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2 years ago

8

Suppose you deposit $1000 in a college fund that pays paid interest quarterly instead of annually. Find the account balance afte

r 5 years

1 answer:

BartSMP [9]2 years ago

3 0

Answer:

$1,161.18 (assuming if interest rate is 3%)

Step-by-step explanation:

Kindly check the attached picture for explanation

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If f (x)=2x +6 and g (x) =x^3, what is ( g*f) (0)

8090 [49]

(g*f)(0)= (x^3)*(2x+6)
(g*f)(0)= (0^3)*(2(0)+6)
(g*f)(0)= (0)*(0+6)
(g*f)(0)= (0)*(6)
(g*f)(0)= 0

4 0

2 years ago

Multiply<br><br> 5( 4n -5)<br> tell me how

rewona [7]

Distribute

5 × 4n + 5 × -5

Simplify 5 × 4n to 20n

20n + 5 × -5

Simplify 5 × -5 to -25

<u>20n - 25</u>

8 0

2 years ago

Complete the pattern<br> 47-------43,41,-------,37,-------

telo118 [61]

Answer:

Next will be 33, 25, and 21

Step-by-step explanation:

3 0

2 years ago

When was the midpoint formula discovered?

photoshop1234 [79]

Answer:

$(\frac{x1+x2}{2}, \frac{y1+y2}{2})$

Step-by-step explanation:

3 0

2 years ago

For a function g(x), the difference quotient is StartFraction 6 Superscript x + h minus 3 Baseline minus 6 Superscript x + 3 Bas

kobusy [5.1K]

Answer:

The average rate of change in that interval is 1.99

Step-by-step explanation:

Here we have that the difference quotient for g(x) is:

$\frac{6^{x + h} - 3 - 6^x + 3}{h}$

Remember that for a general function f(x), the difference quotient is:

$\frac{f(x + h) - f(x)}{h}$

So if we look at the difference quotient for g(x), we can conclude that:

$g(x) = 6^x - 3$

Also remember that the average rate of change in an interval (a, b) is just:

$\frac{g(b) - g(a)}{b - a}$

So here we want the average rate of change of g(x) in the interval from x = -2 to x = 1, this is:

$R = \frac{(6^1 - 3) - (6^{-2} - 3)}{1 - (-2)} = \frac{6 - 6^{-2}}{3} = 1.99$

The average rate of change in that interval is 1.99

8 0

2 years ago