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vova2212 [387]
3 years ago
7

How to solve this problem -5-3x-x-(-16)

Mathematics
2 answers:
dimaraw [331]3 years ago
7 0

Answer:

11-4x

Step-by-step explanation:

add the like fractions

Tcecarenko [31]3 years ago
4 0

Answer:

11 - 4x

Step-by-step explanation:

-5 - 3x - x - (-16)

Reorder the terms so that like terms are next to each other

-5 - (-16) - 3x - x

Combine like terms

11 - 4x

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What we have here in the picture is a graph of a segment. The goal is to find the segment's length, which we can do by using the distance formula:

d^2=(x_2-x_1)^2-(y_2-y_1)^2

Note that the distance between two points is equal to the length of the line segment that joins them.

So now we already have answers for two of the boxes.

The difference in the x-coordinates gives us the horizontal distance between the two points, while the difference in the y-coordinates gives us the vertical distance between the points.

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Which expression matches the following: Jill scored 3 points for each of the 18 multiple
podryga [215]

Answer:18(3)

Step-by-step explanation:

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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
Write an algebra expression to represent each scenario:<br><br> “the sum of 6 and y”
Lina20 [59]

Answer:

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Step-by-step explanation:

simple:)

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