Ask question

Ask question

All categories

FromTheMoon [43]

3 years ago

15

How long does it take for an investment to double in value if it is invested 8% per annum compounded monthly? Compounded continu

ously? It's number 29.

1 answer:

Fynjy0 [20]3 years ago

8 0

2p=p(1+0.08/12)^12t
T=[log(2)/log(1+0.08/12)]/12
T==8.7 years

You might be interested in

Factor this expression. You will need to show your work for full credit. b^2-6b+8

Ostrovityanka [42]

B^2-6b+8 turns into b^2-2b-4b+8 due to the sum product pattern. which then turns into b(b-2) - 4(b-2) and then its (b-4)(b-2)

4 0

2 years ago

can someone do this dot plot for me :(( plss no links i beg you i would do it my self but i dont get the question will give bran

Rasek [7]

Hope this helps have a nice day

8 0

2 years ago

Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]

Anit [1.1K]

Answer:

The answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

Step-by-step explanation:

$\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}$

Solve the L.H.S part:

$\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]$

After calculating the L.H.S part compare the value with R.H.S:

$\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\$

$\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\$

In equation (i) multiply by 3 and subtract by equation (iii):

$\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to c= - \frac{1}{2}$

put the value of c in equation (i):

$\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\$

In equation (ii) multiply by 3 then subtract by equation (iv):

$\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\$

put the value of d in equation (iv):

$\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2$

The final answer is " $\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}$ ".

4 0

3 years ago

What is the slope of any line that is perpendicular to the line = y - 5 = 3/2 * (x + 1)

klio [65]

Answer:

-2/3

Step-by-step explanation:

6 0

2 years ago

Find the inverse of the matrix , if it exists

Montano1993 [528]

Answer:

So, option A is correct.

Step-by-step explanation:

To find the inverse of the matrix, the formula is:

$A^{-1} = 1/|A| * Adj A$

Inverse of the matrix exists if |A| does not equal to 0

|A| = 7 * 9 -( -2 * -10)

|A| = 63 - 20

|A| = 43

AS |A| ≠ 0

so inverse exist.

now finding Adj A

$Adj A$ = $\left[\begin{array}{cc}9&2\\10&7\end{array}\right]$

$\left[\begin{array}{cc}9/43&2/43\\10/43&7/43\end{array}\right]$

So, option A is correct.

8 0

2 years ago