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Ostrovityanka [42]

1 year ago

8

Consider a $2 principal investment with a 10% annual simple interest rate. Enter the simple interest equation that represents th

e situation. Let t represent the time in years. A = __

1 answer:

PtichkaEL [24]1 year ago

5 0

The simple interest equation that represent the situation is A = t/5

<h3>What is simple interest?</h3>

Simple Interest (S.I.) is the method of calculating the interest amount for a particular principal amount of money at some rate of interest. For example, when a person takes a loan of 500 dollars, at a rate of 10 per annum for two years, the person's interest for two years will be S.I. on the borrowed money.

Principal(P) borrowed = $2

Rate(R) in percentage = 10

time in years = t

Simple interest(A) = P X R X t / 100

A = 2 X 10 X t/ 100

After reducing 20/100 to the lowest term

A = t/5

In conclusion the simple interest(A) = t/5

Learn more about simple interest: brainly.com/question/20690803

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If d=3 what is the answer? 7b to the power of three +10

Sati [7]

Answer:

199 if it's 7b^3 + 10

9271 if it's (7b)^3 + 10

Step-by-step explanation:

You just substitute the 3 in for d. If your expression was 7b^3 + 10, then only b is being raised to 3.

So 3^3 is 27, 27*7 is 189, 189+10 is 199.

If the expression was (7b)^3 + 10, then first you multiply 7 and 3 to get 21. Then you raise 21 to the power of 3, which is 9261. Then you add 10, which is 9271.

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3 years ago

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A 9 gram sample of a substance that's used for drug research has a k-value of 0.1459

snow_tiger [21]

<span>N = N0 e^(-kt)

for half life N/N0 = 1/2 and k = 0.1459

1/2 = e^(-0.1459t)

=> -0.1459 t = ln(0.5)

t = - ln(0.5) /0.1459

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6 0

4 years ago

Find the general solution of<br><br> dz/dt=5ze^(sint)cost+2e^(sint)cost

gizmo_the_mogwai [7]

Answer:

The general solution of given differential equation is $\frac{1}{5}ln|5z+2|=e^{\sin t}+C$ .

Step-by-step explanation:

The given differential equation is

$\frac{dz}{dt}=5ze^{\sin t}\cos t+2e^{\sin t}\cos t$

Taking out common factors.

$\frac{dz}{dt}=(5z+2)e^{\sin t}\cos t$

Using variable separable method, we get

$\frac{dz}{5z+2}=e^{\sin t}\cos t dt$

Integrate both sides.

$\int\frac{dz}{5z+2}=\int e^{\sin t}\cos t dt$

$I_1=I_2$ .... (1)

First solve LHS,

$I_1=\int\frac{dz}{5z+2}$

Substitute $5z+2=u$

$5\frac{dz}{du}=1$

$dz=\frac{1}{5}du$

$I_1=\frac{1}{5}\int\frac{du}{u}$

$I_1=\frac{1}{5}ln|u|+C_1$

$I_1=\frac{1}{5}ln|5z+2|+C_1$

Now, solve RHS,

$I_2=\int e^{\sin t}\cos t dt$

Substitute $\sin t=v$ ,

$\cos t\frac{dt}{dv}=1$

$\cos tdt=dv$

$I_2=\int e^{v}dv$

$I_2=e^{v}+C_2$

$I_2=e^{\sin t}+C_2$

Subtitle the values of I₁ and I₂ in equation (1).

$\frac{1}{5}ln|5z+2|+C_1=e^{\sin t}+C_2$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C_2-C_1$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C$

Therefore the general solution of given differential equation is $\frac{1}{5}ln|5z+2|=e^{\sin t}+C$ .

4 0

4 years ago