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3 years ago

Help please.........,

Mathematics

1 answer:

Xelga [282]3 years ago

4 0

y-6 = -4(x+1)

y-6 = -4x-4

add 6 to each side

y=-4x+2

A is the correct answer

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GUYS PLEASE HELP ME PLEASE HELP ME PLEASE PLEASE!!!!!

Pani-rosa [81]

Answer:

2y-8

Step-by-step explanation:

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

Do the parenthesis first

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

Multiply

-y-5+y+2y-3

Combine like terms

2y-8

3 0

3 years ago

You are helping to plan an awards banquet for your school, and you need to rent tables to seat 180 people. Tables come in two si

pshichka [43]

Answer:

An equation modeling the situation is : 4 m + 6 n = 180

Step-by-step explanation:

Here, the total number of guests in the banquet hall = 180

Let us assume number of tables ordered with capacity of

4 people each = m

And number of tables ordered with capacity of 6 people each = n

So, total number of people in m tables = m x ( Capacity of 1 table)

= m x (4) = 4 m

And, total number of people in n tables = n x ( Capacity of 1 table)

= n x (6) = 6 n

So, the total capacity of ( m + n) tables = 4 m + 6 n

Also, the total required capacity = 180

⇒ 4 m + 6 n = 180

Hence, an equation modeling the situation is : 4 m + 6 n = 180

3 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

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

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

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

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

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

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