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3 years ago

Given ADC ACB and BDC BCA prove a squared + b squared = c squared. Use the two Column proof.

Mathematics

1 answer:

Yuri [45]3 years ago

7 0

Answer:

a² + b² = c · (e + d) = c × c = c²

a² + b² = c²

Please see attachment

Step-by-step explanation:

Statement, Reason

ΔADC ~ ΔACB, Given

AC/AD = BA/AC, The ratio of corresponding sides of similar triangles

b/e = c/b

b² = c·e

ΔBDC ~ ΔBCA, Given

BC/BA = BD/BC, The ratio of corresponding sides of similar triangles

a/c = d/a

a² = c·d

a² + b² = c·e + c·d

a² + b² = c · (e + d)

e + d = c, Addition of segment

a² + b² = c × c = c²

Therefore, a² + b² = c²

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3 years ago

Genevieve is 3 years older than twice her brother’s age. Genevieve is 21 years old. Write an equation to find her brother’s age.

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3 years ago

Read 2 more answers

Question regarding logarithms.

Eddi Din [679]

$5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}$

$\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

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3 years ago

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A. C+7<_3 B. C - 3 > 1 C. C - 7< 3

olga_2 [115]

$c + 7 < 3$

8 0

3 years ago

Check out attachment:) I don't understand how do you find out the length ?? pls help with explanation thx

hjlf

Answer:

x=3.89

Step-by-step explanation:

I'll go in depth for you.

Before we figure out what we do, let understand what we know about this triangle.

We know that both triangles have a angle that measure 27°.
We also know EH=5
FG=9
ZG=7
We need to know how to find EZ

Notice how line EG and HF intersect at Angle Z. We know that if two lines intersect at an angle, it form angles called vertical angles. This means that the two angles that are vertical to each other are congruent.

This means that angle Z in both triangles both measure the same.

Now since both triangles have 2 congruent corresponding angles, we can say that the Triangles are Similar due to the Angle-Angle Postulate.

"If two corresponding angles of two triangles are congruent, then the two triangles are similar."

What is mean when Triangles are similar? 

It means that the similar triangles corresponding angles are equal and their sides are in proportion.

The corresponding sides are 

EH and GF

EZ and ZG

HZ and HF.

Our proportion formula for similar triangles is 

Any two sides of the first triangle divided by each other must equal the two corresponding sides of the second triangles divided by each other respectively.

We know FG and ZG so let set up our first fraction

 $\frac{fg}{zg}$ 

The corresponding sides of both are 

EH and EZ respectively so our proportion looks like
 $\frac{fg}{zg} = \frac{eh}{ez}$ 
Plug in the values for each. Let x represent the value of EZ
 $\frac{9}{7} = \frac{5}{x}$ 
Cross Multiply
 $9x = 35$ 
 $x = 3 \frac{8}{9} = 3.89$ 
So x=3.89

3 0

3 years ago