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lawyer [7]
3 years ago
5

Which set of line segments could create a right triangle?

Mathematics
1 answer:
ikadub [295]3 years ago
4 0

Answer:

5, 12, 13

Step-by-step explanation:

I took the exam and I got it right

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Please show steps in how to solve the problem 2 1/6 - 1 5/6
IgorC [24]

Answer: 2/6 or 1/3

Step-by-step explanation: First you want to change the two fractions to improper fractions, so 2 1/6 will become 13/6 and 1 5/6 will become 11/6.

Using those improper fractions, you subtract the numerators since the denominators are the same, so 13 - 11 is 2 for the numberator.

2/6 can be simplified into 1/3 by dividing the numerator and denominator by 2.

8 0
3 years ago
Solve for h v=pie*r^2h
Elanso [62]

Answer:

v=πr^2h

v/πr^2=h

Step-by-step explanation:

4 0
3 years ago
What divided by 4 =8
frez [133]
To find the answer, multiply 8 by 4 to get 32. Now divide 32 by 4 and you should get the answer 8. This will make sure that your answer is correct. And so 32 divided by 4 equals 8.

8*4
Multiply
Final Answer: 32

32/4
Divide
Final Answer: 8
8 0
3 years ago
Use the given transformation to evaluate the given integral, where r is the triangular region with vertices (0, 0), (8, 1), and
Jlenok [28]
We first obtain the equation of the lines bounding R.

For the line with points (0, 0) and (8, 1), the equation is given by:

\frac{y}{x} = \frac{1}{8}  \\  \\ \Rightarrow x=8y \\  \\ \Rightarrow8u+v=8(u+8v)=8u+64v \\  \\ \Rightarrow v=0

For the line with points (0, 0) and (1, 8), the equation is given by:

\frac{y}{x} = \frac{8}{1}  \\  \\ \Rightarrow y=8x \\  \\ \Rightarrow u+8v=8(8u+v)=64u+8v \\  \\ \Rightarrow u=0

For the line with points (8, 1) and (1, 8), the equation is given by:

\frac{y-1}{x-8} = \frac{8-1}{1-8} = \frac{7}{-7} =-1 \\  \\ \Rightarrow y-1=-x+8 \\  \\ \Rightarrow y=-x+9 \\  \\ \Rightarrow u+8v=-8u-v+9 \\  \\ \Rightarrow u=1-v

The Jacobian determinant is given by

\left|\begin{array}{cc} \frac{\partial x}{\partial u} &\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{array}\right| = \left|\begin{array}{cc} 8 &1\\1&8\end{array}\right| \\  \\ =64-1=63

The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v

Therefore, the integration is given by:

63 \int\limits^1_0 \int\limits^{1}_0 {(5u-23v)} \, dudv =63 \int\limits^1_0\left[\frac{5}{2}u^2-23uv\right]^{1}_0 \\  \\ =63\int\limits^1_0(\frac{5}{2}-23v)dv=63\left[\frac{5}{2}v-\frac{23}{2}v^2\right]^1_0=63\left(\frac{5}{2}-\frac{23}{2}\right) \\  \\ =63(-9)=|-576|=576
6 0
3 years ago
Granted DVD ever DVD and bc bc rn even if sm can ms rn get th bee lol you can go to
tester [92]
WhAt dO YoU MeAn???
Huh??
6 0
2 years ago
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