Simplify (shown in photo) A: 2^1/6 B: 2^2/3 C: 2^5/6 D: 2^7/6
1 answer:
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Answer:
The measure of the first angle is 50°.
Step-by-step explanation:
Let the three angles be <em>a, b, </em>and <em>c. </em>
<em />
The measure of the first angle is twice the measure of the second angle. In other words:
The measure of the third angle is 80° more than the measure of the second angle. In other words:
And since the interior angles of a triangle must equal 180°:
Substitute <em>a</em> and <em>c: </em>
<em /><em />
Combine like terms:
Subtract 80 from both sides:
And divide both sides by four:
So, the measure of the second angle is 25°.
Since the measure of the first is twice the second, the measure of the first angle is 50°.
(And since the measure of the third is 80° than the second, the measure of the third angle is 105°.)
Hello,
We know that if we make the same operation on both sides, we don't affect the equation:
For example:
Do you see? we add 2 on both sides and we don't affect the equality
We can do the same with the multiplication, for example
4=4 (i'll multiply both sides by 1/2)
Now, we do exactly the same in your excercise:
![P=2(L+W) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[multiply\,\,by\,\,1/2] \\ \\ P*\bold{ \frac{1}{2}}=2(L+W)*\bold{ \frac{1}{2}} \\ \\ \frac{P}{2}=L+W\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[Subtract\,\,L]\\ \\ \frac{P}{2}-\bold{L}=L+W-\bold{L}\\ \\ \frac{P}{2}-\bold{L}=L-\bold{L}+W \\ \\ \frac{P}{2}-L=W\\ \\ \boxed{ W= \frac{P}{2}-L }](https://tex.z-dn.net/?f=P%3D2%28L%2BW%29%20%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5Bmultiply%5C%2C%5C%2Cby%5C%2C%5C%2C1%2F2%5D%20%5C%5C%20%5C%5C%20P%2A%5Cbold%7B%20%5Cfrac%7B1%7D%7B2%7D%7D%3D2%28L%2BW%29%2A%5Cbold%7B%20%5Cfrac%7B1%7D%7B2%7D%7D%20%5C%5C%20%20%5C%5C%20%5Cfrac%7BP%7D%7B2%7D%3DL%2BW%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5BSubtract%5C%2C%5C%2CL%5D%5C%5C%20%20%5C%5C%20%5Cfrac%7BP%7D%7B2%7D-%5Cbold%7BL%7D%3DL%2BW-%5Cbold%7BL%7D%5C%5C%20%20%5C%5C%20%5Cfrac%7BP%7D%7B2%7D-%5Cbold%7BL%7D%3DL-%5Cbold%7BL%7D%2BW%20%5C%5C%20%20%5C%5C%20%5Cfrac%7BP%7D%7B2%7D-L%3DW%5C%5C%20%5C%5C%20%5Cboxed%7B%20W%3D%20%5Cfrac%7BP%7D%7B2%7D-L%20%7D)
We know that if we make the same operation on both sides, we don't affect the equation:
For example:
Do you see? we add 2 on both sides and we don't affect the equality
We can do the same with the multiplication, for example
4=4 (i'll multiply both sides by 1/2)
Now, we do exactly the same in your excercise: