I need help with this picture ? Please

Mathematics

1 answer:

IgorC [24]4 years ago

7 0

You need to divide

If you want you can do it by steps

20/4= 5

48/4=12

the answer would be 5.12

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1. Find the value of x. A. 90 B. 80 C. 60 D. 45

tia_tia [17]

Answer:

A 90

Step-by-step explanation:

multiple ways to prove this.

e.g. since the angle between the two lines from the center of the circle to the 2 tangent touching points is 90 degrees (that is the meaning of these 90 degrees here as the angle of the circle segment defined by the 2 tangent touching points and the circle center), the tangents have the same "behavior" as tan and cot = the tangents at the norm circle at 0 and 90 degrees. they hit each other outside of the circle again at 90 degrees.

another way

imagine the two right triangles of the tangents crossing point to the circle center and the tangent/circle touching points.

the Hypotenuse of each triangle is cutting the 90 degree angle of the circle segment exactly in half (due to the symmetry principle). so the angle between radius side and Hypotenuse is 90/2 = 45 degrees.

that means also the angle of such a right triangle at the tangent crossing point is 45 degrees (as the sum of all angles in a triangle must be 180, we have the remainder of 180 - 90 - 45 = 45 degrees).

the angles of both right triangles at that point are the same, and so we can add 45+45 = 90 degrees for the total angle at the tangent crossing point.

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:

$63 \int\limits^1_0 \int\limits^{1}_0 {(5u-23v)} \, dudv =63 \int\limits^1_0\left[\frac{5}{2}u^2-23uv\right]^{1}_0 \\ \\ =63\int\limits^1_0(\frac{5}{2}-23v)dv=63\left[\frac{5}{2}v-\frac{23}{2}v^2\right]^1_0=63\left(\frac{5}{2}-\frac{23}{2}\right) \\ \\ =63(-9)=|-576|=576$