V² - 2v = 3
v² - 2v - 3 = 0
v = <u>-(-2) +/- √((-2²) - 4(1)(-3))</u>
2(1)
v = <u>2 +/- √(4 + 12)</u>
2
v =<u> 2 +/- √16</u>
2
v = <u>2 +/- 4</u>
2
v = <u>2 + 4</u> v = <u>2 - 4</u>
2 2
v = <u>6</u> v = <u>-2</u>
2 2
v = 3 v = -1
It should increase but I’m between stays the same or increase, I’m not sure in a 100% sorry...
First of all, this problem is properly done with the Law of Cosines, which tells us

giving us a quadratic equation for b we can solve. But let's do it with the Law of Sines as asked.

We have c,a,A so the Law of Sines gives us sin C

There are two possible triangle angles with this sine, supplementary angles, one acute, one obtuse:


Both of these make a valid triangle with A=20°. They give respective B's:


So we get two possibilities for b:



Answer: 2.3 units and 7.8 units
Let's check it with the Law of Cosines:


There's a shortcut for the quadratic formula when the middle term is 'even.'




Looks good.
We have been given that in ΔJKL, the measure of ∠L=90°, KL = 22 feet, and JK = 54 feet. We are asked to find the measure of angle J to nearest degree.
First of all, we will draw a triangle as shown in the attachment.
We can see from our attachment that side KL is opposite side to angle J and side JK is hypotenuse of right triangle.
We know that sine relates opposite side of right triangle to hypotenuse.


Using inverse sine or arcsin, we will get:


Upon rounding to nearest degree, we will get:

Therefore, the measure of angle J is approximately 24 degrees.
Answer:
-1
Step-by-step explanation:
You can find the average my finding the sum of all temperatures and then dividing it by the number of temperatures ( in this case 7)
So you get (-3-5+3+7-2-2-5)/7=-7/7=-1