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3 years ago

What triangle congruent postulate or thereom would prove the two triangles are congruent?

Mathematics

1 answer:

Vinvika [58]3 years ago

5 0

Answer:

A. ASA

Step-by-step explanation:

There are two angles that equal, and a side in between them. So that is ASA congruence.

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Step-by-step explanation:

The triangles are equal

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3 years ago

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What is the slope for the graph y - 4x =-2 ?

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Answer:

4.000

Step-by-step explanation:

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3 years ago

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3 years ago

The sum of three consecutive numbers is 537. What is the 3rd number?

maks197457 [2]

For this problem your equation will be x+(x+1)+(x+2)=537

Multiply and Simplify:

x+x+1+x+2=537

3x+3=537

Subtract 3 from both sides:

3x+3-3=537-3

3x=534

Divide both sides by 3:

3x/3=537/3

x=179

179 is the middle number so: 178+179+180=537

180 is the third number.

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3 years ago

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$

Another identity:

$\displaystyle \large{\sin^2x+\cos^2x=1}$

Therefore:

$\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}$

Hence proved, this is proof by using identity helping to find the specific identity.

6 0

3 years ago