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Law Incorporation [45]
3 years ago
5

A rectangle is twice as long as it is wide. if its length is increase by 4 cm and its width is decreased by 3 cm, the new rectan

gle formed has an area of 100 cm2 . find the dimensions of the original rectangle.
Mathematics
1 answer:
strojnjashka [21]3 years ago
5 0
Now the width is w.
It's twice as long as wide, so now the length is 2w.

If the length is increased by 4 cm, the length will be 2w + 4.
The width is decreased by 3 cm, so the width will be w - 3.

The are of the new rectangle is 100 cm^2.

area = length * width

area = (2w + 4)(w - 3)

The area of the new rectangle is 100, so we get

(2w + 4)((w - 3) = 100

2w^2 - 6w + 4w - 12 = 100

2w^2 - 2w - 112 = 0

w^2 - w - 56 = 0

(w - 8)(w + 7) = 0

w - 8 = 0   or   w + 7 = 0

w = 8   or   w = -7

A width cannot be negative, so discard w = -7.

w = 8

The width is 8 cm.
The length is twice the width, so the length is 16 cm.
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4 years ago
Which expression has a value of 8, using only four 4s and any of the following: + , – , • , ÷, exponents, and/or parentheses?
nekit [7.7K]
PEMDAS
A. 4*4-4-4=16-4-4=12-4=8 true
B. 4*4*4-4=16*4-4=64-4=60, not 8 false
C. 4*4+4-4=16+0=16, not 8, incorect
D. 4(4+4)-4=4(8)-4=32-4=28, not 8 incorrect


A
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4 years ago
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Furkat [3]

Answer:

D) 2, 4, 8, 16, 32

Step-by-step explanation:

We have given four sets of sequence.

And we have to find out which sequence is not an arithmetic sequence.

For this the given sequences should satisfy the value of common difference(d) and Arithmetic Progression formula.

A.P. Formula,

T_n=a+(n-1)d

Where T_n = nth term of an A.P.

a = first term of an A.P.

n = number of terms.

d = common difference.

'd' is calculated by subtracting fist term from second term.

d = second\ term-first\ term

A) 4, 7, 10, 13, 16

d = 7-4=3

d = 10-7=3

Here d=3 and 5th term is 16.

So we find out the 5th term by using the formula of A.P. To check whether the sequence is in A.P. or not.

T_5=4+(5-1)3=4+4\times\ 3=4+12=16

Here the given sequence fulfills the condition of being in A.P.

Hence the given sequence is an arithmetic sequence.

B) 1, 2, 3, 4, 5

d =2-1=1

d =3-2=1

Here d=1 and 5th term is 5.

T_5=1+(5-1)1=1+4=5

Here the given sequence fulfills the condition of being in A.P.

Hence the given sequence is an arithmetic sequence.

C) 15, 9, 3, -3, -9

d =9-15=-6

d =3-9=-6

Here d=-6 and 5th term is -9.

T_5=15+(5-1)-6=15+4\times -6=15+(-24)=-9

Here the given sequence fulfills the condition of being in A.P.

Hence the given sequence is an arithmetic sequence.

D) 2, 4, 8, 16, 32

d_1=4-2=2

d_2=8-4=4

Here d_1=2\ But\ d_2=4

The common difference between the terms is not same.

In case of d_1.

T_5=2+(5-1)2=2+4\times 2=2+8=10

In case of d_2.

T_5=2+(5-1)4=2+4\times 4=2+16=18

Here the given sequence does not fulfills the condition of being in A.P.

Hence the given sequence is not an arithmetic sequence.

Hence the correct option is D) 2, 4, 8, 16, 32.

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Your original area would be multiplied by 9 (9 times larger)
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