In a 45-45-90 triangle, the two legs are congruent. Let's call them x. The hypotenuse is equal to 1 as we're using the unit circle. The hypotenuse of the triangle is the same as the radius of the unit circle.
a = x
b = x
c = 1
Use those values in the Pythagorean theorem to solve for x.
a^2 + b^2 = c^2
x^2 + x^2 = 1^2
2x^2 = 1
x^2 = 1/2
x = sqrt( 1/2 )
x = sqrt(1)/sqrt(2)
x = 1/sqrt(2)
x = sqrt(2)/2 ... rationalizing the denominator
So this right triangle has legs that are sqrt(2)/2 units long. Once we know the legs of the triangle, we can divide them over the hypotenuse to find the sine and cosine values.
sin(angle) = opposite/hypotenuse
sin(45) = (sqrt(2)/2) / 1
sin(45) = sqrt(2)/2
and
cos(angle) = adjacent/hypotenuse
cos(45) = (sqrt(2)/2) / 1
cos(45) = sqrt(2)/2
------------------------------------------------------
For a 30-60-90 triangle, we would have
a = 1
b = x
c = 2
so,
a^2+b^2 = c^2
1^2+x^2 = 2^2
1+x^2 = 4
x^2 = 4-1
x^2 = 3
x = sqrt(3)
The missing leg is sqrt(3) units long.
Once we know the three sides of the 30-60-90 triangle, you should be able to see that
sin(30) = 1/2
sin(60) = sqrt(3)/2
cos(30) = sqrt(3)/2
cos(60) = 1/2
<h2>
Hello!</h2>
The answer is:
It will take 42.35 minutes to weed the garden together.
<h2>
Why?</h2>
To solve the problem, we need to use the given information about the rate for both Laura and her husband. We know that she can weed the garden in 1 hour and 20 minutes (80 minutes) and her husband can weed it in 1 hour and 30 minutes (90 minutes), so we need to combine both's work and calculate how much time it will take to weed the garden together.
So, calculating we have:
Laura's rate:

Husband's rate:

Now, writing the equation we have:








Hence, we have that it will take 42.35 minutes to weed the garden working together.
Have a nice day!
X^2 + 9x -10 = (x -1)(x +10) = (x +(-1)) (x + 10)
p = -1 and q = 10
answer is A.
Answer:

Step-by-step explanation:

Applying the Laplace transform:
![\mathcal{L}[y'']+5\mathcal{L}[y']+4\mathcal{L}[y']=0](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%27%5D%2B5%5Cmathcal%7BL%7D%5By%27%5D%2B4%5Cmathcal%7BL%7D%5By%27%5D%3D0)
With the formulas:
![\mathcal{L}[y'']=s^2\mathcal{L}[y]-y(0)s-y'(0)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%27%5D%3Ds%5E2%5Cmathcal%7BL%7D%5By%5D-y%280%29s-y%27%280%29)
![\mathcal{L}[y']=s\mathcal{L}[y]-y(0)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%5D%3Ds%5Cmathcal%7BL%7D%5By%5D-y%280%29)
![\mathcal{L}[x]=L](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5Bx%5D%3DL)

Solving for 




Apply the inverse Laplace transform with this formula:
![\mathcal{L}^{-1}[\frac1{s-a}]=e^{at}](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5E%7B-1%7D%5B%5Cfrac1%7Bs-a%7D%5D%3De%5E%7Bat%7D)
![y=3\mathcal{L}^{-1}[\frac1{s+4}]=3e^{-4t}](https://tex.z-dn.net/?f=y%3D3%5Cmathcal%7BL%7D%5E%7B-1%7D%5B%5Cfrac1%7Bs%2B4%7D%5D%3D3e%5E%7B-4t%7D)