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Blababa [14]
3 years ago
10

How much must be deposited today into the following account in order to have $ 65,000 in 8 years for a down payment on a​ house?

Assume no additional deposits are made.
An account with annual compounding and an APR of 4​%
​$
nothing should be deposited today.
​(Do not round until the final answer. Then round to the nearest cent as​ needed.)
Mathematics
1 answer:
svp [43]3 years ago
3 0

Answer:

\$47,494.86

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

t=8\ years\\ A=\$65,000\\ r=4\%=4/100=0.04\\n=1  

substitute in the formula above  

65,000=P(1+\frac{0.04}{1})^{1*8}  

65,000=P(1.04)^{8}  

Solve for P

P=65,000/(1.04)^{8}  

P=\$47,494.86

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Answer:

Step-by-step explanation:

$1.11

7 0
3 years ago
Write a sine and cosine function that models the data in the table. I need steps to both for a, b, c, and d.
andrezito [222]

Answer(s):

\displaystyle y = -29sin\:(\frac{\pi}{6}x + \frac{\pi}{2}) + 44\frac{1}{2} \\ y = -29cos\:\frac{\pi}{6}x + 44\frac{1}{2}

Step-by-step explanation:

\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 44\frac{1}{2} \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-3} \hookrightarrow \frac{-\frac{\pi}{2}}{\frac{\pi}{6}} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{12} \hookrightarrow \frac{2}{\frac{\pi}{6}}\pi \\ Amplitude \hookrightarrow 29

<em>OR</em>

\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 44\frac{1}{2} \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{12} \hookrightarrow \frac{2}{\frac{\pi}{6}}\pi \\ Amplitude \hookrightarrow 29

You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of <em>sine</em>, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the centre photograph displays the trigonometric graph of \displaystyle y = -29sin\:\frac{\pi}{6}x + 44\frac{1}{2},in which you need to replase "cosine" with "sine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the cosine graph [photograph on the left], accourding to the <u>horisontal shift formula</u> above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the <em>sine</em> graph [centre photograph] is shifted \displaystyle 3\:unitsto the right, which means that in order to match the <em>cosine</em> graph [photograph on the left], we need to shift the graph BACKWARD \displaystyle 3\:units,which means the C-term will be negative, and by perfourming your calculations, you will arrive at \displaystyle \boxed{3} = \frac{-\frac{\pi}{2}}{\frac{\pi}{6}}.So, the sine graph of the cosine graph, accourding to the horisontal shift, is \displaystyle y = -29sin\:(\frac{\pi}{6}x + \frac{\pi}{2}) + 44\frac{1}{2}.Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph WILL hit \displaystyle [12, 15\frac{1}{2}],from there to the y-intercept of \displaystyle [0, 15\frac{1}{2}],they are obviously \displaystyle 12\:unitsapart, telling you that the period of the graph is \displaystyle 12.Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the <em>midline</em>. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at \displaystyle y = 44\frac{1}{2},in which each crest is extended <em>twenty-nine units</em> beyond the midline, hence, your amplitude. Now, there is one more piese of information you should know -- the cosine graph in the photograph farthest to the right is the OPPOCITE of the cosine graph in the photograph farthest to the left, and the reason for this is because of the <em>negative</em> inserted in front of the amplitude value. Whenever you insert a negative in front of the amplitude value of <em>any</em> trigonometric equation, the whole graph reflects over the <em>midline</em>. Keep this in mind moving forward. Now, with all that being said, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

I am delighted to assist you at any time.

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2 years ago
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strojnjashka [21]

area of quadrilateral (rhombus)=<u>4square units</u>

<u>:</u>

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8 0
3 years ago
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navik [9.2K]
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A gardener has a rose garden that measures 30 feet by 20 feet. He wants to put a uniform border of pine bark around the outside
babunello [35]

Answer:

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Step-by-step explanation:

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The area of the garden alone, without the border, is:

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