Can someone please help me with this problem

Please help me with this one

Rasek [7]

Answer:

6 (2)^ = 6

6 (2)^2 = 24

a = 24

b = 6

y-intercept: (0,6)

y-intercept =

Step-by-step explanation:

6(2)^x

6 (2)^0 = 6 (a number ^0 = 1, 6 times 1 = 6)

6 (2)^2 = 6 x 4 = 24

8 0

3 years ago

Need help math problem give you 5 stars and a good rating if you do this

Mazyrski [523]

Answer:

Add 9.6 then divide by 3.2

8 0

2 years ago

Which number line shows the correct locations of ALL the given values?

bija089 [108]

A is the only number line that correctly shows the locations of all given values.

Answer=A

***EDIT***

Okay, so if you look at the number line, you notice that the distance between one line and the next is equal to 1/4. When we move to the left, we decrease in value. When we move to the right, we increase in value.

So, knowing that, lets look at A and see if all of the numbers are correctly placed.

1/2 should be two lines to the right of 0 (since distance between one line and the next is equal to 1/4, and 1/4+1/4=2/4 or 1/2)

-4 is a relatively easy value to find, since its line is labeled.

-2 3/4 should be three lines to the right of -2 (since distance between one line and the next is equal to 1/4, and 1/4+1/4+1/4=3/4 and if you go 3/4 to the left of -2 u get -2 3/4)
(-2-3/4=-2 3/4)

1 1/4 should be three lines behind 2 (since distance between one line and the next is equal to 1/4, and 1/4+1/4+1/4=3/4 and if you go 3/4 to the left of 2 you get 1 1/4)
(2-3/4=1 1/4)

6 0

3 years ago

How to divide 10,768÷8

Karolina [17]

The Answer in this question is 1,346

6 0

3 years ago

Determine the values of the constants B and C so that the function given below is differentiable.

laila [671]

For the function to be differentiable, its derivative has to exist everywhere, which means the derivative itself must be continuous. Differentiating gives

$f'(x)=\begin{cases}24x^2&\text{for }x1\end{cases}$

The question mark is a placeholder, and if the derivative is to be continuous, then the question mark will have the same value as the limit as $x\to1$ from either side.

$\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}24x^2=24$
$\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}B=B$

So the derivative will be continuous as long as $B=24$

For the function to be differentiable everywhere, we need to require that $f(x)$ is itself continuous, which means the following limits should be the same:

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}8x^3=8$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}Bx+C=24+C$

$24+C=8\implies C=-16$

So, the function should be

$f(x)=\begin{cases}8x^3&\text{for }x\le1\\24x-16&\text{for }x>1\end{cases}$

with derivative

$f'(x)=\begin{cases}24x^2&\text{for }x$

5 0

3 years ago