Answer:
see below
Step-by-step explanation:
We need to find the diameter of the square
We can find this using the Pythagorean theorem
a^2+b^2 = c^2 where the legs are 7 and 7 and the diameter is c
7^2 +7^2 = c^2
49+49 = c^2
98 = c^2
Taking the square root of each side
sqrt(98) = sqrt(c^2)
9.899494937 = c
Since 9.9 is less than 11 which is the diameter of the circle it will never touch the circle.
Since the longest part of the square is less than the diameter of the circle, the square will fit inside the circle without touching
Is it OK if I explain it before I write down the next term ?
-1 ___ 3 ___ 11 ___ 23 ___ 39 ___ ( )
The first blank is . . . . . " + 4 = " .
The second blank is . . " + 8 = " .
The third blank is . . . . " + 12 = " .
The fourth blank is . . . " + 16 = " .
So the fifth blank is . . . " + 20 = " and the next number is 59 .
Answer:
The angle is "27 and 63".
Step-by-step explanation:
Let A and B are two angles, in which A is " 
" complementary angles and B be is another complementary angle.
Condition of complementary angle
:
Solution:
![\bf \begin{array}{clclll} -6&+&6\sqrt{3}\ i\\ \uparrow &&\uparrow \\ a&&b \end{array}\qquad \begin{cases} r=\sqrt{a^2+b^2}\\ \theta =tan^{-1}\left( \frac{b}{a} \right) \end{cases}\qquad r[cos(\theta )+i\ sin(\theta )]\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bclclll%7D%0A-6%26%2B%266%5Csqrt%7B3%7D%5C%20i%5C%5C%0A%5Cuparrow%20%26%26%5Cuparrow%20%5C%5C%0Aa%26%26b%0A%5Cend%7Barray%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ar%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%5C%5C%0A%5Ctheta%20%3Dtan%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7Bb%7D%7Ba%7D%20%5Cright%29%0A%5Cend%7Bcases%7D%5Cqquad%20r%5Bcos%28%5Ctheta%20%29%2Bi%5C%20sin%28%5Ctheta%20%29%5D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)
now, notice, there are two valid angles for such a tangent, however, if we look at the complex pair, the "a" is negative and the "b" is positive, that means, "x" is negative and "y" is positive, and that only occurs in the 2nd quadrant, so the angle is in the second quadrant, not on the fourth quadrant.
thus
![\bf 12\left[cos\left( \frac{2\pi }{3} \right) +i\ sin\left( \frac{2\pi }{3} \right) \right]](https://tex.z-dn.net/?f=%5Cbf%2012%5Cleft%5Bcos%5Cleft%28%20%5Cfrac%7B2%5Cpi%20%7D%7B3%7D%20%5Cright%29%20%2Bi%5C%20sin%5Cleft%28%20%5Cfrac%7B2%5Cpi%20%7D%7B3%7D%20%5Cright%29%20%5Cright%5D)
Answer:
306
Step-by-step explanation:
