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Sunny_sXe [5.5K]

4 years ago

6

In a right triangle, the length of the hypotenuse is 20 inches and the length of one leg is 15 inches. What is the length of the

other leg in inches?
A. 5√ 7
B. 25
C. √ 7
D. 5

1 answer:

Eduardwww [97]4 years ago

3 0

By Pythagoras Theorem,
Length= (20^2 - 15^2)^0.5= (175)^0.5=5√7

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Antony buys one pound of oranges for $12 and sells it for $14. How can he write the profit in his account book?

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A line passes through the points (6,5) and (2,2) .

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3 years ago

What is the nth term rule of the linear sequence below?<br> 5,8,11,14,17.<br> Tn=

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Tn = 5 +3(n -1)

Step-by-step explanation:

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Tn = a1 +d(n -1)

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8 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

Simon is considering buying a home priced at $104,000. If the current market

krok68 [10]

Answer:

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I hope this helps. If you have any more questions, please feel free to post them and someone will be able to help you, whether it's myself or others. Please leave a like, rating, and if possible, Brainliest. Have a great day!

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3 years ago