This is a 30-60-90 triangle and we can apply rules to easily identify the hypotenuse of this triangle, which is denoted by <em>x</em>.
The length of the longer side of the triangle is given in the problem. To solve the hypotenuse of this triangle, let's solve first for the length of the shorter side of the triangle.
The shorter side can be solved by just dividing the length of the longer side by the square root of 3. Hence, we have
![short=\frac{4}{\sqrt[]{3}}](https://tex.z-dn.net/?f=short%3D%5Cfrac%7B4%7D%7B%5Csqrt%5B%5D%7B3%7D%7D)
Since we already have the values for the length of the shorter side and longer side, we can solve for the hypotenuse using the Pythagorean theorem.
![\begin{gathered} c=\sqrt[]{a^2+b^2} \\ c=\sqrt[]{4^2+(\frac{4}{\sqrt[]{3}})^2} \\ c=\sqrt[]{16+\frac{16}{3}} \\ c=\sqrt[]{\frac{64}{3}} \\ c=\frac{8}{\sqrt[]{3}}\cdot\frac{\sqrt[]{3}}{\sqrt[]{3}} \\ c=\frac{8\sqrt[]{3}}{3} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20c%3D%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%20%5C%5C%20c%3D%5Csqrt%5B%5D%7B4%5E2%2B%28%5Cfrac%7B4%7D%7B%5Csqrt%5B%5D%7B3%7D%7D%29%5E2%7D%20%5C%5C%20c%3D%5Csqrt%5B%5D%7B16%2B%5Cfrac%7B16%7D%7B3%7D%7D%20%5C%5C%20c%3D%5Csqrt%5B%5D%7B%5Cfrac%7B64%7D%7B3%7D%7D%20%5C%5C%20c%3D%5Cfrac%7B8%7D%7B%5Csqrt%5B%5D%7B3%7D%7D%5Ccdot%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B%5Csqrt%5B%5D%7B3%7D%7D%20%5C%5C%20c%3D%5Cfrac%7B8%5Csqrt%5B%5D%7B3%7D%7D%7B3%7D%20%5Cend%7Bgathered%7D)
Hence, the value of hypotenuse for this right triangle is
![x=\frac{8\sqrt[]{3}}{3}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B8%5Csqrt%5B%5D%7B3%7D%7D%7B3%7D)
We are asked about the fraction of pizza each student
gets, so the relevant numbers here is only 14 students and 7 pizzas. SO the
fraction of pizza each student get is the ratio of the two:
fraction = 7 pizzas / 14 students
fraction = 1 / 2
<span>So each student gets 1/2</span>
From the moment the friend passes the bicyclist, his friend covers a distance over time t of (3.63 m/s)*t.
The bicyclist covers a distance of 1/2*(2.11 m/s^2)*t^2. They meet when these distances are equal:
3.63 t = 1.055 t^2 ==> 1.055 t^2 - 3.63 t = 0
==> t = 0 s or t = 3.44 s
Answer:
(h*h)(10) is the same as h(10)*h(10). Let's find the value of h(10) first. To do this, replace every x with 10 like so
Step-by-step explanation:
h(x) = 6-x
h(10) = 6-10
h(10) = -4
So,
h(10)*h(10) = (-4)*(-4)
h(10)*h(10) = 16
The final answer is 16
