In order to answer this one, it's really really really really helpful if you know what the law of cosines says. In fact it's absolutely necessary.
The law of cosines says if you know two sides of a triangle and the angle between them then you can use that information to find the length of the third side.
In the picture you know the lengths of two sides and you know the angle between them. So you can use the law of cosines to find the length of the third sidethat. That's side AC.
The last one, v = -2,000y + 20,000
I don’t really know exactly to what degree you need to put in your solution. But I rounded to the nearest tenth degree.
A = 112 degrees
B= 28 degrees (180-112-40 bc all sides of a triangle must equal 180)
C= 40
a= 27.6 (this is the side opposite of angle A)
b= 14 (side opposite b)
c= 19.2 (side opposite c)
HOW TO SOLVE:
c= law of sines, so c/sin(40) = 14/sin(28), multiply both sides by sin(40) so c can be isolated and solved for. c = 14sin(40)/sin(28). Plug into calculator then get answer. c is approximately 19.2.
a = law of sines again, so a/sin(112) = 14/sin(28). Multiply both sides again by sin(112) then solve. a = 14sin(112)/sin(28). Calculator again. a is around 27.6
Hello!
We are trying to describe the behavior of the graph given in the question.
To help us understand how to solve this question, we would need to understand <u>concavity.</u>
There are two types of concavity:
- Concave <em>up</em>
- Concave <em>down</em>
When a graph is concave up, the slope of the line would look like a "U".
When a graph is concave down, the slope of the line would look like a "U" that is flipped upside down.
In this case, we can see that the graph is concave down.
We can tell that the <em>slope</em> is negative due to the fact that the slope is going <u>down,</u> which results in the graph having a negative slope.
We can also tell that the graph is decreasing due to the fact that the line is doing downward.
Answer:
C). negative and decreasing
Answer: 0.0035
Step-by-step explanation:
Given : The readings on thermometers are normally distributed with a mean of 0 degrees C and a standard deviation of 1.00 degrees C.
i.e. 
and 
Let x denotes the readings on thermometers.
Then, the probability that a randomly selected thermometer reads greater than 2.17 will be :_
![P(X>2.7)=1-P(\xleq2.7)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{2.7-0}{1})\\\\=1-P(z\leq2.7)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.9965\ \ [\text{By z-table}]\ \\\\=0.0035](https://tex.z-dn.net/?f=P%28X%3E2.7%29%3D1-P%28%5Cxleq2.7%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cleq%5Cdfrac%7B2.7-0%7D%7B1%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq2.7%29%5C%20%5C%20%5B%5Cbecause%5C%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-0.9965%5C%20%5C%20%5B%5Ctext%7BBy%20z-table%7D%5D%5C%20%5C%5C%5C%5C%3D0.0035)
Hence, the probability that a randomly selected thermometer reads greater than 2.17 = 0.0035
The required region is attached below .
