Please help!!!!!!!!!!!!!!

If \($f$\)is bounded and \($\displaystyle\lim_{x \rightarrow c} g(x)$\) exists, then the limit of function\($f(x)g(x)$\)as \($x$\) approaches domain\($c$\)also exists and is equal to \($Lf(c)$\)

Let \($M$\)be an upper bound of \($f$\) on the domain . Since \($\displaystyle\lim_{x \rightarrow c} g(x)$\) exists, there exists a number \($L$\) such that for all \($\epsilon > 0$\) there exists a \($\delta > 0$\) such that for \($x \in A$\) with \($0 < |x - c| < \delta |g(x) - L| < \epsilon\).

Now let \(\epsilon > 0Then $|f(x)g(x) - Lf(x)| = |f(x)||g(x) - L| < M\epsilon\)

for all \(x \in A$ with $0 < |x - c| < \delta$\). So \(\displaystyle\lim_{x \rightarrow c} f(x)g(x)$ exists and is equal to $Lf(c)\)

If \($f$\) is bounded and \($\displaystyle\lim_{x \rightarrow c} g(x)$\) exists, then the limit of \(f(x)g(x)$\)as \($x$\)approaches\($c$\)also exists and is equal to \($Lf(c)$\).

The complete question is:

Let \($f$\) and \($g$\) be functions defined on a domain\($A$\). Prove that if \($f$\) is bounded, and \($\displaystyle\lim_{x \rightarrow c} g(x)$\)exists, then \($\displaystyle\lim_{x \rightarrow c} f(x)g(x)$\) exists.

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Answer:

I h ok honestly dont know that I'm so sorry

Jessica has three dice with 10 sides, 13 sides and 5 sides respectively and she rolls all dice once and if Z is number of ones occur then

The expectation value i.e, E(Z) is 0.376 and variance V[Z] = 0.321..

Indicator method :

This is a powerful way to find expected counts. This follows from the observation that the number of "good" outcomes in a trial can be counted by first coding each "good" outcome as one, coding each other outcome as zero, and then adding 1 and 0.

let X₁ be the outcome of a 10 sided die

X₂ be the outcome of 13 sided die

X₃ be the outcome of 5 sided die

let l₁ ,I₂ ,I₃ be three indicator functions such that

I₁ =1 if X₁ = 1

=0 otherwise

I₂ = 1 if X₂ = 1

=0 otherwise

I₃ = 1 if X₃ = 1

=0 otherwise

so Z denotes the number of ones showing when the three dice are rolled once.

so Z=I₁ +I₂ +I₃

so E[Z]=E[I₁]+E[I₂]+E[I₃]

so E[I₁]=1×P[X₁=1] = 1/10

E[I₂]=1×P[X₂=1] = 1/13

E[I₃]=1×P[X₃=1] = 1/5

so, E[Z]=1/10+1/13+1/5=49/130 = 0.3769

the variance = V[Z] = V[I₁]+V[I₂]+V[I₃] since the outcomes of the 3 different die are independent to each other hence no covariance term.

V[I₁]= 1²P[X₁=1]- E²[I₁]=1/10 -(1/10)²= 0.09

V[I₂]= 1²P[X₂=1]- E²[I₂]=1/13-(1/13)²= 12/169 = 0.071

V[I₃]= 1²P[X₃=1]- E²[I₃]=1/5 -(1/5)²= 4/25 = 0.16

so V[Z]= 0.09 + 0.071 + 0.16 = 0.321

Hence, the expectation value i.e, E(Z) is 0.3769..

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The perimeter of the entire sector is 24 feet.

What is circle?

A circle is a geometric shape that consists of all points in a plane that are equidistant from a fixed point called the center.

The perimeter of the entire sector is equal to the sum of the arc length and the two radii.

Since the central angle of the sector measures 1 radian, the arc length of the sector can be calculated using the formula:

arc length = radius x central angle

arc length = 8 ft x 1 radian

arc length = 8 ft

The length of each radius is equal to the radius of the sector, which is given to be 8 ft.

Therefore, the perimeter of the entire sector is:

perimeter = arc length + 2 x radius

perimeter = 8 ft + 2 x 8 ft

perimeter = 24 ft

So, the perimeter of the entire sector is 24 feet.

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Answer:

p=‒1

Step-by-step explanation:

the answer is p=‒1

because ‒1+1/3=‒1+1/4

0/3 =0/4

0=0

i hope this helps you

Answer:

the answer is p=‒1

because ‒1+1/3=‒1+1/4

0/3 =0/4

0=0

Step-by-step explanation:

Answer:

60

Step-by-step explanation:

double 50 is 100 half of 80 is 40 answer is 60

hope this helps

Answer:

In order (Problem 1-4)

B

C

A

D

Step-by-step explanation:

These will give you 100%! Wish I had these before I took the quick check.

Answer:

The answer is b

Step-by-step explanation:

The whole test is

B

C

A

D

C

since we have our two medians, we need to add them up 1st then divide the sum by two;

50+60

    2

= 110

    2

= 55

55 is the median

3,200 + 560 + 16 √

(400 × 8) + (70 × 8) + (2 × 8) √

(400 × 8) × (70 × 8) × (2 × 8) X

3,776 √

3,200 × 560 × 16 √

(400 + 70 + 2) × 8 √

(400 × 70 × 2) × 8 X

28,672,000 X

\( \)

In the image, we can identify the region of the plane in which the two subregions overlap (the part of the graph which has vertical and horizontal lines). We can obtain the values of x and y in that region by means of the inequalities, in this way:

We solve the first inequality for y

This gives us the first constrain for the value of y

Now, to obtain the second constrain we simply need to use the second inequality:

Finally, the last step is to get the constraints for the value of x.

For this, we can directly analyze the image of the region, notice how it covers any value of x. This means that there are no constraints for the value of x.

So, the region is given by:

Finally, the point that is in the region is the one that satisfies the previous inequalities.

1. (2,0)

If x=2, then

So, (2,0) cannot be the correct option

2. (0,2)

This means that (0,2) cannot be the right option

3. (0,3) notice that the same condition (y>2) holds in this situation too (since x=0)

But y=3>2. This point indeed satisfies the conditions! (0,3) is the answer

4. (0,-1) Notice that in this case y=-1 but it should happen that y>2. So this point cannot be in the region

The given integral is: ∫∫R 5/√(2) xy dA, where R is the region in the xy-plane bounded by the curves y = 0, x = √(25 - y^2) and x = 0. The volume of the solid is (5^5/8) cubic units.

Converting to polar coordinates, we have:

x = r cos(θ), y = r sin(θ), and the limits of integration become:

0 ≤ θ ≤ π/2, 0 ≤ r ≤ 5.

Also, the differential of area becomes dA = r dr dθ.

Substituting for x and y, and dA, we have:

∫∫R 5/√(2) xy dA = ∫θ=0π/2 ∫r=0^5 5/√(2) (r cos(θ)) (r sin(θ)) r dr dθ

= 5/√(2) ∫θ=0π/2 ∫r=0^5 r^3 cos(θ) sin(θ) dr dθ

= 5/√(2) ∫θ=0π/2 [sin(θ)/4] ∫r=0^5 r^4 cos(θ) dr dθ

= 5/√(2) ∫θ=0π/2 [sin(θ)/4] [(5^5 cos(θ))/5] dθ

= (5^5/4√(2)) ∫θ=0π/2 sin(θ) cos(θ) dθ

= (5^5/8)

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Answer: 83r^2 - 59 + 60

Step-by-step explanation:

R^2 - 59R - 82r^2 + 60

= 83r^2 - 59 + 60

This is true because a probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment.

The area under the probability distribution is a measure of the probability of all the possible outcomes. Since the sum of all the probabilities of all the possible outcomes must be 1, the total area under the probability distribution must also be 1. This is because the probability of an event happening is the area under the curve of the probability distribution at that event. Thus, the total area under the probability distribution must always equal 1.

This is true because a probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment.

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Answer:

Robin’s median score is 107. Evelyn’s median score is 138. The medians are the same as the means, so the same conclusion would be reached that Robin is winning

Step-by-step explanation:

Answer:

Robin’s median score is 107. Evelyn’s median score is 138. The medians are the same as the means, so the same conclusion would be reached that Robin is winning.

In solving the following fraction, step (A) (4/7) × (-2/-1) is wrong and the correct step should be (4/7) × (-2/1).

So, the wrong step will be:

Then, the next step should be:

But, the steps it is given as:

Which makes the error.

Therefore, in solving the following fraction, step (A) (4/7) × (-2/-1) is wrong and the correct step should be (4/7) × (-2/1).

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Answer:

16.5 should be correct because OS is half the length.

The Fundamental Theorem of Algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This theorem is important as it provides a way to prove the existence of solutions to polynomial equations and provides an analytical tool to find the exact location of the solutions.

This theorem is also known as the algebraic version of the Intermediate Value Theorem as it states that if a polynomial is continuous on a closed interval, then it must take on all values between its maximum and minimum.

The theorem can be easily proven by considering a single-variable polynomial of degree n and transforming it into a polynomial of degree n−1 with the same roots. By repeating this process, the polynomial can be reduced to a constant and hence, it must have at least one root.

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Answer:

Step-by-step explanation:

x = 180 - 72 = 108° (linear pair)

y = 72° (vertically opposite angles)

z = 180 - (90+72) = 180 - 162 = 18°

Answer:

Step-by-step explanation:

we have

\(y=2x2-8x+9\)

Group terms that contain the same variable, and move the constant to the opposite side of the equation

\(y-9=2x2-8x\)

\(y-9=2(x2-4x)\)

\(y-9+8=2(x2-4x+4)\)

\(y-1=2(x2-4x+4)\)

\(y-1=2(x-2)2\)

\(y=2(x-2)2+1\)

the answer is the option C

\(y=2(x-2)2+1\)

The equation  y = 2x² – 8(x + 9) can be rewritten in vertex form as y = 2(x – 2)² + 1 option (C) is correct.

It is defined as the graph of a quadratic function that has something bowl-shaped.

(x - h)² = 4a(y - k)

(h, k) is the vertex of the parabola:

a = √[(c-h)² + (d-k²]

(c, d) is the focus of the parabola:

It is given that:

The equation is:

y = 2x² – 8(x + 9)

y = 2x² - 8x + 8 + 1

y = 2(x² - 4x + 4) + 1

y = 2(x – 2)² + 1

Thus, the equation  y = 2x² – 8(x + 9) can be rewritten in vertex form as y = 2(x – 2)² + 1 option (C) is correct.

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The midsegment of AD (which in this case is DE) always generates two congruent parts (which in this case are AD and DB)

Answer:

60x+300=96x-24

324=36x

x=324/36

x=9

yes it is correct

The solution to the given separable differential equation for u, with the initial condition u(0) = 7.

To solve the separable differential equation for u, we start by rearranging the equation f as:

(1/u) du/dt = e^(3u)/u + 10t/u

We can now integrate both sides of the equation with respect to t and u, separately. Starting with the left-hand side, we have:

∫(1/u) du = ln|u| + C1

where C1 is the constant of integration. For the right-hand side, we can use u-substitution by letting v = 3u, dv/du = 3, and du/dv = 1/3u. Substituting these values into the equation f and simplifying, we have:

(1/3) ∫e^v dv = (1/3) e^v + C2

where C2 is another constant of integration. Substituting v = 3u back into the equation and combining the constants of integration, we get:

ln|u| = e^(3u)/3 + 10t/3 + C

where C = C1 + C2. To solve for u, we exponentiate both sides of the equation:

|u| = e^(e^(3u)/3 + 10t/3 + C)

We can drop the absolute value since u(0) = 7 > 0, and simplify the exponential expression by using the properties of exponents:

u = e^(e^(3u)/3) * e^(10t/3 + C)

Finally, we use the initial condition u(0) = 7 to solve for C:

7 = e^(e^(3(7))/3) * e^(10(0)/3 + C)
7 = e^(e^21/3) * e^C
ln(7/e^(e^21/3)) = C

Substituting this value of C back into the equation for u, we get:

u = e^(e^(3u)/3) * e^(10t/3 + ln(7/e^(e^21/3)))

This is the solution to the given separable differential equation for u, with the initial condition u(0) = 7.

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Answer:

18x + 6

Step-by-step explanation:

In a rectangle, the opposite side lengths are congruent and have the same measurement. In solving the perimeter, add all sides together:

(4x + 7) + (4x + 7) + (5x - 4) + (5x - 4)

Combine like terms (terms with the same variable, as well as same amount of variable):

(4x + 4x + 5x + 5x) + (7 + 7 - 4 - 4)

(8x + 10x) + (14 - 8)

18x + 6

18x + 6 is your simplified expression that represents the perimeter of the rectangle.

~

(b) 5.Let's first express 1 + 2 + ... + n as the sum of an arithmetic series:

1 + 2 + ... + n = n(n+1)/2

Now, we need to find the positive integers n for which 6n is divisible by n(n+1)/2, or equivalently, for which n+1 is divisible by 3.

If n+1 is divisible by 3, then either n+1 = 3, 6, 9, ... or n+1 = 4, 7, 10, .... The first case gives n = 2, 5, 8, ... and the second case gives n = 3, 6, 9, ....

Thus, there are 5 possible values of n: 2, 3, 5, 6, and 9.

An arithmetic series is a sequence of numbers where each term is obtained by adding a constant value to the previous term. This constant value is called the common difference. For example, the sequence 2, 5, 8, 11, 14, ... is an arithmetic series with a common difference of 3. The formula for the nth term of an arithmetic series is: a_n = a_1 + (n-1)d, where a_n is the nth term, a_1 is the first term, n is the position of the term in the sequence, and d is the common difference. The sum of the first n terms of an arithmetic series can be calculated using the formula: S_n = n/2(a_1 + a_n).

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All the correct expression in each steps are,

⇒ Step 1; (m³ + m²n + mn² - m²n - mn² - n³)

( By multiplication distributor )

⇒ Step 2;  m³ - n³

( By combine like terms)

To multiply means to add a number to itself a particular number of times. Multiplication can be viewed as a process of repeated addition.

Given that;

The expression to prove is,

⇒ (m - n) (m² + mn + n²)

Now, We can simplify as;

⇒ (m - n) (m² + mn + n²)

Step 1;

⇒ (m³ + m²n + mn² - m²n - mn² - n³)

( By multiplication distributor )

Step 2;

⇒ m³ - n³

( By combine like terms)

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There are 7 months in the year that have thirty-one days.

These months are January, March, May, July, August, October, and December.

There are seven months in the Gregorian calendar that have thirty-one days: January, March, May, July, August, October, and December.

This pattern of months with 31 days followed by months with fewer days repeats throughout the year.

This pattern was established by the Roman calendar, which had ten months totaling 304 days in a year.

The months of January and February were later added by King Numa Pompilius to align the calendar with the lunar year.

The months of January and February initially had 29 and 28 days respectively, but in 45 BC, Julius Caesar added one day to January and one day to August, which was originally a 30-day month, to make them both 31-day months.

In the Gregorian calendar, which is the most widely used calendar in the world, January, March, May, July, August, October, and December all have 31 days.

The remaining five months have fewer days, with February having 28 days most of the time, and 29 days in a leap year.

Knowing the number of days in each month is important for various reasons, such as planning events, scheduling appointments, and calculating pay periods.

There are several mnemonics used to remember the number of days in each month, such as "30 days hath September, April, June, and November, all the rest have 31, except February, with 28 days clear, and 29 in each leap year."

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For any ε > 0, there exists a δ > 0 such that for any (a, c) in A x C, if ||(a, c) - (a_0, c_0)|| < δ, then ||(f(a), g(c)) - (f(a_0), g(c_0))|| < ε. This shows that the map f x g is continuous.

To show that the map f x g : A x C -> B x D is continuous, we need to demonstrate that it preserves the continuity of the individual functions f and g.

Let (a_0, c_0) be an arbitrary point in A x C, and let ε > 0 be given. Since f is continuous, there exists δ_1 > 0 such that for any point a in A, if ||a - a_0|| < δ_1, then ||f(a) - f(a_0)|| < ε/2.

Similarly, since g is continuous, there exists δ_2 > 0 such that for any point c in C, if ||c - c_0|| < δ_2, then ||g(c) - g(c_0)|| < ε/2.

Now, let δ = min(δ_1, δ_2). Consider any point (a, c) in A x C such that ||(a, c) - (a_0, c_0)|| < δ. This implies that both ||a - a_0|| < δ and ||c - c_0|| < δ, which in turn implies that ||f(a) - f(a_0)|| < ε/2 and ||g(c) - g(c_0)|| < ε/2.

By the triangle inequality, we have: ||(f(a), g(c)) - (f(a_0), g(c_0))|| = ||(f(a) - f(a_0), g(c) - g(c_0))|| <= ||f(a) - f(a_0)|| + ||g(c) - g(c_0)|| < ε/2 + ε/2 = ε.

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