the G.C.F.

\frac{8}{6}=\frac{8\div2}{6\div 2}=\frac{4}{3}

Now, since \frac{2}{1}\neq \frac{4}{3}, hence the ratios are not proportional.

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25 Points ! Write a paragraph proof.<br> Given: ∠T and ∠V are right angles.<br> Prove: ∆TUW ∆VWU
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Answer:

Δ TUW ≅ ΔVWU ⇒ by AAS case

Step-by-step explanation:

* Lets revise the cases of congruent for triangles

- SSS  ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ

- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and  

 including angle in the 2nd Δ  

- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ  

 ≅ 2 angles and the side whose joining them in the 2nd Δ  

- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 angles

 and one side in the 2ndΔ

- HL ⇒ hypotenuse leg of the first right angle triangle ≅ hypotenuse

 leg of the 2nd right angle Δ

* Lets solve the problem

- There are two triangles TUW and VWU

- ∠T and ∠V are right angles

- LINE TW is parallel to line VU

∵ TW // VU and UW is a transversal

∴ m∠VUW = m∠TWU ⇒ alternate angles (Z shape)

- Now we have in the two triangles two pairs of angle equal each

 other and one common side, so we can use the case AAS

- In Δ TUW and ΔVWU

∵ m∠T = m∠V ⇒ given (right angles)

∵ m∠TWU = m∠VUW ⇒ proved

∵ UW = WU ⇒ (common side in the 2 Δ)

∴ Δ TUW ≅ ΔVWU ⇒ by AAS case

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