The area of a rectangle is (length) times (width).
So you have to find a pair of numbers that multiply to produce 6 .
If you only stick to whole numbers, then I don't think there are three
different ones. You're going to need one pair that multiply to 6 and
are not both whole numbers.
After I explain how to solve problems, I hate to give answers. But with
all due respect, I have a feeling that I haven't nudged you enough yet
for you to use my explanation to find the answers on your own.
So here are some answers:
1 and 6
2 and 3
and sets of dimensions that are not both whole numbers, like
0.6 and 10
1.2 and 5
1.25 and 4.8
1.5 and 4
2.4 and 2.5
Answer:
Option b.
Step-by-step explanation:
We can easily solve this problem by using a graphing calculator or plotting tool.
The function is
h(t) = cos (t +1)
Please, see attached picture below.
By looking at the picture, we can tell that the amplitude is equal to 1
The function has a period of T = 2π
We can see there is no vertical shift.
It has a phase shift of +1. (Which moves the graph to the left)
Answer:
Option b.
Answer:
Step-by-step explanation:
Instead, since the divisor is in the form of 
, use what is called Synthetic Division. Remember, in this formula, 
gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:
4| 1 −5 7 −12
↓ 4 −4 12
_______________
1 −1 3 0 →
You start by placing the 
in the top left corner, then list all the coefficients of your dividend [x³ - 5x² + 7x - 12]. You bring down the original term closest to then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder. Finally, your quotient is one degree less than your dividend, so that 1 in your quotient can be an 
, the −1 [
] follows right behind it, and bringing up the rear, comes the 3, giving you the quotient of .
I am joyous to assist you anytime.
