Third option. The letter has neither rotate or symmetry
No. Two angles will have to be the same and the third angle will just be 180° minus twice the smaller angles.
So if we let the equal angles equal a, and the large angle equal be we have:
2a+b=180
So there are infinitely many values that a and b can have.
a=(180-b)/2, so the smaller equal angles have the range (0,90)
b=180-2a, so the larger has a range of (0,180)
<span> If ∆POW ≅∆BAM
</span>congruent <span>sides:
</span>PO = BA
OW = AM
PW = BM
<span>congruent </span>angles:
<P = <B
<O = <A
<W = <M
<span>First of all, write these in a way people can read them in the future. Don't make it so hard to help you.
Secondly:
Given that a♥b = 2a(32− b)
If a♥b = −23, solve for b in terms of a.
A) b = 16a + 32
B) b = 13a + 32
C) b = 12a + 32
D) b = 16a − 32
</span>-23 = 2a(32− b), devide both sides by 2a -23/2a = 32 - b, subtract 32 from both sides -23/2a - 32 = -b, multiply both sides by -1 23/2a + 32 = b
verified:
b = 23/2a + 32
-23 = 2a(32 − (23/2a + 32))
-23 = 2a(32 -23/2a -32)
-23 = 2a(-23/2a)
-23 = -23
Are you sure you copied this correctly?
Answer:
∆ABC ≅ ∆EDF by the SAS Congruence Theorem.
Step-by-step explanation:
<A ≅ <E,
Side length AC ≅ Side length EF
Side length AB ≅ Side length ED,
Thus, this implies that an included angle and two sides of one triangle are congruent to an included angle and 2 corresponding side lengths of the other triangle.
Therefore, we would conclude that:
∆ABC ≅ ∆EDF by the SAS Congruence Theorem.
