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Vikentia [17]
3 years ago
14

If you are investing money, would you rather deposit into an account that earns simple interest or compound interest?

Mathematics
2 answers:
Citrus2011 [14]3 years ago
6 0

Answer:

Compound interest.

Step-by-step explanation:

-In compound interest, your money grows more rapidly as the interest already earned also earns a further interest over time for the term of the investment.

-While in simple interest, only the principal amount invested will earn interest over the investment's term>

#Let's say you invest $100 each in A(earning simple interest) and B(earning compound interest), take the term of  both investments to be 5 yrs with both subject to an annual rate of 10%:

-Get the investment amounts in both accounts at the end of 5 yrs:

\#A\\A=PRT+P\\\\=100\times 0.1\times 5+100\\\\=150\\\\\#B\\A=P(1+i)^n\\\\=100(1+0.1)^5\\\\=161.05

Hence, you notice that your earn more in compound interest keeping all other conditions similar.

bonufazy [111]3 years ago
3 0

Answer:

compound interest

i took the test, and i got this one correct.

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what the picture?

Step-by-step explanation:

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The diagram below is NOT drawn to scale. Are lines j and k parallel? How can you tell?
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3 0
3 years ago
Answer the question in the picture
nekit [7.7K]

Recall the angle sum identities:

\sin(x+y)=\sin x\cos y+\cos x\sin y

\cos(x+y)=\cos x\cos y-\sin x\sin y

Now,

\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}

Divide through numerator and denominator by \cos x\cos y to get

\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}

Next, we use the fact that x,y lie in the first quadrant to determine that

\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2

\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}

So we then have

\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}

\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1

Finally,

\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73

4 0
3 years ago
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