Answer:
Granted that uppercase letters are the angles measurements and lowercase letters are the side lengths. I also was unable to round as I don't know to what decimal you need to round to.
First to find the measure of angle B, we must subtract A and C from 180

B=54
Next, we can use the law of sines in order to find the measures of a and b
As a reminder, The law of sines is

First, lets solve for b

Next lets solve for a

Solution:
<u>Note that:</u>
- Given inequality: 3 < x + 3 < 6
<u>Solving the inequality one by one.</u>
- 3 < x + 3
- => -3 + 3 < x
- => 0 < x
- => x + 3 < 6
- => x < 6 - 3
- => x < 3
The inequality is <u>0 < x < 3.</u>
Answer:
the two minus symbols cancel out making it positive 5
Step-by-step explanation:
Answer:
a) Find the common ratio of this sequence.
Answer: -0.82
b) Find the sum to infinity of this sequence.
Answer: 2.2
Step-by-step explanation:
nth term in geometric series is given by ![4\ th \ term = ar^n-1\\-2.196 = 4r^{4-1} \\-2.196/4 = r^{3} \\r = \sqrt[3]{0.549} \\r = 0.82](https://tex.z-dn.net/?f=4%5C%20th%20%5C%20term%20%3D%20ar%5En-1%5C%5C-2.196%20%3D%204r%5E%7B4-1%7D%20%5C%5C-2.196%2F4%20%3D%20r%5E%7B3%7D%20%5C%5Cr%20%3D%20%5Csqrt%5B3%5D%7B0.549%7D%20%5C%5Cr%20%3D%200.82)
where
a is the first term
r is the common ratio and
n is the nth term
_________________________________
given
a = 4
4th term = -2.196
let
common ratio of this sequence. be r
![4\ th \ term = ar^n-1\\-2.196 = 4r^{4-1} \\-2.196/4 = r^{3} \\r = \sqrt[3]{-0.549} \\r = -0.82](https://tex.z-dn.net/?f=4%5C%20th%20%5C%20term%20%3D%20ar%5En-1%5C%5C-2.196%20%3D%204r%5E%7B4-1%7D%20%5C%5C-2.196%2F4%20%3D%20r%5E%7B3%7D%20%5C%5Cr%20%3D%20%5Csqrt%5B3%5D%7B-0.549%7D%20%5C%5Cr%20%3D%20-0.82)
a) Find the common ratio of this sequence.
answer: -0.82
sum of infinity of geometric sequence is given by = a/(1-r)
thus,
sum to infinity of this sequence = 4/(1-(-0.82) = 4/1.82 = 2.2
Answer:
A) 1:4
Step-by-step explanation:
#We first calculate the campus' area:

#We then calculate the campus' map area in square ft:

Ratio is a comparison between two dimensions:

Hence, the ratio of the area in square miles of the campus to the area in square feet of the map is 1:4