All categories

2 years ago

Simplify this algbraic expression completely 5y-3(y+2)

Mathematics

1 answer:

Sophie [7]2 years ago

3 0

Answer:

2y - 6

Step-by-step explanation:

5y - 3(y + 2)

distribute

5y + (-3 * y) + (-3 * 2)

simplify

5y -3y + ( -3 * 2)

5y - 3y + ( -6)

5y - 3y - 6

combine like terms

5-3 = 2

2y - 6

You might be interested in

What is the value of x and w and how to you solve the problem

Black_prince [1.1K]

Well, X= 180 - 2w

and angle made by raidus to tangent is 90' , so angle w = 106-90 = 16

and X= 180- 32 = 148

w= 16 , x= 148

hope it helped

5 0

3 years ago

Which expression is equivalent to the following complex fraction?

tatyana61 [14]

$\dfrac{\frac{x}{x-3}}{\frac{x^2}{x^2-9}}=\dfrac{x(x^2-9)}{(x-3)x^2}=\dfrac{x(x+3)(x-3)}{(x-3)x\cdot x}=\dfrac{x+3}{x}$

Answer C.

3 0

3 years ago

1/4(x+2)+4=12 <br> x=? Help idk

Gnoma [55]

$\frac{1}{4}(x+2)+4=12$

$\frac{1}{4}(x+2)=8$

$x+2=32$

$x=30$

5 0

3 years ago

Read 2 more answers

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

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

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

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

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

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

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