i think it’s is a for the first one
Check the picture below.
so we know the radius of the semicircle is 2 and the rectangle below it is really a 4x4 square, so let's just get their separate areas and add them up.
![\stackrel{\textit{area of the semicircle}}{\cfrac{1}{2}\pi r^2}\implies \cfrac{1}{2}(\stackrel{\pi }{3.14})(2)^2\implies 3.14\cdot 2\implies 6.28 \\\\\\ \stackrel{\textit{area of the square}}{(4)(4)}\implies 16 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{sum of both areas}}{16+6.28=22.28}~\hfill](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20semicircle%7D%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B2%7D%28%5Cstackrel%7B%5Cpi%20%7D%7B3.14%7D%29%282%29%5E2%5Cimplies%203.14%5Ccdot%202%5Cimplies%206.28%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20square%7D%7D%7B%284%29%284%29%7D%5Cimplies%2016%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsum%20of%20both%20areas%7D%7D%7B16%2B6.28%3D22.28%7D~%5Chfill)
By applying the definition of difference, we find that the <em>average daily maximum</em> temperature in Syracuse is 26.55 °C higher than the <em>average daily minimum</em> temperature.
<h3>What is the difference between the average daily maximum temperature and the average daily minimum temperature?</h3>
Herein we must find the difference bewteen the two temperatures, defined as the subtraction of the <em>minimum</em> temperature from the <em>maximum</em> temperature:
x = 21.85 °C - (- 4.7 °C)
x = 26.55 °C
By applying the definition of difference, we find that the <em>average daily maximum</em> temperature in Syracuse is 26.55 °C higher than the <em>average daily minimum</em> temperature.
To learn more on differences: brainly.com/question/1927340
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Answer:
and
Step-by-step explanation:
The standard equation of a circle is 
where the coordinate (h,k) is the center of the circle.
Second Problem:
- We can start with the second problem which uses this info very easily.
- (h,k) in this problem is (-2,15) simply plug these into the equation.
. - We can also add the radius 3 and square it so it becomes 9. The equation.
- This simplifies to .
First Problem:
- The first problem takes a different approach it is not in standard form. But we can convert it to standard form by completing the square.
first subtract 37 from both sides so the equation is now 
.- by adding
to both the x and y portions of this equation you can complete the squares. 
and 
which equals 49 and 4. - Add 49 and 4 to both sides and the equation is now:
You can simplify the y and x portions of the equations into the perfect squares or factored form 
and 
. - Finally put the whole thing together. .
I hope this helps!
