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3 years ago

!!HURRY IM BEING TIMED 40PTS!!

Mathematics

2 answers:

dusya [7]3 years ago

8 0

Answer:

It's A on Ed

Step-by-step explanation:

Rudik [331]3 years ago

3 0

The average rate of change in a function from a to b is

$\frac{f(b)-f(a)}{b-a}$ .

For the left function, from x = 4 to x = 5, the rate of change is

$\frac{100-64}{5-4} = \frac{36}{1} = 36$ .

For the right function, the rate of change is

$\frac{1024-256}{5-4} = \frac{768}{1} = 768$

Using a ratio to compare the right function's growth rate with that of the left function, you get $768 \div 36 \approx 21.3$ .

The right function is growing approximately 21 times faster than the left function (over the interval from 4 to 5).

You might be interested in

Solve the inequality for x 19>x-5 Simplify your answer as much as possible.

Stolb23 [73]

19>x-5

Add 5 on each side giving you
19+5>x
24>x

x<24

Hope this helps!!

7 0

4 years ago

Does anyone know what the answer please????

natima [27]

Answer:

I think 25 is this is right

5 0

3 years ago

What two numbers multiply to -30 and add to 1?

svp [43]

These numbers would be -5 & 6...

(-5)(6) = -30
-5 + 6 = 1

4 0

3 years ago

A cone has a base with a radius of 9 mm and a height of 13 mm. What is the volume of the cone?

avanturin [10]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - a cone with base radius 9mm and height 13 mm

To calculate - volume of the cone

We know that ,

$Volume \: of \: cone = \frac{1}{3}\pi \: r {}^{2} h \\$

substituting the values in the formula ,

$Volume = \frac{1}{3} \times 3.14 \times 9 \times 9 \times 13 \\ \\ \implies \: 1.05 \times 81 \times 13 \\ \\ \implies \: 1105.65 \: mm {}^{3}$

hope helpful ~

7 0

2 years ago

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

4 years ago