I think that the answer is 0.9
Angle D is 60° because it is a supplementary angle, meaning 180 (straight line) - 120 (angle A) = 60°
First write out all the two digit square numbers
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
Next divide all of them by 7
16/7 = 2 2/7 or 2 and 2 remainders
25/7 = 3 4/7 or 3 and 4 remainders
36/7 = 5 1/7 or 5 and 1 remainder
49/7 = 7
64/7 = 9 1/7 or 9 and 1 remainder
81/7 = 11 4/7 or 11 and 4 remainders
This tells us that the only possible remainders for a two digit square number divided by 7 are 1, 2 and 4
If Doris got 6 and Horace got 3, they must have made a mistake.
Answer:
Step-by-step explanation:
1S. JM bisects ∠ KJL and ∠JMK ≅ ∠JML
1R. Given
2S. ∠KJM ≅ ∠LJM
2R. Angle bisectors form ≅ ∠s
3S. JM≅JM
3R. Reflexive Propriety ( segment is ≅ with itself)
4S. ΔJMK ≅ΔJML
4R. ASA theorem of congruency
5S. JK≅JL
5R. Coresponding parts of congruent Δs are congruent (CPCTC)
![\bf \textit{Logarithm Cancellation Rules} \\\\ \stackrel{\stackrel{\textit{we'll use this one}}{\downarrow }}{log_a a^x = x}\qquad \qquad a^{log_a x}=x~\hfill\stackrel{recall}{ln=log_e}\qquad log_e(e^z)=z \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7BLogarithm%20Cancellation%20Rules%7D%20%5C%5C%5C%5C%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7B%5Cdownarrow%20%7D%7D%7Blog_a%20a%5Ex%20%3D%20x%7D%5Cqquad%20%5Cqquad%20a%5E%7Blog_a%20x%7D%3Dx~%5Chfill%5Cstackrel%7Brecall%7D%7Bln%3Dlog_e%7D%5Cqquad%20log_e%28e%5Ez%29%3Dz%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%20)
and you plug that in your calculator to get about -0.27465307216702742285.
