Answer: The answer is (A) ∠T.
Step-by-step explanation: Given that the polygon ABCDE is congruent to the polygon TVSRK. We are to find the corresponding angle of ∠EAB.
In the two polygons, the corresponding vertices are
A ⇒ T
B ⇒ V
C ⇒ S
D ⇒ R
E ⇒ K.
Therefore, ∠EAB, which is ∠A will correspond to ∠T.
Thus, the correct option is (A) ∠T.
Answer: a) √50
b) n = 1 + 7i
Step-by-step explanation:
first, the modulus of a complex number z = a + bi is
IzI = √(a^2 + b^2)
The fact that n is complex does not mean that n doesn't has a real part, so we must write our numbers as:
m = 2 + 6i
n = a + bi
Im + nI = 3√10
Im + n I = √(a^2 + b^2 + 2^2 + 6^2)= 3√10
= √(a^2 + b^2 + 40) = 3√10
a^2 + b^2 + 40 = 3^2*10 = 9*10 = 90
a^2 + b^2 = 90 - 40 = 50
√(a^2 + b^2 ) = InI = √50
The modulus of n must be equal to the square root of 50.
now we can find any values a and b such a^2 + b^2 = 50.
for example, a = 1 and b = 7
1^2 + 7^2 = 1 + 49 = 50
Then a possible value for n is:
n = 1 + 7i
An acute angle can also be made inside of that type of triangle.
The answer is y=-2x+4. The y-intercept is (0,4) and the slope is -2.
Answer: 6
Step-by-step explanation: you make a number line or you can start on your left from -15,-14,-13,-12,-11,-10,-9, you “hop” 6 times on a number line to get to -9
