Answer:

Step-by-step explanation:
<u>Simplfy in Algebra
</u>
We have the following expression

Simplifying like factors in the denominator and numerator

All the factors are perfect squares except
, thus we rewrite:

Taking the square root of all the perfect square factors:

Answer:
1. A: 0.25; B: 0.03; C: 1.41; D: -0.28
2. A: 0.39; B: 0.06; C: 40.30; D: 21.81
Step-by-step explanation:
For CDF lookups, we used the Excel NORMDIST(x, mean, stdev, TRUE) function. For inverse CDF lookups, we used the NORMINV(x, mean, stdev) function.
Each of these functions works with the area under the curve from -∞ to x, so for cases where we're interested in the upper tail, we subtract the probability from 1, or subtract the x value from twice the mean.
For question 1, we computed the Z values in each case. The NORMDIST function works directly with x, mean, and standard deviation, so does not need the z value.
Answer:
Plug in your variables value.
Step-by-step explanation:
For your case, plug in -6 into the y slot
12(5+2y)= -(6-9y) +4y
12(5+2x-6)= -(6-9x-6)+4y see if that works!
Answer:
20 Quarters = $5, add 10 dimes to make it $6 and 10 x 2 = 20 so your answer is 10 dimes
Step-by-step explanation:
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,
(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)




