Cassie paid $276.45 for 5 months of cell phone service. At that rate, how much will Cassie pay for a full year of cell phone service?

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The distance the duck fly in all is 99 miles

From the question, we have the following parameters that can be used in our computation:

The distance covered in all can be calculated as

Distance = The sum of the product of speed and time

Substitute the known values in the above equation, so, we have the following representation

Distance = 18 * 3 + 15 * 2

Evaluate the products

This gives

Distance = 54 + 45

Evaluate the sum

Distance = 99 miles

Hence, the distance is 99 miles

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Answer:

15.5

Step-by-step explanation:

Pepe= 2/3 x 12 = 8

Paula= 1/4 x 30 = 7.5

8+7.5=15.5

Answer:

Step-by-step explanation:

A.) To show that triangle ABC is similar to triangle DEC, we need to prove that the ratios of the sides of the two triangles are equal.

First, we can write the ratios of the sides of triangle ABC as follows:

AC/AB = 20/16 = 5/4

BC/AB = 24/16 = 3/2

Now, we can write the ratios of the sides of triangle DEC as follows:

CE/CD = 18/15 = 6/5

AC/CD = 20/15 = 4/3

Since the ratios of the sides of the two triangles are equal, it follows that triangle ABC is similar to triangle DEC.

B.) To find the length of DC, we can use the fact that triangle ABC is similar to triangle DEC. Since the ratios of the sides of the two triangles are equal, we can set up a proportion to solve for DC.

First, we can write the ratio of the sides of triangle ABC as follows:

AC/AB = DC/CE

Then, we can substitute the known values for AC, AB, and CE:

20/16 = DC/18

Then, we can cross-multiply to solve for DC:

DC = (20/16) * 18

= (5/4) * 18

= 45/4

= 11.25 cm

Therefore, the length of DC is approximately 11.25 cm.

Answer:

Step-by-step explanation:

To show that triangle ABC is analogous to triangle DEC, we need to prove that the rates of the sides of the two triangles are equal.

First, we can write the rates of the sides of triangle ABC as follows

AC/ AB = 20/16 = 5/4

BC/ AB = 24/16 = 3/2

Now, we can write the rates of the sides of triangle DEC as follows

CE/ CD = 18/15 = 6/5

AC/ CD = 20/15 = 4/3

Since the rates of the sides of the two triangles are equal, it follows that triangle ABC is analogous to triangleDEC.

B.) To find the length of DC, we can use the fact that triangle ABC is analogous to triangle DEC. Since the rates of the sides of the two triangles are equal, we can set up a proportion to break for DC.

First, we can write the rate of the sides of triangle ABC as follows

AC/ AB = DC/ CE

also, we can substitute the known values for AC, AB, and CE

20/16 = DC/ 18

also, we cancross-multiply to break for DC

DC = (20/16) * 18

= (5/4) * 18

= 45/4

= 11.25 cm

Thus, the length of DC is roughly11.25 cm.

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After integration, the required function f is (2x³ - sin (x) + Cx + D).

What is the integration of 'xⁿ' and 'sin (x)'?

[tex]\int {x^{n} } \, dx = \frac{x^{n+1} }{n+1} + C\\\\\\\int {sinx} \, dx = -cosx + C[/tex]

Given, f''(x) = 12x + sin x

Therefore,

[tex]\int {f''(x)} \, dx \\\\=\int {f'(x)} \, dx \\\\\\= \int{12x + sin x} \, dx + C\\\\= 6x^{2} - cosx + C\\[/tex]

Again, f'(x) = 6x - cos (x) + C

Therefore,

[tex]\int {f'(x)} \, dx\\ \\=\int {f(x)} \, dx \\\\= \int {6x^{2} - cosx + C } \, dx \\\\= 2x^{3} - sinx + Cx + D[/tex]

Therefore, the required function is (2x³ - sin (x) + Cx + D).

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∠U is congruent ∠S because all right angles are congruent.

If it is possible to superimpose one geometric figure on the other so that their entire surface coincides, that geometric figure is said to be congruent, or to be in the relation of congruence. When two sides and their included angle in one triangle are equal to two sides and their included angle in another, two triangles are said to be congruent. This concept of congruence appears to be based on the idea of a "rigid body," which may be moved without affecting the internal relationships between its components.

Given

∠U ≅ ∠S

All right angles are congruent.

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The weight of harsh on the moon if he weighs 54 pounds on earth is 9 pounds

Weight varies directly with gravity

Let

Weight of a person = w

Gravity = g

So,

w = k × g

Where,

k = constant of proportionality

If w = 180 pounds and g = 30 pounds

w = k × g

180 = k × 30

180 = 30k

divide both sides by 30

k = 180/30

k = 6

If w = 54 pounds g = ?

w = k × g

54 = 6 × g

54 = 6g

divide both sides by 6

g = 54/6

g = 9 pounds

Therefore, harsh weighs 9 pounds on the moon.

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The input value for the expression is 1 and the remainder is -15.

The Remainder Theorem begins with an unnamed polynomial p(x), where "p(x)" simply means "some polynomial p with variable x". The Theorem then discusses dividing that polynomial by some linear factor x a, where an is simply a number.

Given polynomial is 5x⁴⁵ - 3x¹⁷ +2x⁴ - 19 and is divided by x + 1. Find the value of the polynomial at 1 and then calculate the remainder by synthetic solution.

F(x) = 5x⁴⁵ - 3x¹⁷ +2x⁴ - 19

F(1) = 5 - 3 + 2 - 19

F(1) = -19 + 4

F(1) = -15

The remainder will be calculated by synthetic calculation,

1 __5 ___-3___2_____-19

 |

 | ______5___ 2______4          

      5         2      4          -15

The remainder of the given expression is -15.

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Applying the definition of congruent triangles, the values of x and y are:

x = 2; y = 6

Length of CR = 8 units.

If two triangles are congruent to each other, based on the CPCTC, all their corresponding angles, and their corresponding sides will be equal to each other.

Given that triangles NEW and CAR are congruent to each other, therefore:

NE = CA

NW = CR

EW = AR

Given the following measures:

NE = 11

EW = 12

AR = 4y - 12

CA = 4x + 3

NW = x + y

Therefore:

NE = CA

11 = 4x + 3

Solve for x:

11 - 3 = 4x

8 = 4x

8/4 = x

x = 2

EW = AR

12 = 4y - 12

12 + 12 = 4y

24 = 4y

24/4 = y

y = 6

NW = CR = x + y

Plug in the values of x and y:

CR = 2 + 6

CR = 8 units.

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The difference between Tom and Alfie is  1 kg

Let's call the weight of Tom T, the weight of Harry H, the weight of Freddie F, and the weight of Alfie A. We are given that T = H + 5, H = F + 3, and F = A - 2.

Substituting the second and third equations into the first equation gives us:

T = (A - 2) + 3 + 5

T = A - 2 + 3 + 5

T = A + 1

So the difference between Tom and Alfie is T - A = A + 1 - A = 1 kg.

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The joint relative frequency for those students who don't like to run, but however enjoy cycling, is C. 32%.

The joint frequency for students who like cycling, but do not like running, can be found by the formula :
= Number of students who enjoy cycling but don't enjoy running / Number of students in total

Number of students who enjoy cycling but don't enjoy running = 64

Number of students in total = 200

The joint frequency is :

= 64 / 200 x 100%

= 0.32 x 100 %

= 32 %

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The {-4, 0, 1} represents members of the domain of the graphed function.

What is the domain of the function?

A function is a mathematical object that accepts input, appears to apply a rule to it, and returns the result.

A function can be thought of as a machine that requires in a number, performs some operation(s), and then outputs the result.

The domain of a function is the collection of all its inputs. Its codomain is the set of possible outputs.

The range refers to the outputs which are actually used.

Domain: {-4, 0, 1}.

Simply list the domain as -4 < x < 2, which would imply ALL values between -4 and 2 inclusive.    

Yes, this is a function. No x-values repeat, and it passes the Diagonal Line Test for functions.

Hence, the {-4, 0, 1} represents members of the domain of the graphed function.

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Answer:

sorry i messed up my first calculations, currently recalculating

Step-by-step explanation:

Answer:

Step-by-step explanation:

You want the number of $30 advance tickets and $35 same-day tickets sold, when a total of 55 tickets were sold for $1825.

Let s represent the number of same-day tickets sold. Then 55-s is the number of advance tickets sold, and total revenue is ...

  35s +30(55 -s) = 1825

  5s = 175 . . . . . . Simplify, subtract 1650

  s = 35

  55-s = 30

35 same-day and 30 advance tickets were sold.

The cost of one chair is [tex]x=54.75[/tex] $ and the cost of one table is [tex]y=49.5[/tex] $.

What is the total cost?

The total cost formula is used to combine the variable and fixed costs of providing goods to determine a total. The formula is:

Total cost = (Average fixed cost x average variable cost)

Let cost of one chair is [tex]x[/tex]

Let cost of one table is [tex]y[/tex]

So,

[tex]5x+3y=31[/tex]       ...(1)

[tex]2x+6y=52[/tex]      ......(2)

Solving equation (1) and (2) we get

[tex]a_1x+b_1y+x_1=0\\\\a_2x+b_2y+c_2=0[/tex]

[tex]\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}\\\\\frac{x}{3(31)-6(52)}=\frac{y}{31(2)-52(5)}=\frac{1}{5(2)-2(3)}\\\\\frac{x}{93-312}=\frac{y}{62-260}=\frac{1}{10-6}\\\\\frac{x}{-219}=\frac{y}{-198}=\frac{1}{-4}\\\\x=54.75,y=49.5[/tex]

Hence, the cost of one chair is [tex]54.75[/tex] $ and the cost of one table is [tex]49.5[/tex] $.

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Answer:

You just need to think about this a little. There is a number, that you don't know, x, and if you square it ([tex]x^2[/tex]), that value will be equal to 72 more than the unknown number itself, so you put that into an equation.

Step-by-step explanation:

[tex]x^2=x+72\\x^2-x-72=0\\[/tex]

Apply quadratic formula (you can look it up)

[tex]x_1=\frac{-(-1)+\sqrt{(-1)^2+(-4*1*-72} )}{2(1)} \\\\x_2=\frac{-(-1)-\sqrt{(-1)^2+(-4*1*-72} )}{2(1)} \\\\[/tex]

There are two answers to the quadratic formula.

The rest should be easy.

A fraction means a part of something or a number of parts of something. The number on the bottom shows how many parts something has been divided into ½ means 1 part of something that has been divided into 2 parts. We call this a half. If you add two halves together you get one.

Answer: A fraction is a numerical quantity that is not a whole number

Step-by-step explanation:

Answer: 7

Step-by-step explanation:

x=12-5

x=7

Answer:

x=7

Step-by-step explanation:

12-5=7 or 7+5=12

The polynomial function whose graph passes through (7,11) (11,-12) and (0,4) will be y = -0.614x² + 5.295x + 4.

A function is an assertion, concept, or principle that establishes an association between two variables. Functions may be found throughout mathematics and are essential for the development of significant links.

Assume the polynomial function is quadratic. Then the equation is given as,

y = ax² + bx + c

At (0, 4), we have

4 = a(0)² + b(0) + c

c = 4

Then the equation is written as,

y = ax² + bx + 4

At (7, 11), we have

y = ax² + bx + 4

11 = 49a + 7b + 4             ...1

At (11, -12), we have

- 12 = 121a + 11b + 4        ...2

Equations 1 and 2 are solved by a calculator. Then we have

a = - 0.614 and b = 5.295

The polynomial capability whose diagram goes through (7,11) (11,- 12) and (0,4) will be y = - 0.614x² + 5.295x + 4.

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the equilibrium constant for this reaction in terms of the equilibrium constants is Knet = √Ka/Kb

and for the reactions below:

2 S(s) + 3 O2(g) 2 SO3(g) is √Ka

S(s) + O2(g) SO2(g) is 1/Kb

A reversible reaction mechanism progresses from initial conditions to chemical equilibrium at a constant temperature, with all reactants and products present in constant non-zero amounts. At this temperature, an equilibrium constant "K" exists. It is frequently followed by a subscript to indicate the type of equilibrium constant that it considers depending on the phases of reaction species that it considers. The expression for each type of equilibrium constant differs with the direction and stoichiometry of the balanced reaction equation. This enables existing equations with known "K" values to be exploited in order to derive new "K" values for related systems.

SO2(g) + 1/2 O2(g)SO3 (g) is the desired reaction equation

We consider the first given reaction system at the bottom. We divide its coefficients by two to get the square root of the initial equilibrium constant: 2 S(s) + 3 O2(g) (g) 2 SO3(g) = √Ka

Then we reverse the second given reaction system at the bottom. Its current equilibrium constant is the reciprocal of the predecessor: S(s) + O2(g) SO2(g) is 1/Kb

The above two equations total to yield our target equation. The product of the two values obtained above yields the target equilibrium constant: Knet = √Ka/Kb

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Answer:

$32.70 ÷ 5

Each case is $6.54.

Step-by-step explanation:

$3270 for 5 cases, meaning you would split $32.70 into 5.

$32.70 ÷ 5 phone cases = $6.54

This also means each case is $6.54. To prove this, multiply by 5.

The expression 10% of 40 is 4 and 10% more than 40 is 44

From the question, we have the following parameters that can be used in our computation:

10% of 40

Express "of" as products

So, we have

10% of 40 = 10% * 40

Evaluate

10% of 40 = 4

In this case, we have:

10% more than 40

This means that

10% more than 40 = 40 * (1 + 10%)

Evaluate

10% more than 40 = 44

Hence. 10% more than 40 is 44

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Considering the definition of a line, the equation of the line that passes through the point (10, -9) and has a slope of -2 is y= -2x +11.

A linear equation o line can be expressed in the form y = mx + b

where

In this case, you know:

Substituting the value of the slope m and the value of the point y=mx+b, the value of the ordinate to the origin b is obtained as follow:

-9= -2×10 + b

-9= -20 + b

-9 + 20= b

11= b

Finally, the equation of the line is y= -2x +11.

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Total  no. of cups of baking soda used for the experiment in improper fraction is [tex]\frac{91}{8}[/tex].

Initial no. of cups of baking soda used for an experiment = [tex]1\frac{3}{4}[/tex]

Converting mixed fraction to improper fraction,

[tex]1\frac{3}{4}=\frac{7}{4}[/tex]

No. of times the experiment performed by students = [tex]6\frac{1}{2}[/tex]

Converting mixed fraction to improper fraction,

[tex]6\frac{1}{2} = \frac{13}{2}[/tex]

Total  no. of cups of baking soda  =(Initial no. of cups)* (No. of times)

                                                        [tex]=\frac{7}{4} *\frac{13}{2} \\\\=\frac{91}{8} \\\\[/tex]

                                                        = 11.4 (decimal form)

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Answer:

answer is 2.5

Step-by-step explanation:

The difference between the model and the data will get smaller.

A sample is a percentage of the total population in statistics. You can use the data from a sample to make inferences about a population as a whole.

Given here, she 50 people and finds that only 75% and additional 100 people hence changing the sample size

As our sample size increases, the confidence in our estimate increases, our uncertainty decreases and we have greater precision. Thus with the increase in sample size, the data would move closer to the estimated probability.

Hence The difference between the model and the data will get smaller.

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Answer:

The difference between the data and the model will get smaller.

Step-by-step explanation:

Plato/Edmentum

the total number of license plate is 17576000 and In this the number ways to place 'A' in step 1 is = '1'1, and the number of steps '0' in final steps id '1' thus the total answer is 67600

In mathematics, there are two alternative methods for dividing up a collection of items into subsets: combinations and permutations. Any order may be used by a combination to list the subset's elements. An ordered list of a subset's components is called a permutation.

There are [tex]26^3[/tex] = 17576, ways to perform step 1-3, and there are [tex]10^3[/tex] = 1000, ways to perform remaining steps .

the total number of license plate is 17576000

In this the number ways to place 'A' in step 1 is = '1'1, and the number of steps '0' in final steps id '1'

thus the total answer is 67600

26 X 25 X 24 X 10 X 9 X 8 = 11232000

24 X 10 X 9 X 8 = 17280

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Answer:

natural exponential

natural

quizlet

Answer:

b. natural exponential

c. natural

Step-by-step explanation:

Given function:

[tex]f(x)=e^x[/tex]

The given exponential function is called the:

The base e is called the:

The number "e" occurs naturally in math and the physical sciences.

It is the base rate of growth shared by all continually growing processes, and so is called the natural base.

It is an irrational number and named after the 18th century Swiss mathematician, Leonhard Euler, and so is often referred to as "Euler's number".

The Divergence Test is inconclusive because lim ak = ∞

The Divergence Test is used to determine whether a given infinite series converges or diverges.

In order to use the Divergence Test, we must examine the limit of the terms of the series as n approaches infinity.

In the given series, the limit of the terms as n approaches infinity is not defined. Therefore, the Divergence Test is inconclusive.

The Divergence Test is inconclusive because lim ak = ∞

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Note that at any time during business hours, about 7 shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store. This is solved using Little Law.

Little's Law asserts that the long-term average number of individuals in a stable system, L, is equal to the long-term average effective arrival rate,λ, multiplied by the average duration a customer spends in the system, W.

To compute,

First, let's make sure that all the variables use the same time unit.

Thus, if there are 84 shoppers who are making a purchase per house, then there will be: 84/60 minutes

= 1.4

This means that  λ (rate) = 1.4 and the average time (W) is 5 minutes

Using Little's Law which states that the queuing formula is:

L = λW

Note that:

λ = 1.4 (computed)

W = 5 mintues (given)

Hence, the Average number of people in queue per time is:

L = 1.4 x 5

L = 7 shoppers.

Thus, it is correct to state that at any time during work hours, about 7 shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store.

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Full Question:

If shoppers enter a store at an average rate of r shoppers per minute and each stays in the store for an average time of T minutes, the average number of shoppers in the store, N, at any one time is given by the formula N=rT. This relationship is known as Little's law.

The owner of the Good Deals Store estimates that during business hours, an average of 3 shoppers per minute enter the store and that each of them stays an average of 15 minutes. The store owner uses Little's law to estimate that there are 45 shoppers in the store at any time.

Little's law can be applied to any part of the store, such as a particular department or the checkout lines. The store owner determines that, during business hours, approximately 84 shoppers per hour make a purchase and each of these shoppers spend an average of 5 minutes in the checkout line. At any time during business hours, about how many shoppers, on average, are waiting in the checkout line to make a purchase at the Good Deals Store?

f cannot have a jump discontinuity at [tex]$x_0 \in(a, b)$[/tex] and  [tex]$$ \lim _{x \uparrow x_0} f(x) < \lim _{x \mid x_0} f(x) .$$[/tex]

f cannot have a removable discontinuity at [tex]$$x_0 \in(a, b) $$[/tex] and [tex]\alpha=\lim _{x \rightarrow x_0} f(x) < f\left(x_0\right)[/tex]

Let f be a function defined on (a, b) satisfies intermediate value property.

Claim: f ca not have removable on jump discontinuity.

Suppose f has a jump discontinuity at [tex]$x_0 \in(a, b)$[/tex]

We take [tex]$\theta$[/tex] such that

[tex]$$\lim _{x \rightarrow x_0} f(x) < \theta < \lim _{x \downarrow x_0} f(x) \text { and } \theta \neq f\left(x_0\right)$$[/tex]

Now there exist [tex]$\delta > 0$[/tex] such that [tex]$f(x) < \theta$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0\right)$[/tex] and [tex]$f(x) > \theta$[/tex] for all [tex]$x \in\left(x_0, x_0+\delta\right]$[/tex]

Now [tex]$f\left(x_0-\delta\right)[/tex][tex]< \theta < f\left(x_0+\delta\right)$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0+\delta\right] \backslash\left\{x_2\right\}$[/tex] and [tex]$f\left(x_0\right) \neq \theta$[/tex].

Therefore the point [tex]$\theta$[/tex] has no preimage under f

that is, there does not exists [tex]$y \in\left[x_0-\delta, x_0+\delta\right][/tex] for which

[tex]$$f(y)=\theta[/tex] because [tex]\left\{\begin{array}{l}y=x_0 \Rightarrow f(y) \neq \theta \\y > x_0 \Rightarrow f(y) > \theta \\y < x_0 \Rightarrow f(y) < \theta\end{array}\right.$$[/tex]

Therefore f does not satisfies intermediate value property on [tex]$\left[x_0-\delta, x_0+\delta\right]$[/tex],

Hence f does not satisfies IVP on (a, b) which is not possible because we assume f satisfies IVP on (a, b),

Therefore f can not have a jump discontinuity.

Suppose f has a removable point of discontinuity at [tex]$x_0 \in(a, b)$[/tex],

Let [tex]$\alpha=\lim _{\alpha \rightarrow x_0} f(x)$[/tex],

Let [tex]\alpha < f\left(x_0\right)$[/tex] so [tex]$f\left(x_0\right)-\alpha > 0$[/tex].

Now [tex]$\lim _{x \rightarrow x_0} f(x)=\alpha$[/tex] then [tex]\exists$ \delta > 0$[/tex] such that

[tex]$$\begin{aligned}& |f(x)-\alpha| < \frac{f\left(x_0\right)-\alpha}{2} \text { for all } x \in\left\{x_0-\delta, x_0-\alpha\right]-\left\{x_0\right\} \\& \Rightarrow \quad f(x) < \alpha+\frac{f\left(x_0\right)-\alpha}{2} \text { for all } x \in\left[x_0-\delta, x_0+\delta\right]-\left\{x_0\right\}\end{aligned}$$[/tex]

So [tex]$f(x) < \frac{f\left(x_0\right)+\alpha}{2}$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0\right]-\left\{x_0\right\}$[/tex]

Now [tex]$f\left(x_0\right) > \alpha$[/tex].

And  [tex]$f(x) < \frac{f\left(x_0\right)+\alpha}{2} < f\left(x_0\right)$[/tex] for all [tex]$x \in\left[\left(x_0 \delta, x_0\right)\right.$[/tex]

Let [tex]$\mu=\frac{f\left(x_0\right)+\alpha}{2}$[/tex].

Then there does not exist [tex]$e \in\left[x_0-\delta, c\right]$[/tex] such that [tex]$f(c)=\mu$[/tex]

Because for [tex]$e=x_0 \quad f(e) > \mu$[/tex]

                for [tex]$c < x_0 \quad f(c) < \mu$[/tex].

Therefore f does not satisfy IVP on [tex]$\left[x_0-\delta_1 x_0\right]$[/tex] which contradict our hypothesis,

therefore [tex]$\alpha \geqslant f\left(x_0\right)$[/tex]

Let [tex]$\alpha > f\left(x_0\right)$[/tex]. so [tex]$\alpha-f\left(x_0\right) > 0$[/tex]

[tex]$\lim _{x \rightarrow x_0} f(x)=\alpha$[/tex]

Then [tex]\exists $ \varepsilon > 0$[/tex] such that

[tex]$|f(x)-\alpha| < \frac{\alpha-f\left(x_0\right)}{2}$[/tex] for all [tex]$\left.x \in\left[x_0-\varepsilon_0 x_0+\varepsilon\right]\right\}\left\{x_i\right\}$[/tex]

[tex]$\Rightarrow f(x) > \alpha-\frac{\alpha-f\left(x_0\right)}{2}$[/tex] for all [tex]$x \in\left[x_0-\varepsilon_1, x_0+\varepsilon\right] \backslash\left\{x_0\right\}$[/tex]

[tex]$\Rightarrow f(x) > \frac{\alpha+f\left(x_0\right)}{2}$[/tex] for all [tex]$x \in\left[x_0-\varepsilon_1, x_0\right)$[/tex]

Now [tex]$f\left(x_0\right) < \alpha$[/tex]

The [tex]$f(x) > \frac{f\left(x_0\right)+\alpha}{2} > f\left(x_0\right)$[/tex].

So [tex]$f\left(x_0\right) < \frac{f\left(x_0\right)+\alpha}{2} < f(x)$[/tex] for all [tex]$x \in\left[x_0 \varepsilon, \varepsilon_0\right)$[/tex]

Let [tex]$\eta=\frac{f\left(x_e\right)+\alpha}{2}$[/tex]

Then there does not exist [tex]$d \in\left[x_0-\varepsilon, x_0\right]$[/tex] such that [tex]$f(d)=\xi$[/tex].

Because if [tex]$d=x_0, f(d)=f\left(x_0\right) < \eta$[/tex] if [tex]$d E\left[x_0-\varepsilon, x_0\right)$[/tex]

Then [tex]$f(d) > \eta$[/tex]

Therefore f does not satisfies IVP on [tex]$\left[x_0-\varepsilon, x_0\right]$[/tex] which contradict olio hypothesis.

Therefore [tex]$\alpha \leq f\left(x_0\right)$[/tex] (b) From (a) and (b) it follows [tex]$\alpha=f\left(x_0\right)=\lim _{x \rightarrow x_0} f(x)$[/tex]. Therefore f can not have a removable discontinuous

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