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1 year ago

Pls help the question is here (12points) DUE TO DAY

Mathematics

1 answer:

scZoUnD [109]1 year ago

5 0

Answer:

$\bf Area= 90\;in$

Step-by-step explanation:

Area of ABCD = L * W

$\bf 6\times 12$

$\bf 72 \;in^2$

Area of EFGC:- L * W

$\bf 6 \times 3$

$\bf 18\; in^2$

Now, the area of the combind rectangles:-

ABCD + EFGC

$\bf 72+18$

$\boxed{\bf 90\; in}$

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Consider the half of the sphere of radius 2 centered at the origin which lies above the xy-plane (a hemisphere). Suppose the den

Harlamova29_29 [7]

Answer:

25.133 units

Step-by-step explanation:

Since the density ρ = r, our mass is

m = ∫∫∫r³sinθdΦdrdθ. We integrate from θ = 0 to π (since it is a hemisphere), Φ = 0 to 2π and r = 0 to 2 and the maximum values of r = 2 in those directions. So

m =∫∫[∫r³sinθdΦ]drdθ

m = ∫[∫2πr³sinθdθ]dr ∫dФ = 2π

m = ∫2πr³∫sinθdθ]dr

m = 2π∫r³dr ∫sinθdθ = 1

m = 2π × 4 ∫r³dr = 4

m = 8π units

m = 25.133 units

5 0

2 years ago

The perimeter of a rectangular metal plate is 15 cm. Find the length of the rectangular plate (in decimals) whose width is 228 c

scoray [572]

The Solution.

Representing the problem in a diagram, we have

By formula,

$P=2(L+W)$

In this case,

$P=15\operatorname{cm},L=?,W=\frac{22}{8}cm$

Substituting these values in the formula above, we get

$15=2(L+\frac{22}{8})$

Clearing the bracket, we get

$15=2L+(2\times\frac{22}{8})$ $\begin{gathered} 15=2L+\frac{11}{2} \\ \\ 15=2L+5.5 \end{gathered}$ $\begin{gathered} 15-5.5=2L \\ 9.5=2L \end{gathered}$

Dividing both sides by 2, we get