So ASA is angle side angle, and that means that if you prove that the side, and the side adjacent to that side and the angle between those two sides are all congruent to another triangle's sides and angle, the triangles are both congruent.
The AAS is angle angle side, or something, so say you have a triangle and you prove that two of its angles are congruent along with a side to another triangle's, then it's AAS. I understand where the confusion might be. I guess it's just a matter of what you state first in your proof?
Answer:
<em><u>The answer is 4 </u></em><em><u> </u></em><em><u>because </u></em><em><u>the</u></em><em><u> </u></em><em><u>2</u></em><em><u> </u></em><em><u>over</u></em><em><u> </u></em><em><u>3</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>canceled</u></em><em><u> </u></em><em><u>out</u></em><em><u /></em>
Answer:
<h3>
m∠2 = 67°</h3>
Step-by-step explanation:
m║n ⇒ (9x + 2)° = 74° {Corresponding Angles}
(9x)° = 72°
x = 8
Angles: angle corresponding to angle (5x-1)°, angle 74° and angle 2 add to 180° (straight angle)
(5×8 - 1)° + 74° + m∠2 = 180°
39° + 74° + m∠2 = 180°
113° + m∠2 = 180°
m∠2 = 180° - 113°
m∠2 = 67°
Answer:
16
Step-by-step explanation:
solve for X.
16+4x=10+14
16+4x=24
4x=24-16
4x=8
x=2
8x=8(2)
8x=16
21−=2(2−)=2cos(−1)+2 sin(−1)
−1+2=−1(2)=−1(cos2+sin2)=cos2+ sin2
Is the above the correct way to write 21− and −1+2 in the form +? I wasn't sure if I could change Euler's formula to =cos()+sin(), where is a constant.
complex-numbers
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edited Mar 6 '17 at 4:38
Richard Ambler
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asked Mar 6 '17 at 3:34
14wml
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1 Answer
1
No. It is not true that =cos()+sin(). Notice that
1=1≠cos()+sin(),
for example consider this at =0.
As a hint for figuring this out, notice that
+=ln(+)
then recall your rules for logarithms to get this to the form (+)ln().