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3 years ago

15

What must be given to prove that CJG ~ CEA

1 answer:

Karolina [17]3 years ago

4 0

1) if ∠J=∠E and ∠G=∠A ⇒ ΔCJG ~ Δ<span>CEA
2) </span>if ∠J=∠E and JC:JC=JG:EA ⇒ ΔCJG ~ Δ<span>CEA
3) </span>if JC:JC=JG:EA= CG:CA ⇒ ΔCJG ~ Δ<span>CEA</span>

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2 years ago

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2 years ago

Find the difference between8/15 and 2/3 Show all calculations in your final answer.

kherson [118]

Answer:

$\frac{8}{15}-\left(\frac{-2}{3}\right)=\frac{6}{5}$

Step-by-step explanation:

Let the value of a number 'a' be = 8/15

Let the value of a number 'b' be = 2/3

The difference between the two numbers can be calculated by subtracting the numbers

$a-b=\frac{8}{15}-\left(\frac{-2}{3}\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$=\frac{8}{15}-\frac{-2}{3}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=\frac{8}{15}-\left(-\frac{2}{3}\right)$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a$

$=\frac{8}{15}+\frac{2}{3}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{8+10}{15}$

$=\frac{18}{15}$

$\mathrm{Cancel\:the\:common\:factor:}\:3$

$=\frac{6}{5}$

Thus,

$\frac{8}{15}-\left(\frac{-2}{3}\right)=\frac{6}{5}$

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2 years ago