The same value as cosα
cosα=sin(90-α) and vice versa
sinα=cos(90-α)
The area of the square is greater than the circle: the area of the circle is pi(r)^2 while the area of the square is length times width. The area of the circle is about 12.5 while the area of the square is 16.
Answer:
3 1/3
Step-by-step explanation:
10+z=z+z+z+z
10+z=4z
10=3z
z=3 1/3
Check the picture below, so, that'd be the square inscribed in the circle.
so... hmm the diagonals for the square are the diameter of the circle, and keep in mind that the radius of a circle is half the diameter, so let's find the diameter.
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -2}}\quad ,&{{ 5}})\quad % (c,d) &({{ -8}}\quad ,&{{ -3}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-8-(-2)]^2+[-3-5]^2} \\\\\\ d=\sqrt{(-8+2)^2+(-3-5)^2}\implies d=\sqrt{(-6)^2+(-8)^2} \\\\\\ d=\sqrt{36+64}\implies d=\sqrt{100}\implies d=10](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%28%7B%7B%20-2%7D%7D%5Cquad%20%2C%26%7B%7B%205%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%20-8%7D%7D%5Cquad%20%2C%26%7B%7B%20-3%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdiameter%7D%7Bd%7D%3D%5Csqrt%7B%5B-8-%28-2%29%5D%5E2%2B%5B-3-5%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-8%2B2%29%5E2%2B%28-3-5%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-6%29%5E2%2B%28-8%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B36%2B64%7D%5Cimplies%20d%3D%5Csqrt%7B100%7D%5Cimplies%20d%3D10)
that means the radius r = 5.
now, what's the center? well, the Midpoint of the diagonals, is really the center of the circle, let's check,

so, now we know the center coordinates and the radius, let's plug them in,
Answer:
A. b(w) = 80w +30
B. input: weeks; output: flowers that bloomed
C. 2830
Step-by-step explanation:
<h3>Part A:</h3>
For f(s) = 2s +30, and s(w) = 40w, the composite function f(s(w)) is ...
b(w) = f(s(w)) = 2(40w) +30
b(w) = 80w +30 . . . . . . blooms over w weeks
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<h3>Part B:</h3>
The input units of f(s) are <em>seeds</em>. The output units are <em>flowers</em>.
The input units of s(w) are <em>weeks</em>. The output units are <em>seeds</em>.
Then the function b(w) above has input units of <em>weeks</em>, and output units of <em>flowers</em> (blooms).
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<h3>Part C:</h3>
For 35 weeks, the number of flowers that bloomed is ...
b(35) = 80(35) +30 = 2830 . . . . flowers bloomed over 35 weeks