Answer:
(4+x) + x = 18; 2x=14; x=7; width = 7, length = 11
Step-by-step explanation:
Answer:
![V=5\sqrt{3}\ m^3](https://tex.z-dn.net/?f=V%3D5%5Csqrt%7B3%7D%5C%20m%5E3)
Step-by-step explanation:
we know that
The volume of a trough is equal to
![V=BL](https://tex.z-dn.net/?f=V%3DBL)
where
B is the area of equilateral triangle
L is the length of a trough
step 1
Find the area of equilateral triangle B
The area of a equilateral triangle applying the law of sines is equal to
![B=\frac{1}{2} b^{2} sin(60\°)](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B1%7D%7B2%7D%20b%5E%7B2%7D%20sin%2860%5C%C2%B0%29)
where
![b=2\ m](https://tex.z-dn.net/?f=b%3D2%5C%20m)
![sin(60\°)=\frac{\sqrt{3}}{2}](https://tex.z-dn.net/?f=sin%2860%5C%C2%B0%29%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D)
substitute
![B=\frac{1}{2}(2)^{2} (\frac{\sqrt{3}}{2})](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B1%7D%7B2%7D%282%29%5E%7B2%7D%20%28%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%29)
![B=\sqrt{3}\ m^{2}](https://tex.z-dn.net/?f=B%3D%5Csqrt%7B3%7D%5C%20m%5E%7B2%7D)
step 2
Find the volume of a trough
![V=BL](https://tex.z-dn.net/?f=V%3DBL)
we have
![B=\sqrt{3}\ m^{2}](https://tex.z-dn.net/?f=B%3D%5Csqrt%7B3%7D%5C%20m%5E%7B2%7D)
![L=5\ m](https://tex.z-dn.net/?f=L%3D5%5C%20m)
substitute
![V=(\sqrt{3})(5)](https://tex.z-dn.net/?f=V%3D%28%5Csqrt%7B3%7D%29%285%29)
![V=5\sqrt{3}\ m^3](https://tex.z-dn.net/?f=V%3D5%5Csqrt%7B3%7D%5C%20m%5E3)
D. You start out with $500, it does not use a variable, knocking out a and b. Also, you need at least the needed amount of money, so b or c cannot be the answer. This only leaves d.
Answer:
The first one is OBTUSE, since there is one obtuse angle(greater than 90 and less than 180)
The second one is acute, since all side have angles less than 90 degrees
The third and fourth ones are right triangles, the right triangle usually consists of one angle with 90 degrees(denoted by a square)
5n-10-n=14.8
6n-10=14.8
6n=24.8
24.8. divided by 6
n=4.something