Answer:
<u>Apply the property above:</u>
- 1. ln 618 = p ⇔ e^p = 618
- 2. ln q = 2 ⇔ e^2 = q
- 3. ln 100 = t ⇔ e^t = 100
- 4. ln (e^3) = 3 ⇔ e^3 = e^3
Answer:
3
Step-by-step explanation:
Michael must find the smallest number possible that can divide 60 and 45 then take the two quotients after dividing the 2 types of cookies by the smallest number possible.
2 is not possible as 45 is an odd number
3 can divide both 60 and 45
60/3 = 20 oatmeal cookies in each container.
45/3 = 15 chocolate chip cookies in each container.
Answer:
Let R = radius of big circle
r=radius of small circle
A1=Area of small circle
A2=Area of big circle
R=9r
A1=50cm^2
A1=pi*r*2
A2=pi*R^2
A2=pi*(9r)^2
A2=81*pi*r^2
A2=81*A1
A2=4050cm^2
Hence the big circle has area of 4050cm^2
Step-by-step explanation:
Answers:
a = -6/37
b = -1/37
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Explanation:
Let's start things off by computing the derivatives we'll need

Apply substitution to get

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.
The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero
a-6b = 0
a = 6b
At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)
-6a -b = 1
-6(6b) - b = 1 .... plug in a = 6b
-36b - b = 1
-37b = 1
b = -1/37
Use this to find 'a'
a = 6b
a = 6(-1/37)
a = -6/37