Answer:
1000.
it's just 10 times itself twice (10^2) times 10
Answer:
x = 3
Step-by-step explanation:
Use the Pythagorean theorem:
![x^2+x^2=(3\sqrt2)^2\qquad\text{use}\ (ab)^n=a^nb^n\\\\2x^2=3^2(\sqrt2)^2\qquad\text{use}\ (\sqrt{a})^2=a\\\\2x^2=(9)(2)\\\\2x^2=18\qquad\text{divide both sides by 2}\\\\x^2=9\to x=\sqrt9\\\\\boxed{x=3}](https://tex.z-dn.net/?f=x%5E2%2Bx%5E2%3D%283%5Csqrt2%29%5E2%5Cqquad%5Ctext%7Buse%7D%5C%20%28ab%29%5En%3Da%5Enb%5En%5C%5C%5C%5C2x%5E2%3D3%5E2%28%5Csqrt2%29%5E2%5Cqquad%5Ctext%7Buse%7D%5C%20%28%5Csqrt%7Ba%7D%29%5E2%3Da%5C%5C%5C%5C2x%5E2%3D%289%29%282%29%5C%5C%5C%5C2x%5E2%3D18%5Cqquad%5Ctext%7Bdivide%20both%20sides%20by%202%7D%5C%5C%5C%5Cx%5E2%3D9%5Cto%20x%3D%5Csqrt9%5C%5C%5C%5C%5Cboxed%7Bx%3D3%7D)
Answer:
my best answer for this is B. False.
I calculated as fast as i can.
For part B, you are correct as far as you've gone. There is no algebraic solution to your equation. A graphing calculator or Newton's Method iteration can get you to a solution fairly quickly.
a ≈ 1.114157141
For part C, you need to consider your answer. For a=0, the equation is that of a straight line, so there is no inflection point at x=1. For cos(a)=0, there are an infinite number of possible values of
![a](https://tex.z-dn.net/?f=a)
that will put a point of inflection at x=1. As you have noted, a=π/2 is only one of them in the range 0 < a < 4.
For part D, again you have stopped part way to the answer. Consider what values of
![a](https://tex.z-dn.net/?f=a)
will make
![a \sin(ax)](https://tex.z-dn.net/?f=a%20%5Csin%28ax%29)
strictly greater than 1. There aren't any. The sine function always crosses zero. This part of the question has no solution.