Answer:
7:13
Step-by-step explanation:
There are 14 girls and 12 boys, so there are 26 students in total. Since we are looking for the ratio of girls to students, the ratio will be 14:26. To simplify it, we will divide 14 and 26 by their greatest common factor: 2. We will get the ratio of 7:13.
I hope this helped! :)
Number of spaces would be number of squares - 1
You would have 12 squares and 11 spaces.
Multiply number of each by their size then add the two together:
12 x 7 1/4 = 87 inches
11 x 3 5/16 = 36 7/16 inches
Total = 87 + 36 7/16 = 123 7/16 inches
Answer: 123 7/16 inches
Prove we are to prove 4(coshx)^3 - 3(coshx) we are asked to prove 4(coshx)^3 - 3(coshx) to be equal to cosh 3x
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2 = e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2 = e^(3x) /2 + e^(-3x) /2 = cosh(3x) = LHS Since y = cosh x satisfies the equation if we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work.
i.e. e^(3x)/2 + e^(-3x)/2 = 2
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3, Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3,
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2 = (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3) to be equ
= 4(e^x+e^(-x))^3/8 - 3(e^x+e^(-x))/2
= e^3x /2 +3e^x /2 + 3e^(-x) /2 + e^(-3x) /2 - 3(e^x+e^(-x))/2
= e^(3x) /2 + e^(-3x) /2
= cosh(3x)
= LHS
<span>Therefore, because y = cosh x satisfies the equation IF we replace the "2" with cosh3x, we require cosh 3x = 2 for the solution to work. </span>
i.e. e^(3x)/2 + e^(-3x)/2 = 2
Setting e^(3x) = u, we have u^2 + 1 - 4u = 0
u = (4 + sqrt(12)) / 2 = 2 + sqrt(3), so x = ln((2+sqrt(3))/2) /3,
Or u = (4 - sqrt(12)) / 2 = 2 - sqrt(3), so x = ln((2-sqrt(3))/2) /3,
Therefore, y = cosh x = e^(ln((2+sqrt(3))/2) /3) /2 + e^(-ln((2+sqrt(3))/2) /3) /2
= (2+sqrt(3))^(1/3) / 2 + (-2-sqrt(3))^(1/3)
Answer:
(a). y'(1)=0 and y'(2) = 3
(b). 
(c). 
Step-by-step explanation:
(a). Let the curve is,
So the stationary point or the critical point of the differential function of a single real variable , f(x) is the value 
which lies in the domain of f where the derivative is 0.
Therefore, y'(1)=0
Also given that the derivative of the function y(t) is 3 at t = 2.
Therefore, y'(2) = 3.
(b).
Given function,
Differentiating the above equation with respect to x, we get
![y'(t)=\frac{d}{dt}[k \sin (bt^2)]\\ y'(t)=k\frac{d}{dt}[\sin (bt^2)]](https://tex.z-dn.net/?f=y%27%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Bk%20%5Csin%20%28bt%5E2%29%5D%5C%5C%20y%27%28t%29%3Dk%5Cfrac%7Bd%7D%7Bdt%7D%5B%5Csin%20%28bt%5E2%29%5D)
Applying chain rule,
![y'(t)=k \cos (bt^2)(\frac{d}{dt}[bt^2])\\ y'(t)=k\cos(bt^2)(b2t)\\ y'(t)= kb2t\cos(bt^2)](https://tex.z-dn.net/?f=y%27%28t%29%3Dk%20%5Ccos%20%28bt%5E2%29%28%5Cfrac%7Bd%7D%7Bdt%7D%5Bbt%5E2%5D%29%5C%5C%20y%27%28t%29%3Dk%5Ccos%28bt%5E2%29%28b2t%29%5C%5C%20y%27%28t%29%3D%20kb2t%5Ccos%28bt%5E2%29)
(c).
Finding the exact values of k and b.
As per the above parts in (a) and (b), the initial conditions are
y'(1) = 0 and y'(2) = 3
And the equations were
Now putting the initial conditions in the equation y'(1)=0
2kbcos(b) = 0
cos b = 0 (Since, k and b cannot be zero)
And
y'(2) = 3
![$\therefore kb2(2)\cos [b(2)^2]=3$](https://tex.z-dn.net/?f=%24%5Ctherefore%20kb2%282%29%5Ccos%20%5Bb%282%29%5E2%5D%3D3%24)
