Tan(x)= sin(x)/cos(x), therefore, substitute sin(x)/cos(x) in the expression:
=(cos(x)(sin(x)/cos(x))-1)/cos^2(x)
simplify the 2 cos(x):
=(sin(x)-1)/cos^2(x)
Sin^2(x)+ cos^2(x)=1, sin^2(x)-1=- cos^2(x), substitute in the expression:
= -cos^2(x)/cos^2(x)= -1
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:
x=9
Step-by-step explanation:
It’s 9 trust me I know it’s 9 ok it’s 9 I’m sure it’s 9 ok the answer is 9 got it 9 byee
Answer:
the
Step-by-step explanation:
it's fairly easy actually. You just have to use sin
It all depends on how big the boxes are but i would say about 5 or 6 boxes if they are big boxes
