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3 years ago

Which postulate or theorem can be used to prove that AABC = ADCB? SSS ASA SAS AAS

Mathematics

1 answer:

koban [17]3 years ago

6 0

<h3>Answer : </h3>

SSS congruency criteria can be used here to prove ∆ABC ≅ ∆DCB

<h3>Solution : </h3>

In ∆ABC and ∆DCB

AB = CD (given)

AC = BD (given)

BC = BC (common)

So, by SSS congruency criteria,

∆ABC ≅ ∆DCB

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Find the length of the third side. If necessary, write in simplest radical form.<br> 10<br> 7

oksian1 [2.3K]

Answer:

$\sqrt{51}$

Step-by-step explanation:

10²-7² (C squared minus B squared equals A squared) (basically Pythagorean theorem but reversed)

100-49= 51 (square the numbers and subtract)

$\sqrt{51}$

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3 years ago

PLS ANSWER WILL MARK BRAINLIEST

fomenos

Answer:

All you have to do is to the distributive method.

Step-by-step explanation:

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3 years ago

Read 2 more answers

X^2 - y^2 = 100

luda_lava [24]

Answer:

Step-by-step explanation:

Let's start by rewriting the second equation in terms of "x":

$x+y=5$

Subtract y from both sides:

$x=5-y$

Now, substitute "5-y" for "x" in the first equation:

$(5-y)^2-y^2=100$

Note that:

$(a-b)^2=a^2-2ab+b^2$

$25-10y+y^2-y^2=100$

Cancel out like terms:

$25-10y=100$

Subtract 25 from both sides:

$-10y=75$

Divide both sides by -10

$y=\frac{75}{-10}=\frac{15}{-2}=-\frac{15}{2}$

Now, substitute this value back into either of the equations to solve for x.

$x+y=5\\x-\frac{15}{2}=5\\$

Add 15/2 to both sides:

$x=5+\frac{15}{2}\\x=\frac{10}{2}+\frac{15}{2}\\x=\frac{25}{2}$

Now, find the difference:

$x-y=\frac{25}{2}-(-\frac{15}{2})=\frac{25}{2}+\frac{15}{2}=\frac{40}{2}=20$

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3 years ago

Is x=12 a solution to the given inequality x > 5

Ludmilka [50]

Answer: True

Step-by-step explanation:

Substitute 12 for x in the inequality

x > 5

12 > 5

12 is greater than 5 so x = 12 is a solution to x > 5

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2 years ago

Inga [223]

I don’t know the answer to this question

7 0

3 years ago