a golf driving range charges a flat fee of $20 to practice and then $5.75 for a bucket of balls what equation that modela C

2. Find the<br> slope of the<br> line between<br> (8, 6) and<br> (20, -14)

Natalka [10]

Answer:

slopee = 5/3

Step-by-step explanation:

The slope of a line is defined by the rise (the change in y-value) divided by the run (the change in x-value). Here, the change in y-value is 6 - (-14) = 6 + 14 = 20. The change in x-value is 20 - 8 = 12. Therefore, the slope is 20/12 or (simplified) 5/3.

3 0

2 years ago

What is the balance after 15 years in a savings account that earns 2% interest compounded bimonthly when the initial deposit is

Ierofanga [76]

Answer:

$1348.07

Step-by-step explanation:

Hello!

<h3>Compound Interest Formula: $A = P(1 + \frac rn)^{nt}$ </h3>

A = Account Balance
P = Principle/Initial Amount
r = Rate of Interest (decimal)
n = Number of times compounded (per year)
t = Number of Years

<h3>Given Information</h3>

Account Balance = ?
Principle Amount = $1000
Rate of Interest = 0.02

Why is the Rate 0.02?

This is because we are gaining money, so the multiplier should be greater than 1. We already added 1, which is 100% so you simply add the 0.02 for the extra 2%.

Number of times compounded per year = 6

This is because it is being compounded bi-monthly, or once every 2 months. 12 months divided by 2 months is 6 months, so 6 times a year.

Number of years = 15

<h2>Solve </h2>

Solve by plugging in the given values into the formula.

$A = P(1 + \frac rn)^{nt}$
$A = 1000(1 + \frac {0.02}{6})^{6*15}$
$A = 1000(1 + 0.00333...)^{90}$
$A = 1000(1.00333...)^{90}$
$A = 1000(1.145743)$
$A \approx 1457.43$

This is really close to the first option, and since there is rounding involved with the repeating decimal, the first option should be correct.

The answer is $1348.07.

3 0

2 years ago

Determine the equation of the polynomial, p(x),in both factored and standard form.

melamori03 [73]

We can see that the graph touches $x=-1$ without crossing the x-axis (i.e. it is a double solution), and then there's another zero at $x=2$ (this time it's a crossing zero, so a single solution).

This leads, up to multiple, to the polynomial

$p(x)=a(x+1)^2(x-2)$

If we impose the passing through $(0,4)$ we have

$p(0)=4=a(1)(-2) \iff -2a=4 \iff a=-2$

So, the polynomial is

$p(x)=-2(x+1)^2(x-2)=-2 x^3 + 6 x + 4$

Finally, to solve $p(x)$ , simply look at the graph, searching for the points, where the graph is below the x-axis. You can see that this happens only if $x>2$ , so that's the solution to your question.