Step-by-step explanation:
see the pic for the answer
Answer:
y-3
Problem:
What is the remainder when the dividend is xy-3, the divisor is y, and the quotient is x-1. ?
Step-by-step explanation:
Dividend=quotient×divisor+remainder
So we have
xy-3=(x-1)×(y)+remainder
xy-3=(xy-y)+remainder *distributive property
Now we just need to figure out what polynomial goes in for the remainder so this will be a true identity.
We need to get rid of minus y so we need plus y in the remainder.
We also need minus 3 in the remainder.
So the remainder is y-3.
Let's try it out:
xy-3=(xy-y)+remainder
xy-3=(xy-y)+(y-3)
xy-3=xy-3 is what we wanted so we are done here.
Answer:
D. m∠A=43, m∠B=55, a=20
Step-by-step explanation:
Given:
∆ABC,
m<C = 82°
AB = c = 29
AC = b = 24
Required:
m<A, m<C, and a (BC)
SOLUTION:
Find m<B using the law of sines:








m<B = 55°
Find m<A:
m<A = 180 - (82 + 55) => sum of angles in a triangle.
= 180 - 137
m<A = 43°
Find a using the law of sines:


Cross multiply


(approximated)