For the first day we have the following function:
f (n) = (0.3 * 10) n + 8
For the second day we have the following function:
f (n) = (0.4 * 10) n + 5
You spent the same amount of money as the day before:
(0.3 * 10) n + 8 = (0.4 * 10) n + 5
3n + 8 = 4n + 5
n = 8-5
n = 3 items
We evaluate each function for n = 3
f (3) = (0.3 * 10) * 3 + 8 = 17 $
f (3) = (0.4 * 10) * 3 + 5 = 17 $
The total amount of money is:
17 + 17 = 34 $
Answer:
you purchase 6 items
you spend at the craft store during the sale $ 34

8 0

3 years ago

Which is the equation of a hyperbola centered at the origin with x-intercept +\- 3 and asymptote y=2x

Radda [10]

Answer:

${\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}$

Step-by-step explanation:

The hyperbola has x-intercepts, so it has a horizontal transverse axis.

The standard form of the equation of a hyperbola with a horizontal transverse axis is $\dfrac{(x - h)^{2}}{a^{2}} - \dfrac{(y - k)^{2}}{b^{2}} = 1$

The center is at (h,k).

The distance between the vertices is 2a.

The equations of the asymptotes are $y = k \pm \dfrac{b}{a}(x - h)$

1. Calculate h and k. The hyperbola is symmetric about the origin, so

h = 0 and k = 0

2. For 'a': 2a = x₂ - x₁ = 3 - (-3) = 3 + 3 = 6

a = 6/2 = 3

3. For 'b': The equation for the asymptote with the positive slope is

$y = k + \dfrac{b}{a}(x - h) = \dfrac{b}{a}x$

Thus, asymptote has the slope of

$\begin{array}{rcl}m& =& \dfrac{b}{a}\\\\2& =& \dfrac{b}{3}\\\\b& =& \mathbf{6}\end{array}$

4. The equation of the hyperbola is

$\large \boxed{\mathbf{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}}$

The attachment below represents your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.

7 0

3 years ago

Veronica has been saving dimes and quarters she has 110 coins in all and the total value is 18.50 how many dimes and quarters do

Lady_Fox [76]

This problem can be solved by the chicken rabbits method or you can just do simple algebra.

I.) Chicken and rabbits method

First assume all 110 coins are dimes and none are quarters.

We will have a total value of 11 dollars

Now for each dime we switch out for a quarter, we adds 15 cents to the total value.

18.50-11=7.50 dollars

There are 750/15=50 group of 15 cents in the 7 dollars and 50 cents.

This also meant that we need to switch out 50 dimes for 50 quarters.

So we have 50 quarters.

That first method is very good and very quick once you get the hang of it, now I'm going to show you the algebraic way to solve this.

Let's say there are x dimes and y quarters.

Set up equation

x+y=110

10x+25y=1850

Now solve multiply first equation by 10

10x+10y=1100

subtract

15y=750

y=50

Now we set the numbers of quarters to y so the answer is 50 quarters.

I personally recommend using algebra whenever you can because the practice is very important and you will eventually get really fast at setting up and solving equations. The first method is faster in this case but the second is more generalize, hope it helps.

6 0

3 years ago