Is the dolphin deeper than point C or point E?

What is the average rate of change of f over the interval [3,4]?

masya89 [10]

Answer:

4

Step-by-step explanation:

The average rate of change of f(x) in the closed interval [ a, b ] is

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

Here [ a, b ] = [ 3, 4 ] , then

f(b) = f(4) = - 1 ← from table

f(a) = f(3) = - 5 ← from table , then

average rate of change = $\frac{-1-(-5)}{4-3}$ = $\frac{-1+5}{1}$ = 4

3 0

3 years ago

The original loan amount is referred to as the______.

marishachu [46]

The answer is principal.
Hope this helped c:

5 0

3 years ago

Read 2 more answers

How many times greater was the temperature in Miami (45) than the lowest in San Diego (25)

Greeley [361]

Answer:

The temperature in Miami is 9/5 times the temperature in San Diego.

Step-by-step explanation:

Ratios

To compare the temperature in Miami (45) with the temperature in San Diego (25), we use the division or ratio between both numbers:

$\displaystyle \frac{45}{25}$

Simplify the fraction dividing by 5:

$\displaystyle \frac{9}{5}$

Since the relation between them is 9/5, it means the temperature in Miami is 9/5 times the temperature in San Diego.

3 0

3 years ago

It costs 15 dollars to send 3 packages through a certain shipping company consider the number of packages pero dollar?

docker41 [41]

solution:

= $15 for 3 packages

= 5 packages per dollar (15/3)

5 0

3 years ago

Factor completely. <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra

Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

$x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)$

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize $x^2-\frac13$ as yet another difference of squares:

$x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)$

And if you're working over the field of complex numbers, you can go even further. For instance,

$x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)$

But I think you'd be fine stopping at the first result,

$x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}$

6 0

3 years ago

Is the dolphin deeper than point C or point E?​

Is the dolphin deeper than point C or point E?