The first thing we are going to do to graph our equation is solve them for
:
For our first equation:
For our second equation:
Now, we just need to use the graphing utility to graph our equations:
We can conclude that the graph of the equations is:
Congrats on making it to integrals!
Basically you need to integral your function because integral rate in respect to time = total amount.
Also your bounts are (2001-1990,2006-1990)=(11,16)
Thus we take integral like:
int(11,16)(928.5e^(0.0249x))=
(11,16)928/0.0249e^(0.0249(x))-928/0.0249e^(0.0249(x))
(You can check this by taking its derivative and seeing if you get the original function)
928/0.0249e^(0.0249(16))-928/0.0249e^(0.0249(11))=6498.1
Answer:
The answer is C
Step-by-step explanation:
The rewritten form of the expression given in the task content in which case, the properties of the logarithm are used is; log2(z³/y²) + log9(y⁴x¹²).
<h3>What is the equivalent expression to the logarithmic expression given in the task content?</h3>
It follows from the task content that the logarithmic expression given in the task content is;
log2z + 2log2z + 4log9y + 12log9x - 2log2y
In which case, all terms are to their respective logarithmic bases.
Hence, it follows that; we have;
log2z³ - log2y² + log9y⁴ + log9x¹²
= log2(z³/y²) + log9(y⁴x¹²)
Read more on logarithm laws;
brainly.com/question/10208274
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1)
LHS = cot(a/2) - tan(a/2)
= (1 - tan^2(a/2))/tan(a/2)
= (2-sec^2(a/2))/tan(a/2)
= 2cot(a/2) - cosec(a/2)sec(a/2)
= 2(1+cos(a))/sin(a) - 1/(cos(a/2)sin(a/2))
= 2 (1+cos(a))/sin(a) - 2/sin(a)) (product to sums)
= 2[(1+cos(a) -1)/sin(a)]
=2cot a
= RHS
2.
LHS = cot(b/2) + tan(b/2)
= [1 + tan^2(b/2)]/tan(b/2)
= sec^2(b/2)/tan(b/2)
= 1/sin(b/2)cos(b/2)
using product to sums
= 2/sin(b)
= 2cosec(b)
= RHS
