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3 years ago

15

The length of a rectangle is 2 feet less than twice the width. The area of the rectangle is 180 feet squared. Find the length an

d width of the rectangle.

1 answer:

Ksenya-84 [330]3 years ago

3 0

Answer:

Width=10

Length=18

Step-by-step explanation:

The length is 2 feet less then twice the width

Let the width be x.

Then the length is 2x -2 based on the given information. Use this to make an equation since it is known that the area is 180feet squared.

x(2x-2)=180

2x^2 - 2x - 180 = 0

2( x^2 - x - 90) = 0

We now have a factorable quadratic.

2(x+9)(x-10)=0

So by zero product property, x=-9, x=10 Since a measure cannot be negative, -9 is an extraneous solution.

So x=10

Then the width is 10 and the length is 18

Hope this helps! Let me know if it is correct!

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and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

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Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

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$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

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and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

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