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2 years ago

If ef bisects angle ceb,angle cef=7x+31 and angle feb=10x-3

Mathematics

1 answer:

Nastasia [14]2 years ago

7 0

Given : Angle < CEB is bisected by EF.

< CEF = 7x +31.

< FEB = 10x-3.

We need to find the values of x and measure of < FEB, < CEF and < CEB.

Solution: Angle < CEB is bisected into two angles < FEB and < CEF.

Therefore, < FEB = < CEF.

Substituting the values of < FEB and < CEF, we get

10x -3 = 7x +31

Adding 3 on both sides, we get

10x -3+3 = 7x +31+3.

10x = 7x + 34

Subtracting 7x from both sides, we get

10x-7x = 7x-7x +34.

3x = 34.

Dividing both sides by 3, we get

x= 11.33.

Plugging value of x=11.33 in < CEF = 7x +31.

We get

< CEF = 7(11.33) +31 = 79.33+31 = 110.33.

< FEB = < CEF = 110.33 approximately

< CEB = < FEB + < CEF = 110.33 +110.33 = 220.66 approximately

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1 year ago

Read 2 more answers

A 3pack of boxes of juice costs 1.09. A 12 pack of boxes cost 4.49. A case of 24 boxes costs 8.78. Which is the best buy?Explain

Leona [35]

1.09÷3= .3633 each
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3 years ago

$700 is invested at 4% per annum. How long will it take for the amount to reach $784?

Blababa [14]

Answer:

3yrs

Step-by-step explanation:

Given parameters:

Capital = $700

Percentage interest = 4% per annum

Unknown:

Time it will take for amount to be = $784

Solution:

To find the simple interest on an amount;

I = $\frac{PRT}{100}$

P is the principal

R is the rate

T is the time

I is the interest

Insert the parameters:

I = $\frac{700x 4 x T}{100}$

I = 28T

So;

At the end of the duration:

Principal + Interest = Amount

700+28T = 784

28T = 784 - 700

28T = 84

T = 3yrs

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1 year 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:

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

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2 years ago