Let from the height at which ball was thrown = h units
It is given that , ball rebounds the same percentage on each bounce.
Let it rebounds by k % after each bounce.
Height that ball attains after thrown from height h(on 1 st bounce)= ![h + \frac{h k}{100}=h \times (1+\frac{k}{100})](https://tex.z-dn.net/?f=h%20%2B%20%5Cfrac%7Bh%20k%7D%7B100%7D%3Dh%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29)
Height that ball attains after thrown from height h (on 2 n d bounce)= ![h \times (1+\frac{k}{100})+h \times (1+\frac{k}{100})\times \frac{k}{100}=h \times (1+\frac{k}{100})^2](https://tex.z-dn.net/?f=h%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29%2Bh%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29%5Ctimes%20%5Cfrac%7Bk%7D%7B100%7D%3Dh%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29%5E2)
Similarly, the pattern will form geometric sequence.
S= ![h +h \times (1+\frac{k}{100})+h \times (1+\frac{k}{100})^2+h \times (1+\frac{k}{100})^3+.........](https://tex.z-dn.net/?f=h%20%2Bh%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29%2Bh%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29%5E2%2Bh%20%5Ctimes%20%281%2B%5Cfrac%7Bk%7D%7B100%7D%29%5E3%2B.........)
So, Common Ratio = ![\frac{\text{2nd term}}{\text{1 st term}}=1 +\frac{k}{100}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B2nd%20term%7D%7D%7B%5Ctext%7B1%20st%20term%7D%7D%3D1%20%2B%5Cfrac%7Bk%7D%7B100%7D)
Common Ratio= 1 + the percentage by which ball rebounds after each bounce
the percentage by which ball rebounds after each bounce= negative integer= k is negative integer.
let area of triangle be x
let area of rectangle be y
Then,
2x = y
Y=-3x+2:
Already in slope intercept form
Graph by plotting a point on (0,2)
Then move 3 points up and 1 to the right and plot the point and keep doing that a couple times
-3x-y=-2:
Add -3x on both sides
The result would be
-y=3x-2
Divide everything by negative 1
Y=-3x+2
Both equations are the same so they are the same line
Answer:
1
Step-by-step explanation:
Hey there!
This situation can be represented by the following equation:
y = 20 + 20x
This is because she starts without any time passing with 20 pounds of cans and collects 20 more pounds every week.
So when x = 0, y = 20.
The variable y is the number of pounds of cans collected, while x is the number of weeks passed.
If we want to find the number of pounds of cans she has after 10 weeks, we have to substitute x = 10 like this:
y = 20 + 20(10)
Multiply 20 x 10.
y = 20 + 200
Add 20 + 200.
y = 220
Your answer is 220 pounds, the second option shown!
Hope this helps!