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patriot [66]
3 years ago
7

There are 6 rectangular flower gardens each measuring 18 feet by 17 feet in a rectangular city park measuring 80 feet by 160 fee

t.
How many square feet of the park are not used for flower gardens?
Mathematics
1 answer:
Aneli [31]3 years ago
4 0

The area which is not in used for the flower gardens is 10964 sq ft.

Step-by-step explanation:

Given,

Length (l) of the rectangular park = 160 ft

Width (b) of the rectangular park = 80 ft

Length (l) of each rectangular flower garden = 18 ft

Width (b) of each rectangular flower garden = 17 ft

To find the area which is not in used.

Formula

The area of rectangle = l×b

Now,

The area of the rectangular park = 160×80 sq ft = 12800 sq ft

The area of each rectangular flower garden = 18×17 sq ft = 306 sq ft

The area of 6 rectangular flower garden =  306×6 sq ft = 1836 sq ft

So,

The area which is not in used = 12800-1836 sq ft = 10964 sq ft.

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4 years ago
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y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\

\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\

This shows that y' = y is true when y = e^x

-----------------------

  • Note 1: A more general solution is y = Ce^x for some constant C.
  • Note 2: It might be tempting to say the general solution is y = e^x+C, but that is not the case because y = e^x+C \to y' = e^x+0 = e^x and we can see that y' = y would only be true for C = 0, so that is why y = e^x+C does not work.
6 0
3 years ago
The second term in a geometric sequence is 50 . The fourth term in the same sequence is 112.5 . What is the common ratio in this
Alja [10]
Geometric an = ar^(n - 1)

Second term =ar² ⁻ ¹ = ar = 50

Fourth term =ar⁴ ⁻ ¹ =  ar³ = 112.5

ar = 50 ..........(a)

<span>ar³ = 112.5......(b)
</span>
Equation (b)/(a)

ar³ / <span>ar = 112.5/50
</span>
r² = 2.25

r = √2.25

r = 1.5

Common ratio = 1.5
7 0
3 years ago
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