Ask question

Ask question

All categories

2 years ago

14

If $5000 is deposited at the beginning of every six months into an account that earns 8% compounded semiannually find the amount

after 10 years

1 answer:

mixas84 [53]2 years ago

7 0

Answer:

Step-by-step explanation:

This is a geometric sum

s=a(r^n-1)/(r-1), here a=5000, n=20, r=1.04

s=5000(1.04^20-1)/(0.04)

s=$148890.39

You might be interested in

What is the area of a regular hexagon with a side length of 4 m? Enter your answer in the box. Round only your final answer to t

hichkok12 [17]

The area of a hexagon is
A= a^2 (3√3)/2
we replace a with 4
A=41.57

5 0

2 years ago

Read 2 more answers

10 to the power of 2 divided by 4 divided by 3 plus 2

mylen [45]

The answer would be = <span>10.3333333
</span>

5 0

3 years ago

Last year, Mr. Jones made $30,000. His boss just informed him that he will be receiving at least an 11.2% raise for this year. H

Lady bird [3.3K]

Answer:

$33 336

Step-by-step explanation:

Increase = 30 000 × 0.112

Increase = $3336

New salary = 30 000 + 3336

New salary = $33 336

7 0

3 years ago

DIA [1.3K]

True. If you're just changing the form, yes. But remember when you change one side of an equation, you must change the other in the same way.

7 0

3 years ago

Read 2 more answers

Please answer question three and four if you can :)<br> Show full working out ty;)

nikklg [1K]

Step-by-step explanation:

3

Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].

Therefore, by mid-point formula:

$D = ( \frac{2 + 5}{2}, \: \: \frac{ - 1 + 2}{2} ) = ( \frac{7}{2}, \: \: \frac{ 1}{2} ) \\ \therefore D= (3.5, \: \: 0.5) \\ \& \: A=(-1,\:\:4)...(given) \\\\ now \: by \: distance \: formula \\ \\ Length \: of \: segment \: AD \\ = \sqrt{( - 1 - 3.5)^{2} + {(4 - 0.5)}^{2} } \\ = \sqrt{(4.5)^{2} + {(3.5)}^{2} } \\ = \sqrt{20.25 + 12.25 } \\ = \sqrt{32.5} \\ \red{ \boxed{\therefore Length \: of \: segment \: AD = 5.7 \: units}}$

4 (a)

Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:

$\frac{y-y_1}{y_1-y_2} =\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-1}{1-(-11)} =\frac{x-2}{2 - (-2) } \\\\\therefore \frac{y-1}{1+11} =\frac{x-2}{2 +2} \\\\\therefore \frac{y-1}{12} =\frac{x-2}{4} \\\\\therefore \frac{y-1}{3} =\frac{x-2}{1} \\\\\therefore y-1= 3(x - 2)\\\\\therefore y= 3x - 6+1\\\\\therefore y= 3x - 5\\\\ \huge \purple {\boxed {\therefore 3x - y-5=0}} \\$

is the equation of line AB.

Now we have to check whether C(4, 7) lie on line AB or not.

Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.

$LHS = 3x - y-5\\=3\times 4-7-5\\= 12-12\\=0\\= RHS$

Hence, point C (4, 7) lie on the straight line AB.

4(b)

Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.

4 0

3 years ago