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ohaa [14]
2 years ago
8

Which graphs show functions with direct variation? select three options.

Mathematics
1 answer:
Nikolay [14]2 years ago
4 0

Answer:

the daily graphs

Step-by-step explanation:

the news paper

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4 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}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

\neq 0

ie., RHS\neq 0

Therefore LHS\neq RHS

ie.,\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  

Hence proved

5 0
3 years ago
HELP PLEASE 15 POINTS
vodomira [7]
The answer is c hopefully it helps you
7 0
3 years ago
Solve the system of equations by using substitution.<br> x+8y=-2<br> x-3y=20
zysi [14]

Answer:

(14, -2)

Step-by-step explanation:

To solve by using substitution, begin by solving for a variable in the first equation. Let's solve for x:

x+8y=-2\\x=-2-8y

Now we know what <em>x</em> equals. Let's substitute this into the second equation:

x-3y=20\\(-2-8y)-3y=20

We can then simplify:

-2-8y-3y=20 Given equation

-2-11y=20 Combine y terms

-11y=22 Add 2

y=-2 Divide by -11

So, we now know the value of y = -2.

To find the value of X, we can substitute the value of Y into one of the equations. Let's use the first one:

x+8y=-2

x+8(-2)=-2 Substitute for y

x-16=-2 Distribute 8

x=14 Add 16

So, we now know the value of x = 14.

Therefore, we know a solution to the system of equations is (14, -2).

4 0
3 years ago
How do I do this? <br><br>f(a)=4a+5; Find f(2)
Mariana [72]
F(2)=4(2)+5
f(2)=8+5
f(2)=13
6 0
3 years ago
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