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3 years ago

7

The perimeter of a rectangle is 74 cm. the length of the rectangle is 4 more than twice the width. find the length and width of

the rectangle.

1 answer:

adell [148]3 years ago

8 0

L=2W+4
L+L+W+W=74
substitute L with 2W+4: 2W+4+2W+4+W+W=74
6W=74-8=66
W=11
L=2W+4=22+4=26

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Find the next number in the following sequences

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

kkurt [141]

Answer:

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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

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

The nth term can be written as

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

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

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

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Answer:

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Step-by-step explanation:

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substitute equation 2 into equation 1

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5x = 12 - 4.5

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divide both sides by the coefficient of x which is 5

5x/5 = 7.5/5

x = 1.5

put x = 1.5 into equation 2

y=x+1.5..................................... equation 2

y = 1.5 + 1.5

y = 3

therefore the value of x = 1.5 and y = 3 respectively.

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3 years ago