I need help with this <br> (-7)(-5) =

Tony wants to save $10,000 in 6 years. Assuming a 4% interest rate, what is the minimum he must save each month to reach his goa

arlik [135]

We have been given that Tony wants to save $10000 in 6 years.

That means future value S = $10000

Time t= 6 years

Interest rate = 4% yearly = 0.04 yearly

n=12 months per year

Now we have to find monthly payment to recieve $10000 in 6 years. so we need to apply monthly payment formula which is

$S=R(\frac{(1+\frac{r}{n})^{nt}-1}{\frac{r}{n}})$

$10000=R(\frac{(1+\frac{0.04}{12})^{12*6}-1}{\frac{0.04}{12}})$

$10000=R(\frac{(1+0.00333333333333)^{72}-1}{0.00333333333333})$

$10000=R(\frac{(1.00333333333333)^{72}-1}{0.00333333333333})$

$10000=R(\frac{1.27074187908-1}{0.00333333333333})$

$10000=R(\frac{0.27074187908}{0.00333333333333})$

$10000=R(81.2225637241)$

$\frac{10000}{81.2225637241}=R$

$123.11849739=R$

which is approx $123.

Hence final answer is C) $123.

4 0

3 years ago

There are a total of 100 balloons. If the ratio of yellow balloons to blue balloons is 8 to 12, how many of them are yellow

Ivanshal [37]

If you have 8 yellow balloons for every 12 blue balloons, then if you have 20 balloons in total, 8 out of 20 of those balloons will be yellow. Then you can set up this ratio:

$\frac{8yellow}{20total} = \frac{Xyellow}{100total}$

Then, you can use either cross multiplication or multiply the numerator and denominator by the same number to help solve for x. Personally, I prefer cross multiplication, so that's what I'll show you here:

$\frac{8yellow}{20total} = \frac{Xyellow}{100total} \\ 8yellow*100total=20total*Xyellow$

Finally, solve for x:

$8yellow*100total=20total*Xyellow \\ 8*100=20*x \\ 800=20x \\ 800/20=20x/20 \\ x=40yellow$

If you had 100 balloons, you will have 40 yellow balloons.

Hope this helps!!

6 0

2 years ago

A rectangle is inside a circle with a 5 cm radius.

o-na [289]

Answer:

Do it your self

Step-by-step explanation:

Gross

8 0

2 years ago

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0

3 years ago

50 POINTS

Zolol [24]

Answer:

90

Step-by-step explanation:

In a box plot, the point farthest to the line is the greatest or maximum number of the problem(in this case E). You look for the greatest number out of them all and that would be your answer(90).

Can i pretty please with a cherry on top get brainliest??

:>

6 0

2 years ago

Read 2 more answers

I need help with this (-7)(-5) =