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4 years ago

What would this be? Please help.

Mathematics

2 answers:

jeka944 years ago

6 0

Answer:

Step-by-step explanation:

Since ray BE is a bisector , you can set 2x+20 = 4x-6

subtract 2x from both sides 20=2x-6

add 6 to both sides 26 = 2x

Divide both sides by 2 13 = x

Since angle ABE = 2x + 20, plug 13 in for x.

Angle ABE = 2(13) +20 =26 + 20=46

Ratling [72]4 years ago

4 0

Answer:

The answer to your question is ∠ABE = 46

Step-by-step explanation:

Data

BE is a bisector of ∠ABE

∠ABE = 2x + 20

∠EBC = 4x - 6

∠ABE = ?

Process

1.- Bisector is a line that divides an angle into two equal angles.

∠ ABE = ∠EBC

2.- Substitution

2x + 20 = 4x - 6

3.- Solve for x

2x - 4x = -6 - 20

- 2x = -26

x = -26/-2

x = 13

4.- Find the measure of ∠ABE

∠ABE = 2(13) + 20

∠ABE = 26 + 20

∠ABE = 46

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Please help :c the first word is find

xeze [42]

Answer:

The values of $r_{2}$ and $\alpha_{2}$ are 2 and 150º.

Step-by-step explanation:

The complete statement is:

Find $\alpha_{2}$ and $r_{2}$ such that $\sin \theta - \sqrt{3}\cdot \cos \theta = r_{2}\cdot \cos (\theta - \alpha_{2})$ .

We proceed to use the following trigonometric identity:

$\cos (\theta - \alpha_{2}) = \cos \theta \cdot \cos \alpha_{2} +\sin \theta \cdot \sin \alpha_{2}$ (1)

$\sin \theta -\sqrt{3}\cdot \cos \theta = r_{2}\cdot \cos \theta \cdot \cos \alpha_{2}+r_{2}\cdot \sin \theta \cdot \sin \alpha_{2}$

By direct comparison we derive these expressions:

$r_{2}\cdot \sin \alpha_{2} = 1$ (2)

$r_{2}\cdot \cos \alpha_{2} = -\sqrt{3}$ (3)

By dividing (2) by (3), we have the following formula:

$\tan \alpha_{2} = -\frac{1}{\sqrt{3}}$

$\tan \alpha_{2} = -\frac{\sqrt{3}}{3}$

The tangent function is negative at second and fourth quadrants. That is:

$\alpha_{2} = \tan^{-1} \left(-\frac{\sqrt{3}}{3} \right)$

There are at least two solutions:

$\alpha_{2,1} = 150^{\circ}$ , $\alpha_{2,2} = 330^{\circ}$

And the value of $r_{2}$ :

$r_{2}^{2}\cdot \sin^{2}\alpha_{2} + r_{2}^{2}\cdot \cos^{2}\alpha_{2} = 4$

$r_{2}^{2} = 4$

$r_{2} = 2$

The values of $r_{2}$ and $\alpha_{2}$ are 2 and 150º.

5 0

3 years ago

Solving systems of equations by substitution

zepelin [54]

Equation 1) 2x - 3y = -1
Equation 2) y = x - 1

We want both equations to be in standard form (Ax+By=C) so we need to fix the second equation.

For the second equation, simply subtract x from both sides, so that x & y are on the same side of the equal sign :)

2) -x + y = -1
1) 2x - 3y = -1

Now, we have to multiply ALL of equation 2 by 2 so that both equations 1 & 2 have "2x" in them :)

2) 2(-x + y = -1)

Simplify.

2) -2x + 2y = -2
1) 2x - 3y = -1

Now, add these equations together.

-y = -3

Divide both sides by -1.

y = 3

Now, plug in 3 for y in the first equation.

1) 2x - 3y = -1

2x - 3(3) = -1

Simplify.

2x - 9 = -1

Then, add 9 to both sides.

2x = -1 + 9

Simplify.

2x = 8

Divide both sides by 2.

x = 4

So, x = 4 & y = 3

~Hope I helped!~

6 0

3 years ago

- 3 3/4 divided by 3

yarga [219]

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