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Kay [80]
3 years ago
9

Suppose that $2000 is invested at a rate of 4%, compounded semiannually. Assuming that no withdrawals are made, find the total a

mount after 5 years. Do not round any intermediate computations, and round your answer to the nearest cent. ​
Mathematics
1 answer:
creativ13 [48]3 years ago
5 0

Answer:

Step-by-step explanation:

A = P(1+r/n)^ nt

A = 2000(1+.04/2)^(5*2)

A = 2000(1.02)^{10} = $2437.99

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Please Help). Use the figure below to answer the questions. A) Describe in words a sequenc of transformations that maps □ABC to
lesya [120]

Answer:

See below

Step-by-step explanation:

The initial ordered-pairs are (x, y)

We have a rotation of 90 degrees counterclockwise with respect to origin

Note. Previously the points were

A(-3, 4), B(-3, 0), C(-1, 3)

After the rotation, we have

A(-4, -3), B(0, -3), C(-3, -1)

Thus, (x, y) \rightarrow (-y, x)

Then shifting horizontally to the right 2 units, we get ΔA'B'C'

Thus,

(x, y) \rightarrow (-y, x) \rightarrow (x+2, y)

7 0
3 years ago
Solve for b: 2(b + 3) = b – 4
Reil [10]

Answer:

b = -10

Step-by-step explanation:

2(b + 3) = b – 4

2b + 6 = b - 4

-b             -b

b + 6 = -4

     -6    -6

b = -10

8 0
3 years ago
Read 2 more answers
How many mL are there in 3,700 microliters?
rewona [7]

Answer:

3.7 mL

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
-4(3 - 2x) + 2x = 2x - 8
aniked [119]
C 1/2 is the correct answer
6 0
3 years ago
Let v1 =(-6,4) and v2=(-3,6) compute the following what i sthe angle between v1 and v2
Agata [3.3K]
\bf ~~~~~~~~~~~~\textit{angle between two vectors }
\\\\
cos(\theta)=\cfrac{\stackrel{\textit{dot product}}{u \cdot v}}{\stackrel{\textit{magnitude product}}{||u||~||v||}} \implies 
\measuredangle \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||~||v||}\right)\\\\
-------------------------------

\bf \begin{cases}
v1=\ \textless \ -6,4\ \textgreater \ \\
v2=\ \textless \ -3,6\ \textgreater \ \\
------------\\
v1\cdot v2=(-6\cdot -3)+(4\cdot 6)\\
\qquad \qquad 42\\
||v1||=\sqrt{(-6)^2+4^2}\\
\qquad \sqrt{52}\\
||v2||=\sqrt{(-3)^2+6^2}\\
\qquad \sqrt{45}
\end{cases}\implies \measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{52}\cdot \sqrt{45}} \right)
\\\\\\
\measuredangle \theta =cos^{-1}\left(  \cfrac{42}{\sqrt{2340}} \right)\implies \measuredangle \theta \approx 29.74488129694^o
8 0
3 years ago
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