Answer:
14 ft.
Step-by-step explanation:
Please consider the complete question.
Ron is designing a new slide for the amusement park in town. The slide makes an angle of 39° with the ground. The top of the slide
is to be 9 feet off the ground. What should be the length of the slide he needs to make to the nearest foot?
We can see from the attachment that slide forms a right triangle with respect to ground, where, h is length of the slide.
We will use sine to solve for h as sine relates opposite side of right triangle with hypotenuse.
![\text{sin}=\frac{\text{Opposite}}{\text{Hypotenuse}}](https://tex.z-dn.net/?f=%5Ctext%7Bsin%7D%3D%5Cfrac%7B%5Ctext%7BOpposite%7D%7D%7B%5Ctext%7BHypotenuse%7D%7D)
![\text{sin}(39^{\circ})=\frac{9}{h}](https://tex.z-dn.net/?f=%5Ctext%7Bsin%7D%2839%5E%7B%5Ccirc%7D%29%3D%5Cfrac%7B9%7D%7Bh%7D)
![h=\frac{9}{\text{sin}(39^{\circ})}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B9%7D%7B%5Ctext%7Bsin%7D%2839%5E%7B%5Ccirc%7D%29%7D)
![h=\frac{9}{0.62932039105}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B9%7D%7B0.62932039105%7D)
![h=14.30114](https://tex.z-dn.net/?f=h%3D14.30114)
![h\approx 14](https://tex.z-dn.net/?f=h%5Capprox%2014)
Therefore, Ron needs to make the slide approximately 14 feet long.
Answer:
2nd one down
Step-by-step explanation:
Put simply, it divides two sides of a triangle equally. The midpoint of a side divides the side into two equal segments. As you can see in the picture below, DE is the midsegment of the triangle ABC. Point D divides segment AB into two equal parts, and point E divides segment CB into two equal parts.
The length of the line segment AB where
![A(x_A;\ y_A)](https://tex.z-dn.net/?f=A%28x_A%3B%5C%20y_A%29)
and
![B(x_B;\ y_B)](https://tex.z-dn.net/?f=B%28x_B%3B%5C%20y_B%29)
![|AB|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}](https://tex.z-dn.net/?f=%7CAB%7C%3D%5Csqrt%7B%28x_B-x_A%29%5E2%2B%28y_B-y_A%29%5E2%7D)
We have:
![A(1;\ 6);\ B(11;-4)](https://tex.z-dn.net/?f=A%281%3B%5C%206%29%3B%5C%20B%2811%3B-4%29)
subtitute
![|AB|=\sqrt{(1-11)^2+(6-(-4))^2}=\sqrt{(-10)^2+10^2}=\sqrt{100+100}\\\\=\sqrt{100\cdot2}=\sqrt{100}\cdot\sqrt2=10\sqrt2\approx14.1](https://tex.z-dn.net/?f=%7CAB%7C%3D%5Csqrt%7B%281-11%29%5E2%2B%286-%28-4%29%29%5E2%7D%3D%5Csqrt%7B%28-10%29%5E2%2B10%5E2%7D%3D%5Csqrt%7B100%2B100%7D%5C%5C%5C%5C%3D%5Csqrt%7B100%5Ccdot2%7D%3D%5Csqrt%7B100%7D%5Ccdot%5Csqrt2%3D10%5Csqrt2%5Capprox14.1)
Answer: b. 14.1
Answer:
10 rounds.
Step-by-step explanation:
Let x represent number of rounds around the field.
We have been given that the distance around our playing field is 300 meters long. So distance covered in x rounds would be
meters.
We are also told that your coach wants you to run 3 kilometers. Let us convert 3 km in meters.
1 km = 1000 m
3 km = 3*1000 m = 3000 m
Now, we will equate total distance covered in x rounds by 3000 as:
![300x=3000](https://tex.z-dn.net/?f=300x%3D3000)
![\frac{300x}{300}=\frac{3000}{300}](https://tex.z-dn.net/?f=%5Cfrac%7B300x%7D%7B300%7D%3D%5Cfrac%7B3000%7D%7B300%7D)
![x=10](https://tex.z-dn.net/?f=x%3D10)
Therefore, you need to run 10 times around the field.