Number one. 897 2. Eight hundred eighty four and fourth seven hundredths. Number three. 17.818. 4. Two thousand forty four and four tenths. Number five. 22.766 6. Twelve thousand five hundred and twelve Number seven. 23.62

7 0

2 years ago

I’m completely confused.

Leviafan [203]

F(-2) = 7
f(0) = 6
f(6) = 3
f represents the function and the number in the parentheses is the input for the function, y on the graph is the output

6 0

3 years ago

Read 2 more answers

A rubber ball dropped on a hard surface takes a sequence of bounces, each one 1/6 as high as the preceding one. If this ball is

Brums [2.3K]

Answer:

$\frac{259}{54}\text{ or }4.8\text{feet}$

Step-by-step explanation:

GIVEN: A rubber ball dropped on a hard surface takes a sequence of bounces, each one $\frac{1}{6}$ as high as the preceding one.

TO FIND: If this ball is dropped from a height of $12$ feet, how far will it have traveled when it hits the surface the fifth time.

SOLUTION:

once the ball is dropped on hard surface it bounces $\frac{1}{6}$ of preceding one and comes down the same distance.

When the ball is dropped from $12$ feet height

after first hit $=12\times\frac{1}{6}\text{ up}+12\times\frac{1}{6}\text{ down}=4$

new height $=12\times\frac{1}{6}=2\text{feet}$

after second hit $=2\times\frac{1}{6}\text{ up}+2\times\frac{1}{6}\text{ down}=\frac{2}{3}$

new height $=2\times\frac{1}{6}=\frac{1}{3}\text{ feet}$

after third hit $=\frac{1}{3}\times\frac{1}{6}\text{ up}+\frac{1}{3}\times\frac{1}{6}\text{ down}=\frac{1}{9}$

new height $=\frac{1}{3}\times\frac{1}{6}=\frac{1}{18}\text{ feet}$

after fourth hit $=\frac{1}{18}\times\frac{1}{6}\text{ up}+\frac{1}{18}\times\frac{1}{6}\text{ down}=\frac{1}{54}$

adding all distance $=4+\frac{2}{3}+\frac{1}{9}+\frac{1}{54}$

$=\frac{259}{54}$ feet

Hence the ball will travel $\frac{259}{54}$ feet before it hits the surface fifth time.

4 0

3 years ago

What is the GCF of h4 and h8​

What is the GCF of h4 and h8