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4 years ago

Which pair of angles must be congruent

1 answer:

8 0

I thin kthe answer your looking for is Vertical angles. Hope i helped! Brainliest?
and i know its ccorect becuase because the interior angles of vertical angles are congruent.

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Please help.<br> Algebra.

Arlecino [84]

Answer:

Step-by-step explanation:

4) C looks pretty good

5) D looks great

:) ask if you'd like to know how I figured that out

6 0

3 years ago

Pls help me!!!!!!!!!!!!!!!

Volgvan

The answer is A because you have to subtract the pink ribbon by the red ribbon.

3 0

3 years ago

Consider the function represented by the equation y – x – 4 = 0. What is the equation written in function notation, with x as th

pochemuha

Answer: First option.

Step-by-step explanation:

Given the equation $y-x-4=0$ , you need to substitute $f(x)=y$ into this equation. Then:

$f(x)-x-4=0$

Remember that the Addition property of equality states that:

$If\ a=b\\then\ a+c=b+c$

Then, you can add "x" and 4 to both sides.

Therefore, you get the equation written in function notation, with "x" as the independent variable:

$f(x)-x-4+(4)+(x)=0+(4)+(x)\\f(x)=x+4$

7 0

3 years ago

Any 10th grader solve it <br>for 50 points

kkurt [141]

Answer:

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$ is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., $a_{1}=A$ and common difference=D

The nth term can be written as

$a_{n}=A+(n-1)D$

pth term of given arithmetic progression is a

$a_{p}=A+(p-1)D=a$

qth term of given arithmetic progression is b

$a_{q}=A+(q-1)D=b$ and

rth term of given arithmetic progression is c

$a_{r}=A+(r-1)D=c$

We have to prove that

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0$

Now to prove LHS=RHS

Now take LHS

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)$

$=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)$

$=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)$

$=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}$

$=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}$

$=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}$