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Alika [10]
3 years ago
12

Which of the following sets could be the sides of a right triangle?

Mathematics
1 answer:
coldgirl [10]3 years ago
7 0

Answer: Choice B) {3, 5, sqrt(34)}

=====================================

Explanation:

We can only have a right triangle if and only if a^2+b^2 = c^2 is a true equation. The 'c' is the longest side, aka hypotenuse. The legs 'a' and 'b' can be in any order you want.

-----------

For choice A,

a = 2

b = 3

c = sqrt(10)

So,

a^2+b^2 = 2^2+3^2 = 4+9 = 13

but

c^2 = (sqrt(10))^2 = 10

which is not equal to 13 from above. Cross choice A off the list.

-----------

Checking choice B

a = 3

b = 5

c = sqrt(34)

Square each equation

a^2 = 3^2 = 9

b^2 = 5^2 = 25

c^2 = (sqrt(34))^2 = 34

We can see that

a^2+b^2 = 9+25 = 34

which is exactly equal to c^2 above. This confirms the answer.

-----------

Let's check choice C

a = 5, b = 8, c = 12

a^2 = 25, b^2 = 64, c^2 = 144

So,

a^2+b^2 = c^2

25+64 = 144

89 = 144

which is a false equation allowing us to cross choice C off the list.

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Answer:

The number is 23.

Step-by-step explanation:

5n + 7 = 122               Subtract 7 on both sides

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        n = 23

<h3><u><em>Hope this helps!!! </em></u></h3><h3><u><em>Please mark this as brainliest!!! </em></u></h3><h3><u><em>Thank You!!! </em></u></h3><h3><u><em>:)</em></u></h3>
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Let F⃗ =2(x+y)i⃗ +8sin(y)j⃗ .
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Answer:

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Step-by-step explanation:

The objective is to find the line integral of F around the perimeter of the rectangle with corners (4,0), (4,3), (−3,3), (−3,0), traversed in that order.

We will use <em>the Green's Theorem </em>to evaluate this integral. The rectangle is presented below.

We have that

           F(x,y) = 2(x+y)i + 8j \sin y = \langle 2(x+y), 8\sin y \rangle

Therefore,

                  P(x,y) = 2(x+y) \quad \wedge \quad Q(x,y) = 8\sin y

Let's calculate the needed partial derivatives.

                              P_y = \frac{\partial P}{\partial y} (x,y) = (2(x+y))'_y = 2\\Q_x =\frac{\partial Q}{\partial x} (x,y) = (8\sin y)'_x = 0

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                                    Q_x -P_y = 0 -2 = - 2

Now, by the Green's theorem, we have

\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA = \int \limits_{-3}^{4} \int \limits_{0}^{3} (-2)\,dy\, dx \\ \\\phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-2y) \Big|_{0}^{3} \; dx\\ \phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-6)\; dx = -6x  \Big|_{-3}^{4} = -42

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