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ra1l [238]
3 years ago
5

Andrew wants to divide his rectangular garden into two parts diagonally. The garden has a length of 40 feet and width 30 feet.

Mathematics
1 answer:
vampirchik [111]3 years ago
8 0
30 times 40= 120
120 divided by 2= 60.
The answer is 60 feet.

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A square tile has an area of 324 square inches. what is the perimeter of the square tile in inches?
kotykmax [81]

The formula of an area of a square: A=a^2


a - side


We have: A=324\ in^2


substitute:


a^2=324\to a=\sqrt{324} \to a=18\ in


The fromula of a perimeter of a square is: P=4a


substitute:


P=4\cdot18=72\ in^2

7 0
3 years ago
I will give brainlest
anastassius [24]
The answer is 6(12+7) because the highest common factor of 72 and 42 is 6. (12•6=72 and 6•7=42)
8 0
3 years ago
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A storage space in a basement is in the shape of a rectangular prism. The height of the storage space is 5 feet. The length is s
Vsevolod [243]

brainly.com/question/16672466

same question buts its answered on the link above

5 0
3 years ago
Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2
exis [7]

The Jacobian for this transformation is

J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}

with determinant |J| = 12, hence the area element becomes

dA = dx\,dy = 12 \, du\,dv

Then the integral becomes

\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv

where R' is the unit circle,

\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1

so that

\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}

Then

\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}

3 0
2 years ago
Luigi recorded the temperature in his garden at diffternt times of the same day
motikmotik

Answer:

wdym

Step-by-step explanation:

6 0
3 years ago
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