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lions [1.4K]
3 years ago
15

Michael opened a savings account at an annual interest rate of 5%. At the end of 3 years, the account balance is $4630.50. If Mi

chael did not add any other amounts to this account, how much was his initial deposit?
Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
3 0

Answer:

$4026.52

Step-by-step explanation:

Given data

rate= 5%

time= 3years

A=$4630.50

P=?

we know that

A=P(1+rt)

$4630.50=P(1+0.05*3)

$4630.50=P(1+0.15)

$4630.50=P(1.15)

P=4630.50/1.15

P=$4026.52

The intitial amount is $4026.52

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Without using trignometery tables find value of , cos70/sin20+cos57.cosec33 - 2cos60
Helen [10]
Here  is some thoughts on this:

cos 57  = sin (90-57) = sin 33 

so cos 57 . cosec 33 = cos 57 / sin 33 = sin 33 / sin33 = 1

2 cos 60 = 2 * 1/2  = 1

so the last 2 parts work out to 0 

now we have to find cos 70 / sin 20

sin 20 = cos 70  so this comes to 1


so finally the answer is 1
4 0
3 years ago
someone please answer this but no links because i have a school computer and it blocks the links so please dont send links
Otrada [13]

Answer:

just typed the whole in mathaway.com

Step-by-step explanation:

it's a better app for quadratic equation and you can use it on a school computer

7 0
3 years ago
Is (10,4),(-2,4),(-1,1),(5,6) a function?
lys-0071 [83]
No it is not a function.
7 0
1 year ago
If f(x) = 3x + 10 and g(x) = 2x - 4, find (f- g)(x).
densk [106]

Answer:

<em>(f-g)(x)=x+14</em>

Step-by-step explanation:

<u>Operations with Functions</u>

Given two functions f(x) and g(x), the following operations are defined:

(f+g)(x)=f(x)+g(x)

(f-g)(x)=f(x)-g(x)

(f*g)(x)=f(x)*g(x)

(f/g)(x)=f(x)/g(x)

We have f(x)=3x+10 and g(x)=2x-4. Use the second formula to find

(f-g)(x)=(3x+10)-(2x-4)

Operating:

(f-g)(x)=3x+10-2x+4

Joining like terms:

(f-g)(x)=x+14

5 0
3 years ago
The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t),
pishuonlain [190]

Answer:

S(3)=22

Step-by-step explanation:

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t).

\dfrac{dS}{dt}=k(30-S(t))\\ \dfrac{dS}{dt}+kS(t)=30k\\$The integrating factor: e^{\int k dt}=e^{kt}\\$Multiply all through by the integrating factor\\ \dfrac{dS}{dt}e^{kt}+kS(t)e^{kt}=30ke^{kt}

(Se^{kt})'=30ke^{kt} dt\\$Integrate both sides\\ Se^{kt}=\dfrac{30ke^{kt}}{k}+C$ (C a constant of integration)\\Se^{kt}=30e^{kt}+C\\$Divide both sides by e^{kt}\\S(t)=30+Ce^{-kt}

When t=0, S(t)=15

15=30+Ce^{-k*0}\\C=15-30\\C=-15

When t = 2, S(t)=20

20=30-15e^{-2k}\\20-30=-15e^{-2k}\\-10=-15e^{-2k}\\e^{-2k}=\dfrac23\\$Take the natural log of both sides$\\-2k=\ln \dfrac23\\k=-\dfrac{\ln(2/3)}{2}

Therefore:

S(t)=30-15e^{\frac{\ln(2/3)}{2}t}\\$When t=3$\\S(t)=30-15e^{\frac{\ln(2/3)}{2} \times 3}\\S(3)=21.8 \approx 22

8 0
3 years ago
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