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777dan777 [17]
3 years ago
12

Help on 20, would be appreciated if there was a explanation as well

Mathematics
1 answer:
ra1l [238]3 years ago
6 0
For a two column proof, we want to start with the given information. From there, we will use various definitions, postulates, and theorems to fill in the rest.

Our two sets of given information are that Plane <em>M </em>bisects Line <em>AB </em>and that Line <em>PA</em> is congruent to Line <em>PB</em>.

We know from the definition of a bisector that it splits a line in two equal parts. Therefore, Line <em>AO</em> must be congruent to Line <em>BO</em>.

Now, we have two sides of a triangle that we have proved to be congruent to each other. From the image given in the original problem, we see that both triangles share Line <em>OP. </em>Line <em>OP</em> is congruent to Line <em>OP</em> through the reflexive property.

We now have proven that all three sides of the one triangle are congruent to the corresponding sides on the other triangle. Therefore, the triangles are congruent through the SSS theorem. 

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165

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A traditional, physical bank with online options is a type of online bank.
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Describe how to convert from vertex form to standard form. (10 points
Anton [14]

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  "simplify" it

Step-by-step explanation:

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7 0
3 years ago
Add the two expressions.<br><br> 3z−4 and 2z + 5<br><br> Enter your answer in the box.
frosja888 [35]
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3 0
3 years ago
Read 2 more answers
Show Work Please Thank You
Misha Larkins [42]

Answer:

\displaystyle{x = \dfrac{\pi}{12}, \dfrac{\pi}{6}}

Step-by-step explanation:

We are given the trigonometric equation of:

\displaystyle{\sin 4x = \dfrac{\sqrt{3}}{2}}

Let u = 4x then:

\displaystyle{\sin u = \dfrac{\sqrt{3}}{2}}\\\\\displaystyle{\arcsin (\sin u) = \arcsin \left(\dfrac{\sqrt{3}}{2}\right)}\\\\\displaystyle{u= \arcsin \left(\dfrac{\sqrt{3}}{2}\right)}

Find a measurement that makes sin(u) = √3/2 true within [0, π) which are u = 60° (π/3) and u = 120° (2π/3).

\displaystyle{u = \dfrac{\pi}{3}, \dfrac{2\pi}{3}}

Convert u-term back to 4x:

\displaystyle{4x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}}

Divide both sides by 4:

\displaystyle{x = \dfrac{\pi}{12}, \dfrac{\pi}{6}}

The interval is given to be 0 ≤ 4x < π therefore the new interval is 0 ≤ x < π/4 and these solutions are valid since they are still in the interval.

Therefore:

\displaystyle{x = \dfrac{\pi}{12}, \dfrac{\pi}{6}}

8 0
1 year ago
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