Answer:
of the mural is sky blue.
Step-by-step explanation:
Let x be the full portion of mural.
Since,
of mural is painted in shades of blue.
Therefore, the blue portion=![\frac{4}{5}x](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B5%7Dx)
Also, Three eighths of the blue area is sky blue.
Therefore, the portion of the mural is sky blue=![\frac{3}{8}\times\frac{4}{5}x](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B8%7D%5Ctimes%5Cfrac%7B4%7D%7B5%7Dx)
The portion of the mural is sky blue=![\frac{12}{40}x](https://tex.z-dn.net/?f=%5Cfrac%7B12%7D%7B40%7Dx)
Simplify, by dividing numerator and denominator by 4, we get
The portion of the mural is sky blue=![\frac{3}{10}x](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B10%7Dx)
Answer:
Newton's third law of motion states that every action has an equal and opposite reaction. In softball when the catcher catches the ball the two forces present are the mitt on the ball and the ball on the mitt. The two are equal and in opposite directions.
Answer:
2^4 or 16
Step-by-step explanation:
We know that a^b / a^c = a^(b-c)
2^9 / 2^5
2^(9-5)
2^4 or 16
Rewriting the equation as a quadratic equation equal to zero:
x^2 - x - 30 = 0
We need two numbers whose sum is -1 and whose product is -30. In this case, it would have to be 5 and -6. Therefore we can also write our equation in the factored form
(x + 5)(x - 6) = 0
Now we have a product of two expressions that is equal to zero, which means any x value that makes either (x + 5) or (x - 6) zero will make their product zero.
x + 5 = 0 => x = -5
x - 6 = 0 => x = 6
Therefore, our solutions are x = -5 and x = 6.
Answer: See Below
<u>Step-by-step explanation:</u>
NOTE: You need the Unit Circle to answer these (attached)
5) cos (t) = 1
Where on the Unit Circle does cos = 1?
Answer: at 0π (0°) and all rotations of 2π (360°)
In radians: t = 0π + 2πn
In degrees: t = 0° + 360n
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![6)\quad sin (t) = \dfrac{1}{2}](https://tex.z-dn.net/?f=6%29%5Cquad%20sin%20%28t%29%20%3D%20%5Cdfrac%7B1%7D%7B2%7D)
Where on the Unit Circle does
<em>Hint: sin is only positive in Quadrants I and II</em>
![\text{Answer: at}\ \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)](https://tex.z-dn.net/?f=%5Ctext%7BAnswer%3A%20at%7D%5C%20%20%5Cdfrac%7B%5Cpi%7D%7B6%7D%5C%20%2830%5Eo%29%5C%20%5Ctext%7Band%20at%7D%5C%20%5Cdfrac%7B5%5Cpi%7D%7B6%7D%5C%20%28150%5Eo%29%5C%20%5Ctext%7Band%20all%20rotations%20of%7D%5C%202%5Cpi%20%5C%20%28360%5Eo%29)
![\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n](https://tex.z-dn.net/?f=%5Ctext%7BIn%20radians%3A%7D%5C%20t%20%3D%20%5Cdfrac%7B%5Cpi%7D%7B6%7D%20%2B%202%5Cpi%20n%20%5Cquad%20%5Ctext%7Band%7D%5Cquad%20%5Cdfrac%7B5%5Cpi%7D%7B6%7D%20%2B%202%5Cpi%20n)
In degrees: t = 30° + 360n and 150° + 360n
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![7)\quad tan (t) = -\sqrt3](https://tex.z-dn.net/?f=7%29%5Cquad%20tan%20%28t%29%20%3D%20-%5Csqrt3)
Where on the Unit Circle does ![\dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)](https://tex.z-dn.net/?f=%5Cdfrac%7Bsin%7D%7Bcos%7D%20%3D%20%5Cdfrac%7B-%5Csqrt3%7D%7B1%7D%5C%20or%5C%20%5Cdfrac%7B%5Csqrt3%7D%7B-1%7D%5Cquad%20%5Crightarrow%20%5Cquad%20%281%2C-%5Csqrt3%29%5C%20or%5C%20%28-1%2C%20%5Csqrt3%29)
<em>Hint: sin and cos are only opposite signs in Quadrants II and IV</em>
![\text{Answer: at}\ \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)](https://tex.z-dn.net/?f=%5Ctext%7BAnswer%3A%20at%7D%5C%20%20%5Cdfrac%7B2%5Cpi%7D%7B3%7D%5C%20%28120%5Eo%29%5C%20%5Ctext%7Band%20at%7D%5C%20%5Cdfrac%7B5%5Cpi%7D%7B3%7D%5C%20%28300%5Eo%29%5C%20%5Ctext%7Band%20all%20rotations%20of%7D%5C%202%5Cpi%20%5C%20%28360%5Eo%29)
![\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n](https://tex.z-dn.net/?f=%5Ctext%7BIn%20radians%3A%7D%5C%20t%20%3D%20%5Cdfrac%7B2%5Cpi%7D%7B3%7D%20%2B%202%5Cpi%20n%20%5Cquad%20%5Ctext%7Band%7D%5Cquad%20%5Cdfrac%7B5%5Cpi%7D%7B3%7D%20%2B%202%5Cpi%20n)
In degrees: t = 120° + 360n and 300° + 360n