Answer:
Step-by-step explanation:
(a - b)² = a² + 2ab + b²
Area of square frame =side²
= (2b - 1)²
= (2b)² - 2*2b * 1 + 1²
= 4b² - 4b + 1
Interest is 100*(2/100)
$2
Answer:
A
Step-by-step explanation:
We are given that:
![\tan(\theta)=1.3](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%3D1.3)
And we want to find:
![\cot(90^\circ -\theta)](https://tex.z-dn.net/?f=%5Ccot%2890%5E%5Ccirc%20-%5Ctheta%29)
Remember that tangent and cotangent are co-functions. In other words, they follow the cofunction identities:
![\tan(\theta)=\cot(90^\circ-\theta)\text{ and } \cot(90^\circ-\theta)=\tan(\theta)](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%3D%5Ccot%2890%5E%5Ccirc-%5Ctheta%29%5Ctext%7B%20and%20%7D%20%5Ccot%2890%5E%5Ccirc-%5Ctheta%29%3D%5Ctan%28%5Ctheta%29)
Therefore, since tan(θ) = 1.3 and cot(90° - θ) = tan(θ), then cot(90° - θ) must also be 1.3.
Our answer is A.
It’s C because
3*2 - 2* -1 =8
6+2=8
8=8
Answer:
○ C
Explanation:
Accourding to one of the circle equations,
the centre of the circle is represented by
Moreover, all negative symbols give you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so you must pay cloce attention to which term gets which symbol. Another thing you need to know is that the radius will ALWAYS be squared, so no matter how your equation comes about, make sure that the radius is squared. Now, in case you did not know how to define the radius, you can choose between either method:
Pythagorean Theorem
![\displaystyle a^2 + b^2 = c^2 \\ \\ 6^2 + 8^2 = r^2 \hookrightarrow 36 + 64 = r^2 \hookrightarrow \sqrt{100} = \sqrt{r^2} \\ \\ \boxed{10 = r}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%5E2%20%2B%20b%5E2%20%3D%20c%5E2%20%5C%5C%20%5C%5C%206%5E2%20%2B%208%5E2%20%3D%20r%5E2%20%5Chookrightarrow%2036%20%2B%2064%20%3D%20r%5E2%20%5Chookrightarrow%20%5Csqrt%7B100%7D%20%3D%20%5Csqrt%7Br%5E2%7D%20%5C%5C%20%5C%5C%20%5Cboxed%7B10%20%3D%20r%7D)
Sinse we are dealing with <em>length</em>, we only desire the NON-NEGATIVE root.
Distanse Equation
![\displaystyle \sqrt{[-x_1 + x_2]^2 + [-y_1 + y_2]^2} = d \\ \\ \sqrt{[-5 - 3]^2 + [3 + 3]^2} = r \hookrightarrow \sqrt{[-8]^2 + 6^2} = r \hookrightarrow \sqrt{64 + 36} = r; \sqrt{100} = r \\ \\ \boxed{10 = r}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Csqrt%7B%5B-x_1%20%2B%20x_2%5D%5E2%20%2B%20%5B-y_1%20%2B%20y_2%5D%5E2%7D%20%3D%20d%20%5C%5C%20%5C%5C%20%5Csqrt%7B%5B-5%20-%203%5D%5E2%20%2B%20%5B3%20%2B%203%5D%5E2%7D%20%3D%20r%20%5Chookrightarrow%20%5Csqrt%7B%5B-8%5D%5E2%20%2B%206%5E2%7D%20%3D%20r%20%5Chookrightarrow%20%5Csqrt%7B64%20%2B%2036%7D%20%3D%20r%3B%20%5Csqrt%7B100%7D%20%3D%20r%20%5C%5C%20%5C%5C%20%5Cboxed%7B10%20%3D%20r%7D)
Sinse we are dealing with <em>distanse</em>, we only desire the NON-NEGATIVE root.
I am joyous to assist you at any time.