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otez555 [7]
3 years ago
7

how much would $200 invested at 6% interest compounded annually be worth after 5 years? round your answer to the nearest cent.

Mathematics
2 answers:
nika2105 [10]3 years ago
5 0
Formula for compound interest

Amount, A = P(1 + r)ⁿ

P= Principal = $200,  r = rate = 6% = 0.06,  n = number of years = 5

Amount, A  = 200(1 + 0.06)⁵

                     = 200*(1.06)⁵

                     = 267.645..

So it would be worth ≈ $267.65 after 5 years of compounding.
Semenov [28]3 years ago
3 0
F=P(1+\frac{r}{n})^{nt}
F = future amount (what you're looking for)
P = present amount = 200
r = interest rate = 0.06
n = number of compoundings a year = 1
t = number of years = 5
F=200(1+\frac{0.06}{1})^{1\cdot 5}
F = $267.65
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Please help steps thx
ExtremeBDS [4]

Answer:

240 cubic metres

Step-by-step explanation:

I'm not really sure for Q17 but I do know Q18.

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hope it helps!!!

8 0
2 years ago
Over which interval is the graph of f(x) = one-halfx2 + 5x + 6 increasing?
kaheart [24]

Answer:

\large\boxed{x\in[-5,\ \infty)}

Step-by-step explanation:

f(x)=\dfrac{1}{2}x^2+5x+6\\\\\text{The coeficient of}\ x^2\ \text{is the positive number. Therefore the parabola is op}\text{en up}.\\\\\text{If a parabola op}\text{en up, then the graph increasing in interval}\ (h,\ \infty),\\\text{and decreasing in interval}\ (-\infty,\ h).\ \text{Where}\ h\ \text{is a first coordinate of a vertex.}\\\\\text{For}\ y=ax^2+bx+c,\ h=\dfrac{-b}{2a}.\\\\\text{We have}\ a=\dfrac{1}{2}\ \text{and}\ b=5.\ \text{Substitute:}\\\\h=\dfrac{-5}{2\left(\frac{1}{2}\right)}=-\cdot\dfrac{5}{1}=-5.

8 0
3 years ago
Read 2 more answers
Convert y + 3 = -3(x + 5) to standard form.
emmasim [6.3K]
Hi there!

y  + 3 =  - 3(x + 5)
First we need to work out the parenthesis, which can for instance be done using rainbow technique.

y + 3 =  - 3x  -  15
Now subtract 3 from both sides.

y =  - 3x  - 18
And finally add 3x to both sides.

3x + y =  - 18
The 4th answer choice is correct.



6 0
3 years ago
What is the equation of the quadratic graph with a focus of (1 ,3) and a directrix of y=1?
zimovet [89]
Check the picture below.

now, we know the directrix is at y = 1, and the focus point is at 1,3, well, notice the picture, the distance between those fellows is just 2 units.

the vertex is half-way between those fellows, therefore, the vertex will be at 1,2.

the distance "p", from the vertex to either the directrix or focus, is really just 1 unit.  Since the focus point is above the directrix, is a vertical parabola, and it opens upwards, like in the picture, and since it opens up, the "p" value is positive, or +1.

\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
4p(x- h)=(y- k)^2
\\\\
\boxed{4p(y- k)=(x- h)^2}
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ( h, k)\\\\
 p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
h=1\\
k=2\\
p=1
\end{cases}\implies 4(1)(y-2)=(x-1)^2\implies 4(y-2)=(x-1)^2
\\\\\\
y-2=\cfrac{1}{4}(x-1)^2\implies y=\cfrac{1}{4}(x-1)^2+2

4 0
3 years ago
Three vertices of a rectangle have coordinates (3,4), (5,−4), and (−7,−7). Determine the fourth vertex of the rectangle and then
Ket [755]
The fourth vertex would be at point (-9,1).
The slope from (5,-4) to (3,4) is -8/2 or -4.
So I did that with (-7,-7) to the fourth vertex to get the answer.
7 0
3 years ago
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