list 5 examples and application where geometric sequences occur in everyday life.justify your answer are geometric sequences

1 answer:

7 0

A ball bouncing is an example of a finite geometric sequence. Each time the ball bounces it's height gets cut down by half. If the ball's first height is 4 feet, the next time it bounces it's highest bounce will be at 2 feet, then 1, then 6 inches and so on, until the ball stops bouncing.

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Use a graph to solve the equation. Check your solution.

nikklg [1K]

1) x=1
2) x=2.9
3) x=-10
4) x=?
5) IDK

4 0

3 years ago

What is the written form of the fraction 1/3 ? Patty stirred 1/3 cup of milk into the macaroni and cheese. A. one-third B. one t

yKpoI14uk [10]

One third is the written form.
Also, just a bit of extra information! The decimal of 1/3 is 0.33(recurring)

Rachel :)

3 0

3 years ago

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Find the volume and area for the objects shown and answer Question

klio [65]

Step-by-step explanation:

You must write formulas regarding the volume and surface area of the given solids.

$\bold{\#1\ Rectangular\ prism:}\\\\V=lwh\\SA=2lw+2lh+2wh=2(lw+lh+wh)\\\\\bold{\#2\ Cylinder:}\\\\V=\pi r^2h\\SA=2\pi r^2+2\pi rh=2\pir(r+h)\\\\\bold{\#3\ Sphere:}\\\\V=\dfrac{4}{3}\pi r^3\\SA=4\pi r^2$

$\bold{\#4\ Cone:}\\\\V=\dfrac{1}{3}\pi r^2h\\\\\text{we need calculate the length of a slant length}\ l\\\text{use the Pythagorean theorem:}\\\\l^2=r^2+h^2\to l=\sqrt{r^2+h^2}\\\\SA=\pi r^2+\pi rl=\pi r^2+\pi r\sqrt{r^2+h^2}=\pi r(r+\sqrt{r^2+h^2})\\\\\bold{\#5\ Rectangular\ Pyramid:}\\\\V=\dfrac{1}{3}lwh\\\\$

$\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}$

$\\\\h_2^2=\left(\dfrac{w}{2}\right)^2+h^2\to h_2=\sqrt{\left(\dfrac{w}{2}\right)^2+h^2}=\sqrt{\dfrac{w^2}{4}+h^2}=\sqrt{\dfrac{w^2}{4}+\dfrac{4h^2}{4}}\\\\h_2=\sqrt{\dfrac{w^2+4h^2}{4}}=\dfrac{\sqrt{w^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{w^2+4h^2}}{2}$

$SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}$

6 0

3 years ago

2x + (3)(x) + (3)(4)

S_A_V [24]

Answer:

if you were looking for the solution i think it is Solution

5 + 1 2

Step-by-step explanation:

4 0

3 years ago

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Jesse drives to work. The distance Jesse’s car travels (d) on a number of gallons of gas (g) is given by the equation d = 15g. T

kirill115 [55]

I can only half answer this question, as it doesn't tell what form of distance is used

the equation is...

d=15g

Now lets solve for 1 gallon of gas

d=15g
d=15(1)
d=15

This means that for every 1 gallon of gas, 15 (insert missing unit of distance here) can be driven.

15/1=15

Answer:The constant of proportionality is 15.

4 0

3 years ago

list 5 examples and application where geometric sequences occur in everyday life.justify your answer are geometric sequences ​

list 5 examples and application where geometric sequences occur in everyday life.justify your answer are geometric sequences