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liberstina [14]
3 years ago
10

An architect designs a rectangular flower garden such that the width is exactly​ two-thirds of the length. If 340 feet of antiqu

e picket fencing are to be used to enclose the​ garden, find the dimensions of the garden.
Mathematics
1 answer:
m_a_m_a [10]3 years ago
5 0
I believe it is about 13 feet
You might be interested in
A:36<br> B:39<br> C:54<br> D:108
nordsb [41]
The answer is 36 my friend too the test and I helped him
3 0
2 years ago
A polygon has a perimeter of 36 inches. Each side is exactly 6 inches long. Which of the following figures does this describe?
Nikolay [14]
Ist b hope that helps have a good day I’ll see
5 0
3 years ago
How do you find the whole number from the percentage increase?<br><br><br>(Give an example)
hjlf
Say the percentage increase is 25%. If you have the whole number of 30 then you would multiply 30 by .25. You should therefore get 7.5 once both are multiplied. That means you should add 7.5 to 30, and your final answer will be 37.5.
4 0
3 years ago
Find midpoint of line segment joining j(6 17 and i(9 16)
Serga [27]
Answer: (15/2, 33/2) or (7.5, 16.5)
depending on if you want fractions or decimals

explanation:
equation for midpoint... (avg x, avg y)
x-coordinate: (6+9)/2 = 15/2 = 7.5
y-coordinate: (17+16)/2 = 33/2 = 16.5
4 0
3 years ago
1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves
erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx

\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx

\ln|y| = x - \ln|x+1| + C

When x = 0, we have y = 2, so we solve for the constant C :

\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)

Then the particular solution to the DE is

\ln|y| = x - \ln|x+1| + \ln(2)

We can go on to solve explicitly for y in terms of x :

e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}

(b) The curves y = x² and y = 2x - x² intersect for

x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1

and the bounded region is the set

\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}

The area of this region is

\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}

7 0
2 years ago
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