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vlabodo [156]
3 years ago
5

Please help with 1 and 2

Mathematics
1 answer:
ZanzabumX [31]3 years ago
5 0

Answer:

1. B. a=14, b=2

2. D. a=1, b=1.4

Step-by-step explanation:

The exponential growth function can be represented as

y=a\cdot b^x,

where b is the growth factor.

1. When the function has equation

g(x)=14\cdot 2^x,

then

a=14,\\ \\b=2

The initial amount is the value of the function at x=0:

g(0)=14\cdot 2^0=14\cdot 1=14

The growth factor is b=2

2. When the function has equation

f(t)=1.4^t,

then

a=1,\\ \\b=1.4

The initial amount is the value of the function at t=0:

f(0)=1.4^0=1

The growth factor is b=1.4

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sergiy2304 [10]
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3 years ago
If tanA+sinA=m and tanA-sinA=n.prove that m^2-n^2=4√mn
mash [69]

Let a=\tan A and b=\sin A. Then

m^2-n^2=(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab

\implies m^2-n^2=4\tan A\sin A

and

mn=(a+b)(a-b)=a^2-b^2

\implies4\sqrt{mn}=4\sqrt{\tan^2A-\sin^2A}

The expression under the square root can be rewritten as

\tan^2A-\sin^2A=\dfrac{\sin^2A}{\cos^2A}-\sin^2A=\sin^2A\left(\dfrac1{\cos^2A}-1\right)=\sin^2A(\sec^2A-1)

Recall that

\sin^2A+\cos^2A=1\implies\tan^2A+1=\sec^2A

so that

\tan^2A-\sin^2A=\sin^2A\tan^2A

and assuming \sin A>0 and \tan A>0, we end up with

4\sqrt{\tan^2A-\sin^2A}=4\tan A\sin A

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m^2-n^2=4\sqrt{mn}

as required.

5 0
3 years ago
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