Answer:
Yes
Step-by-step explanation:
1) Cos 300 = Cos (360 - 60) = Cos 60 {Cos 360 - theta = Cos theta}
= 1/2
Sin 300 = Sin (360 - 60) = -Sin 60 = 
3) 3π/4 = 3*180/4 = 135
Cos 135 = Cos (90 + 45) = -Cos 45 = 
Sin 135 = Sin (90 + 45) = Sin 45 = 
<h3><u>The value of the greater number is 15.</u></h3><h3><u>The value of the smaller number is 7.</u></h3>
x = 2y + 1
3x = 5y + 10
Because we have a value for x we can plug this value in to find the value of y.
3(2y + 1) = 5y + 10
Distributive property.
6y + 3 = 5y + 10
Subtract 5y from both sides.
y + 3 = 10
Subtract 3 from both sides.
y = 7
We can plug this value back into the original equation to find the value of x.
x = 2(7) + 1
x = 15
Answer:
1/4
Step-by-step explanation:
Let the equation be:
y = ax^2 + bx + c.
Then, substitue the three points into the equation.
First point: 0 = a0^2 + b0 + c.
So c = 0.
Second point: -2 = a(-1)^2 + b(-1) + c.
So a - b + c = -2.
Third point: 6 = a*1^2 + b*1 + c.
So a + b + c = 6.
We know that c=0 already, so we substitute c=0 into the last two equations and we would get:
a - b = -2
a + b = 6
We add the two equations and we get:
2a = 4
a = 2
Then, we substitute a=2 into a-b=-2 and we get:
-b = -4
b = 4
Now we know a = 2, b = 4, and c = 0
Then, the equation of the parabola would be:
2x^2 + 4x
Check the picture below.
let's recall that in a Kite, the diagonals meet at 90° angles, therefore, we know the height of each of those 4 triangles, is 2.5 and 6, now, since the pair of triangles above are 45-45-90 triangles, we can use the 45-45-90 rule, as you see there, so, if the height is 2.5, then the base is also 2.5.
so, we really have 2 pair of triangles whose base is 2.5 and height of 2.5, and another pair of triangles whose base is 2.5 and height is 6, let's add their areas.
![\bf \stackrel{\textit{area of 2 triangles above}}{2\left[\cfrac{1}{2}(2.5)(2.5) \right]}~~+~~\stackrel{\textit{area of 2 triangles below}}{2\left[ \cfrac{1}{2}(2.5)(6) \right]}\implies 6.25+15\implies 21.25](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%202%20triangles%20above%7D%7D%7B2%5Cleft%5B%5Ccfrac%7B1%7D%7B2%7D%282.5%29%282.5%29%20%5Cright%5D%7D~~%2B~~%5Cstackrel%7B%5Ctextit%7Barea%20of%202%20triangles%20below%7D%7D%7B2%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%282.5%29%286%29%20%5Cright%5D%7D%5Cimplies%206.25%2B15%5Cimplies%2021.25)
