Given:
square with sides measuring 7 cm.
3 triangles attached to three sides of the square. A line bisecting one triangle is measured at 4 cm.
Area of a square = s² = (7cm)² = 49 cm²
Area of a triangle = hb/2 = (4cm*7cm)/2 = 14 cm²
Area of the 3 triangles = 14 cm² x 3 = 42 cm²
Total area of the logo = 49 cm² + 42 cm² = 91 cm²
Answer:
Therefore,
![r=\sqrt[3]{\frac{3V}{4\pi }}](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%5Cpi%20%7D%7D)
is the required r
Step-by-step explanation:
Given:
Volume of inside of the sphere is given as

where r is the radius of the sphere
To Find:
r =?
Solution:
We have
......Given
![3\times V=4\pi r^{3} \\\\\therefore r^{3}=\frac{3V}{4\pi } \\\\\therefore r=\sqrt[3]{\frac{3V}{4\pi }} \textrm{which is the expression for r}](https://tex.z-dn.net/?f=3%5Ctimes%20V%3D4%5Cpi%20r%5E%7B3%7D%20%5C%5C%5C%5C%5Ctherefore%20r%5E%7B3%7D%3D%5Cfrac%7B3V%7D%7B4%5Cpi%20%7D%20%5C%5C%5C%5C%5Ctherefore%20r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%5Cpi%20%7D%7D%20%5Ctextrm%7Bwhich%20is%20the%20expression%20for%20r%7D)
Therefore,
![r=\sqrt[3]{\frac{3V}{4\pi }}](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%5Cpi%20%7D%7D)
is the required r
Why trees are helpful........................................
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Very easy just state facts about trees in a format like this. Google why trees are helpful and some facts should pop up. A good paragraph should have at least five (5) good strong sentences.
Answer:
Step-by-step explanation:
The position function is
and if we are looking for the time(s) that the ball is 10 feet above the surface of the moon, we sub in a 10 for s(t) and solve for t:
and
and factor that however you are currently factoring quadratics in class to get
t = .07 sec and t = 18.45 sec
There are 2 times that the ball passes 10 feet above the surface of the moon, once going up (.07 sec) and then again coming down (18.45 sec).
For part B, we are looking for the time that the ball lands on the surface of the moon. Set the height equal to 0 because the height of something ON the ground is 0:
and factor that to get
t = -.129 sec and t = 18.65 sec
Since time can NEVER be negative, we know that it takes 18.65 seconds after launch for the ball to land on the surface of the moon.