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2 years ago

Complete the proof....

Mathematics

1 answer:

abruzzese [7]2 years ago

3 0

Answer:

The correct options are;

1. Definition of supplementary angles

2. m∠1 + m∠2 = m∠1 + m∠3

3. m∠2 = m∠3

4. Definition of Congruent Angles

Step-by-step explanation:

The two column proof is presented as follows;

Statement ${}$ Reason

1. ∠1 and ∠2 are supplementary ${}$ Given

${}$ ∠1 and ∠3 are supplementary

2. m∠1 + m∠2 = 180° ${}$ Definition of supplementary angles

${}$ m∠1 + m∠3 = 180°

3. m∠1 + m∠2 = m∠1 + m∠3 ${}$ Transitive Property

4. m∠2 = m∠3 ${}$ Subtraction Property of Equality

5. ∠2 ≅ ∠3 ${}$ Definition of Congruent Angles

Given that angles ∠1 and ∠2 are supplementary angles and angles ∠1 and ∠3 are are also supplementary angles, then the sums of m∠1 + m∠2 and m∠1 + m∠3 are equal, therefore, ∠2 and ∠3 have equal quantitative value and therefore ∠2 = ∠3 and by definition, ∠2 ≅ ∠3.

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yes it is greater:)

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3 years ago

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Answer:

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7 0

3 years ago

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

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

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

5 0

3 years ago

Solve for each system of equations by substitution or elimination.

Tems11 [23]

QUESTION 1

The given system of equations are:

$y=x^2+5x+1$

$y=x^2+2x+1$

We equate the two equations to get:

$x^2+5x+1 =x^2+2x+1$

$x^2 - {x}^{2} + 5x - 2x = 1 - 1$

$3x = 0$

$x = 0$

When x=0,

$y=0^2+2(0)+1 = 1$

The solution is (0,1)

QUESTION 2

The given equations are:

$y=x^2-2x - 1$

and

$y= -x^2-2x-1$

We equate both equations to get:

$x^2-2x - 1 = -x^2-2x-1$

Group similar terms,

$x^2 + {x}^{2} -2x + 2x= -1 + 1$

$2 {x}^{2} = 0$

$x = 0$

We put x=0 into any of the equations to find y.

$y= -0^2-2(0)-1 = - 1$

The solution is (0,-1).

QUESTION 3

The given equations are:

$y=x^2+2x+1$

and

$y=x^2+2x-1$

We equate both equations:

$x^2+2x+1 = x^2+2x-1$

Group similar terms:

$x^2 -x^2+2x = -1 - 1$

$0 = -2$

This is not true.

Hence the system has no solution.

6 0

3 years ago

For 20 Points. 3 Qs.<br> I also gave you guys a help sheet. PLEASE HELP.

Eduardwww [97]

Wat is help sheet, didn't help,

instructions unclear, got my calculator stuck in the ceiling fan

remember some simple exponential laws
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first one
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2nd one
$\frac{7^{5}}{7^{3}} =7^{ 5-3 } = 7^{2} = 49$

third one
$(3^{-4})(3^{2})=3^{-4+2}=3^{-2}= \frac{1}{3^{2}} = \frac{1}{9}$

dunno wat to do with 'help sheet'

4 0

3 years ago