This problem looks really scary am I right? Well, actually it isn't too bad.
Let me explain something really quick.
When two like terms are over each other they equal 1. For instance 3/3. In the case of two numbers with powers they can also cancel out if they are the same number to a power. Here is why:
3² / 3² = (3*3) / (3*3) = 9/9 = 1
What if the powers are unequal?
3²/3 = (3*3) / 3 = 9 / 3 = 3
In the case where the powers are unequal (in the case of the example we have a power of 2 and a power of 1) we simply SUBTRACT the powers. In the case where the bottom is bigger you'll have the following problem:
3²/3³ = (3*3) / (3*3*3) = 9 / 27 = 1/3
If you have trouble understanding the subtraction simply write out the multiplication. Then if you see numbers which are equal on the top on the bottom cross them out till there are no equal terms on the top or bottom. In the case above you can cross out the 2 threes at the top and are left with 1 three at the bottom.
For our problem: (Note the ^ symbol is the same as saying being brought to the power of... so it's equivalent to the problem you see on your screen!)
5^4 / 5^6 = (5*5*5*5) / (5*5*5*5*5*5) = 1 / (5*5) = 1 / 25
Our answer is:
1/25
Answer:
4.
A = arcsin(BC/AB) = arcsin(4/10) = ~23.58 deg
B = arccos(BC/AB) = arccos(4/10)= ~66.42 deg
AC = sqrt(AB^2-BC^2) = sqrt(10^2-4^2) = 9.16
5.
A = 90-B=90-20 =70
AB = AC/sinC = 30/sin(20deg) = ~87.71
BC = sqrt(AB^2-AC^2) = sqrt(87.71^2-30^2) = ~82.42
6.
AB = AC/cosA =125/cos(50deg) = ~194.46
B = 90-A = 90-50 =40
BC = sqrt(AB^2-AC^2) = sqrt(194.46^2-125^2) = ~148.96
7.
AB = AC/cosA =8/cos(20deg) = ~8.51
B=90-A =90-20=70
BC = sqrt(AB^2-AC^2) = sqrt(8.51^2-8^2) =~2.90
What are problems 10 and 11?
