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Vsevolod [243]
3 years ago
5

HELP PLZ I BEG U ASAP How many 4-digit numbers are neither multiples of 2 nor multiples of 5?

Mathematics
1 answer:
ruslelena [56]3 years ago
5 0
In this problem, we must do a permutation.

We'll answering this using something called the "counting principle" (probability).

Our 4-digit number can start with any digit that is not zero (if we assume examples like 0123 or 0009 are not 4 digit numbers, hence assume that leading 0s are not digits), further more, the other restriction in the number will be that it does not end in an even number or in 5; this means there are only 4 possible last digits in the number (i.e., 1,3,7, and 9).

Then, by the counting principle the answer is (9*10*10*4)=3600.
There are 4000 4-digit numbers that are not multiples of 2 nor 5.
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Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices
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Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt

Where P, Q are scalar functions

We want to compute

\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy

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a) Directly

Let us break down C into 4 paths \large C_1,C_2,C_3,C_4 which represents the sides of the rectangle.

\large C_1 is the line segment from (0,0) to (5,0)

\large C_2 is the line segment from (5,0) to (5,1)

\large C_3 is the line segment from (5,1) to (0,1)

\large C_4 is the line segment from (0,1) to (0,0)

Then

\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize \large C_1 as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

\large \displaystyle\int_{C_1}xydx+x^2dy=0

 Now the second integral. We parametrize \large C_2 as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25

The third integral. We parametrize \large C_3 as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2

The fourth integral. We parametrize \large C_4 as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

\large \displaystyle\int_{C_4}xydx+x^2dy=0

So

\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2

Now, let us compute the value using Green's theorem.

According with this theorem

\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx

where A is the interior of the rectangle.

so A={(x,y) |  0≤ x≤ 5,  0≤ y≤ 1}

We have

\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x

so

\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2

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horrorfan [7]

The LCD of the rational expressions are 30x^5, (3x - 1)(x + 6) and (x + 3)(x + 5)^2

<h3>How to determine the LCD of the rational expressions?</h3>

<u>Expression 1</u>

The rational expressions are given as:

3/10x^2 and x + 7/15x^5

Write out the denominators

10x^2 and /15x^5

Expand each of the denominator.

10x^2 = 2 * 5 * x * x

15x^5 = 3 * 5 * x* x * x * x * x

Multiply all common factors without repetition

So, the LCD of the denominators are

LCD = 2 * 3 * 5 * x* x * x * x * x

Evaluate

LCD = 30x^5

<u>Expression 2</u>

The rational expressions are given as:

9/3x - 1 and 2x/x + 6

Write out the denominators

3x - 1 and x + 6

Expand each of the denominator.

10x^2 = 3x - 1

15x^5 = x + 6

Multiply all common factors without repetition

So, the LCD of the denominators are

LCD = (3x - 1)(x + 6)

<u>Expression 3</u>

The rational expressions are given as:

8x/(x + 5)^2 and 4x + 1/x^2 + 8x + 15

Write out the denominators

(x + 5)^2 and x^2 + 8x + 15

Expand each of the denominator.

(x + 5)^2 = (x + 5) * (x + 5)

x^2 + 8x + 15 = (x + 3) * (x + 5)

Multiply all common factors without repetition

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LCD = (x + 3)(x + 5)^2

Hence, the LCD of the rational expressions are 30x^5, (3x - 1)(x + 6) and (x + 3)(x + 5)^2

Read more about LCD at:

brainly.com/question/1025735

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