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3 years ago

I got 30 but it isn't correct. help please.

Mathematics

2 answers:

Rashid [163]3 years ago

6 0

The answer that i got is a

dalvyx [7]3 years ago

5 0

(When calculating don't forget it says 1cm=3 feet so I multiplied my numbers by 3 and then calculated the areas)The rectangle area is Base x height so 6x12=72 and the triangle area is 1/2xbasexheight so it's 6x6x1/2=18. 72+18=90 and note it says he only drew half of the side of the house so multiply 90x2=180

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What is the area of the firgure?

qaws [65]

Answer:

768

Step-by-step explanation:

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3 years ago

The diameter of your bicycle wheel is 25 inches. How far will you move in two turns of your wheel? Use 3.14 for pi. A.40in. B.75

Kisachek [45]

Answer is: 150 inches

Explanation:

d==> 25 inches

π==>3

solution ==>

circumference ==> π*d

==>3*25 inches

==>75 inches

it will cover 75 cm in 1 turn.

so, in two turns, it will cover ==>2*75 inches

==>150 inches

hence, correct answer is 150 inches.

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4 years ago

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Alex787 [66]

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2 years ago

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mel-nik [20]

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6 0

3 years ago

Use the Divergence Theorem to evaluate the following integral S F · N dS and find the outward flux of F through the surface of t

Xelga [282]

Answer:

Result;

$\int\limits\int\limits_S { \textbf{F}} \, \cdot \textbf{N} d {S} = 32\pi$

Step-by-step explanation:

Where:

F(x, y, z) = 2(x·i +y·j +z·k) and

S: z = 0, z = 4 -x² - y²

For the solid region between the paraboloid

z = 4 - x² - y²

div F

For S: z = 0, z = 4 -x² - y²

We have the equation of a parabola

To verify the result for F(x, y, z) = 2(x·i +y·j +z·k)

We have for the surface S₁ the outward normal is N₁ = -k and the outward normal for surface S₂ is N₂ given by

$N_2 = \frac{2x \textbf{i} +2y \textbf{j} + \textbf{k}}{\sqrt{4x^2+4y^2+1} }$

Solving we have;

$\int\limits\int\limits_S { \textbf{F}} \, \cdot \textbf{N} d {S} = \int\limits\int\limits_{S1} { \textbf{F}} \, \cdot \textbf{N}_1 d {S} + \int\limits\int\limits_{S2} { \textbf{F}} \, \cdot \textbf{N}_2 d {S}$

Plugging the values for N₁ and N₂, we have

$= \int\limits\int\limits_{S1} { \textbf{F}} \, \cdot \textbf{(-k)}d {S} + \int\limits\int\limits_{S2} { \textbf{F}} \, \cdot \frac{2x \textbf{i} +2y \textbf{j} + \textbf{k}}{\sqrt{4x^2+4y^2+1} } d {S}$

Where:

F(x, y, z) = 2(xi +yj +zk) we have

$= -\int\limits\int\limits_{S1} 2z \ dA + \int\limits\int\limits_{S2} 4x^2+4y^2+2z \ dA$

$= -\int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 2z \ dA + \int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 4x^2+4y^2+2z \ dA$

$= \int\limits^2_{-2} \int\limits^{\sqrt{4-y^2}} _{-\sqrt{4-y^2}} 4x^2+4y^2 \ dxdy$

$= \int\limits^2_{-2} \frac{(16y^2 +32)\sqrt{-(y^2-4)} }{3} dy$

= 32π.

6 0

4 years ago