PLEASE HELP THISIS GETTING TIMED

A cell phone plan has a monthly cost that is shown in the table below. What is the correct statement regarding the average rate

FromTheMoon [43]

A, its the only option that states that the monthly bill increases by $1.00 every ten minutes of talk time

5 0

3 years ago

Please help if u cannot skip

posledela

The statement that is true about the function is D. it is discontinuous and non-differentiable at x = 3.

<h3>How to determine which statement is true?</h3>

To determine which statement is true, we need to know the conditions for continuity and differentiablity of a function.

<h3>Conditions for continuity and differentiablity of a function.</h3>

For a function f(x) to be continuous at a point x = a, then both the left hand limit of f(x) and the right hand limit of f(x) as x → a must be equal. That is $\lim_{x \to a^{-} } f(x) = \lim_{x \to a^{+} } f(x)$ . So, $\lim_{x \to a^{} } f(x)$ must exist since $\lim_{x \to a^{-} } f(x) = \lim_{x \to a^{+} } f(x) = \lim_{x \to a^{} } f(x)$

Also, for a function to be differentiable at a point x = a, it must also exist at x = a

So, since f(x) = {x² - 1 if -1 ≤ x ≤ 3 and x²/3 if 3 < x ≤ 8}

From the equality on the first condition,we see that f(x) is exists at x = 3 but is not continuous since f(x) changes to another function when x > 3. So,left hand limit of f(x) and the right hand limit of f(x) as x → 3 are not equal.

That is $\lim_{x \to 3^{-} } f(x) \neq \lim_{x \to 3^{+} } f(x)$ . Thus, the function is discontinuous at x = 3.

For differentiability, both conditions must be met. Since only one condition is met, it is non-differentiable.

So, the function is discontinuous and non-differentiable at x = 3.

So, the statement that is true about the function is D. it is discontinuous and non-differentiable at x = 3.

Learn more about continuity of a function here:

brainly.com/question/24177259

#SPJ1

4 0

2 years ago

Need help please with the step by step explanation

Stells [14]

Answer:

exponential decay (0,8)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

(a)Find all integer solutions to the equation 105x + 83y = 1.

Mashcka [7]

Answer:

(a) (34+83t,-43-105t) where t is an integer

(b) (272+83t,-344-105t) where t is an integer.

(c) 62

Step-by-step explanation:

a)

We are going to perform Euclidean's Algorithm.

Let's begin with seeing how many times 83 goes int 105.

105=83(1)+22 (eq1)

83=22(3)+17 (eq2)

22=17(1)+5 (eq3)

17=5(3)+2 (eq4)

5=2(2)+1 (eq5)

Now let's go backwards through those equations.

5-2(2)=1 (eq5 rewritten so that the remainder was by itself)

5-2[17-5(3)]=1 (replaced the 2 in ( ) with eq4 solved for the remainder)

5-2(17)+5(6)=1 (distributive property was performed)

-2(17)+5(7)=1 (combined my 5's)

-2(17)+7(5)=1 (multiplication is commutative)

-2(17)+7(22-17)=1 (used eq3)

-2(17)+7(22)-7(17)=1 (distribute property was performed)

-9(17)+7(22)=1 (combined my 17's)

-9(83-22(3))+7(22)=1 (used eq2)

-9(83)+22(27)+7(22)=1 (distributive property was performed)

83(-9)+22(34)=1 (multiplication is commutative and combined my 22's)

83(-9)+34(105-83)=1 (used eq1)

105(34)+83(-43)=1 (after distributive property and reordering)

So we have a point on the line being (x,y)=(34,-43).

We can use the slope to figure out all the other integer pairs from that initial point there.

The slope of ax+by=c is -a/b.

So the slope of 105x+83y=1 is -105/83.

So every time we go down 105 units we go right 83 units

This says we have the following integer pairs on our line:

(34+83t,-43-105t) where t is an integer.

Let's verify:

Plug it in!

105[34+83t]+83[-43-105t]

105(34)+105(83)t+83(-43)-83(105)t

105(34)+83(-43)

1

We are good!

(b)

We got from part (a) that 105(34)+83(-43)=1.

Multiply both sides we get 8 on the right hand side:

105(34*8)+83(-43*8)=8

Simplify:

105(272)+83(-344)=8

So the integer pairs is (272+83t,-344-105t) where t is an integer.

Let's verify:

105[272+83t]+83[-344-105t]

105(272)+105(83)t+83(-344)-83(105)t

105(272)+83(-344)

8

(c)

Let u=83^(-1) mod 105.

Then 83u=1 mod 105.

This implies:

83u-1=105k for some integers k.

Add 1 on both sides:

83u=105k+1

Subtract 105k on both sides:

83u-105k=1

Reorder:

105(-k)+83u=1.

We found all (x,y) integer pairs such that 105x+83y=1.

We go (34+83t,-43-105t) where t is an integer.

So k=-34-83t while u=-43-105t.

Since we want to find an integer t such that u is between 0 and 104, we could solve 0<-43-105t<104.

Add 43 on all sides:

43<-105t<147

Divide all sides by -105:

-43/105>t>-147/105

-147/105<t<-43/105

This says t is approximately between -1.4 and -0.4 . This includes only the integer -1.

When t=-1, we have u=-43-105(-1)=-43+105=62.

3 0

4 years ago

Please help me out if appreciate it :)

12345 [234]

Answer:

51

Step-by-step explanation:

they are congruent so they are equal to eachotjer

4 0

2 years ago