First multiply 180 and 25 which is 4500.
Then divide by 17 which should be approximately 264.71
Answer:
√23
Step-by-step explanation:
When you are given two side lengths of a right triangle, you use the Pythagorean theorem to find the third side: a² + b² = c², where c is the hypotenuse (the longest side).
All you have to do is plug the given information in:
Remember, 13 is the hypotenuse for this triangle.
12² + b² = 13²
Simplify:
144 + b² = 169
Subtract 144 from both sides:
b² = 169-144
b² = 23
Square root both sides:
b = √23
And that's your answer!
Please answer please please thank you thank goodness
Answer:
1/2+2x or 1/2+2+x
Step-by-step explanation:
i don't know but this what i think i hope this helps
The given function is
![h(x)=-16x^2+300x+20](https://tex.z-dn.net/?f=h%28x%29%3D-16x%5E2%2B300x%2B20)
According to this function, the starting height of the rocket is 20 feet because that's the initial condition of the problem stated by the independent term.
Additionally, we find the maximum height by calculating the vertex of the function V(h,k).
![h=-\frac{b}{2a}](https://tex.z-dn.net/?f=h%3D-%5Cfrac%7Bb%7D%7B2a%7D)
Where a = -16 and b = 300.
![\begin{gathered} h=-\frac{300}{2(-16)}=\frac{300}{32}=\frac{150}{16}=\frac{75}{8} \\ h=9.375 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20h%3D-%5Cfrac%7B300%7D%7B2%28-16%29%7D%3D%5Cfrac%7B300%7D%7B32%7D%3D%5Cfrac%7B150%7D%7B16%7D%3D%5Cfrac%7B75%7D%7B8%7D%20%5C%5C%20h%3D9.375%20%5Cend%7Bgathered%7D)
Then, we find k by evaluating the function
![\begin{gathered} k=-16(9.375)^2+300(9.375)+20 \\ k=-1406.25+2812.5+20=1426.25 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20k%3D-16%289.375%29%5E2%2B300%289.375%29%2B20%20%5C%5C%20k%3D-1406.25%2B2812.5%2B20%3D1426.25%20%5Cend%7Bgathered%7D)
Hence, the maximum height is 1426.25 feet.
At last, to know the time need to hit the ground, we just use h=9.375 and we multiply it by 2
![t=2\cdot9.375=18.75](https://tex.z-dn.net/?f=t%3D2%5Ccdot9.375%3D18.75)
Hence, the rocket hits the ground after 18.75 seconds.