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AVprozaik [17]
2 years ago
6

Suppose that when your friend was born, your friends parents deposited $2000 in an account paying 5.9% interest. What will the a

ccount balance be after 15 years?
Mathematics
1 answer:
irina1246 [14]2 years ago
3 0

Answer:

FV= $4,725.74

Step-by-step explanation:

Giving the following information:

Initial investment (PV)= $2,000

Number of periods (n)= 15 years

Interest rate (i)= 5.9% = 0.059

<u>To calculate the Future Value, we need to use the following formula:</u>

<u></u>

FV= PV*(1 + i)^n

FV= 2,000*(1.059^15)

FV= $4,725.74

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Urgent!!! please help!!
serious [3.7K]

Answer:

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7 0
3 years ago
Find the distance between the points given.<br><br> (-1, -1) and (1, 3)<br><br> √5<br> √(17)<br> 2√5
Pani-rosa [81]

Answer:

2√5

Step-by-step explanation:

d = √(x2 - x1)² + (y2 - y1)²

= √[1 - (-1)]² + [3 - (-1)]

= √(2)² + (4)²

= √(4) + (16)

= √20

= 2√5

5 0
2 years ago
Yolanda's club has 50 members. It's rules require that 60% of them must be present for any vote. At least how many members must
HACTEHA [7]
If you do 60% then you put the percent two times over the the left so 0.60 then you would do 0.60 times 50 so that would be 30. 30 of the members must vote. Hopefully that helped?
7 0
3 years ago
Help with matrices please? Any wrong/not applicable answers will be reported and BLOCKED
marin [14]

m x H = \left[\begin{array}{ccc}-25&37.5&-12.5\\\9\end{array}\right]

Step-by-step explanation:

Step 1; Multiply 5 with this matrix  \left[\begin{array}{ccc}-1&2\\4&8\\\end{array}\right] and we get a matrix \left[\begin{array}{ccc}-5&10\\20&40\\\end{array}\right]

Multiply the fraction  \frac{2}{5} with the matrix  \left[\begin{array}{ccc}-1&2\\4&8\\\end{array}\right] and we get \left[\begin{array}{ccc}-\frac{2m}{5} &\frac{4m}{5} \\\frac{8m}{5} &\frac{16m}{5} \\\end{array}\right]

Step2; Now equate corresponding values of the matrices with each other.

-5 = \frac{-2m}{5} and so on. By equating we get the value of m as \frac{25}{2}

Step 3; Add the matrices to get the value of matrix m.

Adding the three matrices on the RHS we get  \left[\begin{array}{ccc}2&9&-9\\\end{array}\right].

Step 4; Adding the matrices on the LHS we get the resulting matrix as H +

\left[\begin{array}{ccc}4&6&-8\\\9\end{array}\right]. Equating the matrices from step 3 and 4 we get the value of H as \left[\begin{array}{ccc}-2&3&-1\\\9\end{array}\right]

Step 5; Now to find the value of m x H we need to multiply the value of \frac{25}{2} with the matrix \left[\begin{array}{ccc}-2&3&-1\\\9\end{array}\right]

Step 6; Multiplying we get the matrix m x H = [ -25  \frac{75}{2}  \frac{-25}{2} ]

8 0
2 years ago
Anyone know the answer to this algebra problem?
ikadub [295]

Answer:  \bold{a=1\qquad b=\dfrac{1}{16}\qquad c=\dfrac{1}{64}\qquad d=1\qquad e=\dfrac{4}{9}\qquad f=\dfrac{16}{81}}

<u>Step-by-step explanation:</u>

\begin{array}{c|l}\underline{\quad x\quad}&\underline{\quad 4^{-x}\qquad \qquad}\\-1&4^{-(-1)}=4^1=4\\\\0&4^{-(0)}=4^0=1\\\\2&4^{-(2)}=\dfrac{1}{4^2}=\dfrac{1}{16}\\\\4&4^{-(4)}=\dfrac{1}{4^4}=\dfrac{1}{64}\end{array}

\begin{array}{c|l}\underline{\quad \bigg x \quad}&\underline{\quad \bigg(\dfrac{2}{3}\bigg)^x\qquad \qquad}\\\\-1&\bigg(\dfrac{2}{3}\bigg)^{-1}=\dfrac{3}{2}\\\\0&\bigg(\dfrac{2}{3}\bigg)^{0}=1\\\\2&\bigg(\dfrac{2}{3}\bigg)^{2}=\dfrac{2^2}{3^2}=\dfrac{4}{9}\\\\4&\bigg(\dfrac{2}{3}\bigg)^{4}=\dfrac{2^4}{3^4}=\dfrac{16}{81}\end{array}

3 0
2 years ago
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