A) The constant of proportionality in this proportional relationship is ![k = \frac{y}{x}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7By%7D%7Bx%7D)
B) The equation to represent this proportional relationship is y = 0.2x
<h3><u>Solution:</u></h3>
Given that,
The amount Naomi pays each month for international text messages is proportional to the number of international texts she sends that month
Therefore,
This is a direct variation proportion
![\text{ amount Naomi pays each month } \propto \text{ number of international texts she sends that month}](https://tex.z-dn.net/?f=%5Ctext%7B%20amount%20Naomi%20pays%20each%20month%20%7D%20%5Cpropto%20%5Ctext%7B%20number%20of%20international%20texts%20she%20sends%20that%20month%7D)
Let "y" be the amount that Naomi pays each month
Let "x" be the number of international texts she sends that month
Therefore,
![y \propto x](https://tex.z-dn.net/?f=y%20%5Cpropto%20x)
y = kx -------- eqn 1
Where, "k" is the constant of proportionality
Thus the constant of proportionality in this proportional relationship is:
![k = \frac{y}{x}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7By%7D%7Bx%7D)
<em><u>Last month, she paid $3.20 for 16 international texts</u></em>
Therefore,
y = 3.20
x = 16
Thus from eqn 1,
![3.20 = k \times 16\\\\k = \frac{3.20}{16}\\\\k = 0.2](https://tex.z-dn.net/?f=3.20%20%3D%20k%20%5Ctimes%2016%5C%5C%5C%5Ck%20%3D%20%5Cfrac%7B3.20%7D%7B16%7D%5C%5C%5C%5Ck%20%3D%200.2)
Substitute k = 0.2 in eqn 1
y = 0.2x
The equation would then be y = 0.2x
Answer:
Range = {6, 9}.
Step-by-step explanation:
Just plug in the 3 numbers:
Range = (-2)^2+5 , (-1)^2 + 5 and 1^1 + 5
= 9, 6 and 6.
X=7
si quiere el procedimiento dejeme saber por favor
-Diana Perez
To solve this problem you must apply the proccedure shown below:
1. You must apply the following formula:
![FV=(PV)(e^{it})](https://tex.z-dn.net/?f=%20FV%3D%28PV%29%28e%5E%7Bit%7D%29%20)
Where
is the future value,
is the present value,
is the interest rate and
is the time in years.
2. You have that the bank will double your money in
years. Therefore:
![FV=2PV](https://tex.z-dn.net/?f=%20FV%3D2PV%20)
3. Substitute values into the formula and solve for
, as following:
![2PV=(PV)(e^{(i)(20)} \\ 2=e^{(i)(20)}](https://tex.z-dn.net/?f=%202PV%3D%28PV%29%28e%5E%7B%28i%29%2820%29%7D%20%5C%5C%202%3De%5E%7B%28i%29%2820%29%7D%20%20)
4. By applying natural logarithm, you have:
![ln(2)=(i)(20) \\ i= \frac{ln(2)}{20} \\ i=0.0346](https://tex.z-dn.net/?f=ln%282%29%3D%28i%29%2820%29%20%5C%5C%20i%3D%20%5Cfrac%7Bln%282%29%7D%7B20%7D%20%20%5C%5C%20i%3D0.0346)
%
The answer is:
%
Answer: x>-6
Explanation: -2x -6>-18
-2x >-12
-x>6
X>-6