HELP PLEASE. In. AA80 m24 = 10° and m28- 145°. Select the triangle that is similar to AABC. O A. AMNP, in which m_M= 145º and MzN= 15 O B. AXYZ, in which m2X= 10º and m2 7 = 25° O = C. AJKL, in which meJ= 10° and mL = 135 o D. APQR, in which m_P= 25° and MR = 135°​

B. ∆XYZ, in which m∠X = 10° and m∠Z = 25°

Step-by-step explanation:

∆ABC

m∠A = 10°

m∠B = 145°

m∠C = 180° - (m∠A + m∠B )

= 180° - (10° + 145°)

= 180° - 155°

m∠C = 25°

∆XYZ

m∠X = 10°

m∠Z = 25°

m∠Y = 180° - (m∠X + m∠Z)

= 180° - (10° + 25°)

= 180° - 35°

m∠Y = 145°

so,

m∠A = m∠X = 10°

m∠C = m∠Z = 25°

m∠B = m∠Y = 145°

I'm just ordinary human, CMIIW

answer: can you me with homework?

step-by-step explanation:

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this