ở đây có ai giỏi kế toán tài chính 1 và kế toán tài chính 2 không ạ ????

Answer:

0

Step-by-step explanation:

The slope of the line is m=(9-9)/(-2-5)=0/-7=0

Answer:

Step-by-step explanation:

[tex]6 \div \frac{3}{8} = 6 \times \frac{8}{3} \\ [/tex]

[tex] = \frac{6 \times 8}{3} \\ \\ = \frac{48}{3} =1 6[/tex]

Split up the boundary of C (which I denote ∂C throughout) into the parabolic segment from (1, 1) to (0, 0) (the part corresponding to y = x ²), and the line segment from (1, 1) to (0, 0) (the part of ∂C on the line y = x).

Parameterize these pieces respectively by

r(t) = x(t) i + y(t) j = t i + t ² j

and

s(t) = x(t) i + y(t) j = (1 - t ) i + (1 - t ) j

both with 0 ≤ t ≤ 1.

The circulation of F around ∂C is given by the line integral with respect to arc length,

[tex]\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T \,\mathrm ds[/tex]

where T denotes the tangent vector to ∂C. Split up the integral over each piece of ∂C :

• on the parabolic segment, we have

T = dr/dt = i + 2t j

• on the line segment,

T = ds/dt = -i - j

Then the circulation is

[tex]\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(\mathbf i+2t\,\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i-\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (7t^3+10t^5)\,\mathrm dt - 12 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{-\frac7{12}}[/tex]

Alternatively, we can use Green's theorem to compute the circulation, as

[tex]\displaystyle\int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \iint_C\frac{\partial(5y^2)}{\partial x} - \frac{\partial(7xy)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = -7\int_0^1\int_{x^2}^x x\,\mathrm dx \\\\ = -7\int_0^1 xy\bigg|_{y=x^2}^{y=x}\,\mathrm dx \\\\ =-7\int_0^1(x^2-x^3)\,\mathrm dx = -\frac7{12}[/tex]

The flux of F across ∂C is

[tex]\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N \,\mathrm ds[/tex]

where N is the normal vector to ∂C. While T = x'(t) i + y'(t) j, the normal vector is N = y'(t) i - x'(t) j.

• on the parabolic segment,

N = 2t i - j

• on the line segment,

N = - i + j

So the flux is

[tex]\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(2t\,\mathbf i-\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i+\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (14t^4-5t^4)\,\mathrm dt - 2 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{\frac{17}{15}}[/tex]

Answer:

Step-by-step explanation:

[tex]\boxed{\sf a(b+c)=ab+ac}[/tex]

[tex]\\ \sf\longmapsto 2396\times 78 +2396\times 22[/tex]

[tex]\\ \sf\longmapsto 2396(78+22)[/tex]

[tex]\\ \sf\longmapsto 2396(100)[/tex]

[tex]\\ \sf\longmapsto 239600[/tex]

Answer:

Step-by-step explanation:

10/13

Answer: 10/13

Explanation:

= 7/13 + 3/13

Its like fraction

= 10/13

Its already in lowest terms

Must click thanks and mark brainliest

Answer:

Step-by-step explanation:

(17). g(x) = x³ + 4x

f(x) = 4x + 1

( f × g )( x ) = ( x³ + 4x )( 4x + 1 ) = 4 [tex]x^{4}[/tex] + x³ + 16x² + 4x

(19). f(t) = 4t - 4

g(t) = t - 2

( 4f + 3g )( t ) = 4(4t - 4) + 3(t - 2) = 16t - 16 + 3t - 6 = 19t - 22

(21). h(t) = t + 3

g(t) = 4t + 1

h(t - 2) + g(t - 2) = ( t - 2 ) + 3 + 4( t - 2 ) + 1 = t + 4t - 2 + 3 - 8 + 1 = 5t - 6

Answer:

10th term

Step-by-step explanation:

The equation of the arithmetic sequence is an=-7+(n-1)*1=-8+n, plugging in 2 and solving for n we have

2=-8+n, n=10

Answer:

spread over a wide range

cannot be evaded

Everyone can contribute

convenient

Step-by-step explanation:

[tex]\\ \sf\longmapsto 4x+4x+5x-7+5x+2+96=540[/tex]

[tex]\\ \sf\longmapsto 4x+4x+5x+5x-7+2+96=540[/tex]

[tex]\\ \sf\longmapsto 8x+10x-5+96=540[/tex]

[tex]\\ \sf\longmapsto 18x+91=540[/tex]

[tex]\\ \sf\longmapsto 18x=540-91=449[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{449}{18}[/tex]

[tex]\\ \sf\longmapsto x\approx2.4[/tex]

Answer:

x=25

Step-by-step explanation:

4x+5x-8+96+5x+2+4x=540

18x+90=540

18x=540-90=450

x=450/18=25

Answer:

56 half dollars

Step-by-step explanation:

A half dollar is 1/2 of a dollar so divide 28 by 1/2

28 / 1/2

Copy dot flip

28 * 2/1

56

56 half dollars

[tex]\\ \sf\longmapsto \left\{x|x \epsilon I,x\leqslant 3\right\}[/tex]

In listing

[tex]\\ \sf\longmapsto \left\{\dots,0,1,2,3\right\}[/tex]

3.99-1 = 2.99 they lose 2.99 per burger

Multiply the number of people by the amount lost per burger:

2.99 x 86,047 = 257,280.53

They would lose: $257,280.53

Answer:

86043.01

Step-by-step explanation:

The vertex can be written as:

(-b/2a, b^2/(4*a) - b^2/2a + c)

For a general parabola:

y = a*x^2 + b*x + c

We can write the vertex as:

(h, k)

The x-value of the vertex is the value of the axis of symmetry.

Then we have:

h = x = -b/2a

Now we need to find the y-value of the vertex.

To do that, we just replace the variable "x" by the x-value of the vertex in our equation, so we get:

k = y = a*(-b/2a)^2 + b*(-b/2a) + c

k = b^2/(4*a) - b^2/2a + c

Then the coordinates of the vertex are:

(h, k) = (-b/2a, b^2/(4*a) - b^2/2a + c)

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Answer:

12/5

Step-by-step explanation:

since we are dealing with the angle at Z, we consider Z also to be the center of a circle.

then 10 is the radius of this circle.

24 is tan Z in this circle with radius 10.

tan Z in the standard circle with radius 1 is simply 1/10 of the tan Z in the circle with radius 10.

so, tan Z = 24/10 = 12/5

Answer:

a) -3/13

b) -1/8

Step-by-step explanation:

a)  - (21 / 7) / (91 / 7) = 3/13

b) (32 / 32 ) / - (256 / 32) = -1/8​

Answer:

C

Step-by-step explanation:

Let the first term be a and the common ratio be r.

ATQ, ar^4=24 and ar^6=144, r=sqrt(6) and a=24/(sqrt(6))^2=24/36=2/3

Answer:

Option B, parallel

Step-by-step explanation:

for the first line,

[3-(-1)]/[11-9]

= 4/2 = 2

for the second line,

(0-(-4))/(-4-(-6))

= 4/2 = 2

Both has same slope so they're parallel but it doesn't seem like they are the same line

Answer:

D

Step-by-step explanation:

since

cos(2x) = 1 - 2×sin²(x)

and for x near 0, sin(x) is very similar to x.

Answer:

Option E, 4, 4√3

Step-by-step explanation:

s = 4

q = 4√3

Step-by-step explanation:

slope of given line:

m=2

as lines are parallel so slopes will be equal:

required slope m=2

By using point slope form:

y-y1=m(x-x1)

y-5=2(x-5)

y-5=2x-10

y=2x-10+5

y=2x-5

Note:if you need to ask any question please let me know.

To solve this question, we have to understand the sum of all angles of a polygon and identify the polygon, which is classified according to the number of sides, getting that, since the polygon has 13 sides, it is a tridecagon.

-----------------------------

Sum of angles:

The sum of angles of a polygon of n sides is given by:

[tex]S_n = 180(n-2)[/tex]

-----------------------------

Regular polygon, with interior angles of 152.31∘.

In a regular polygon, all of the n angles have the same measure, which means that the sum of the angles is:

[tex]S_n = 152.31n[/tex]

-----------------------------

Finding n:

To classify the polygon, we have to find n, which we do equaling the two equations for [tex]S_n[/tex]. Then

[tex]180(n-2) = 152.31n[/tex]

[tex]180n - 152.31n = 360[/tex]

[tex]27.69n = 360[/tex]

[tex]n = \frac{360}{27.69}[/tex]

[tex]n = 13[/tex]

-----------------------------

Since the polygon has 13 sides, it is a tridecagon.

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9514 1404 393

Answer:

  see below

Step-by-step explanation:

  [tex]\text{A. point: }X\\\\\text{B. ray through Y: }\overrightarrow{XY}\\\\\text{C. line through Z: }\overleftrightarrow{XZ}\\\\\text{D. plane: plane } XYZ[/tex]

__

Additional comment

When you don't have the benefit of typesetting, you can refer to the geometry by name: ray XY, line XZ,

Answer:

Hello,

Step-by-step explanation:

Q1:

[tex]\left\{\begin{array}{ccc}x&=&t+\dfrac{1}{t} \\\\y&=&t-\dfrac{1}{t} \\\end{array}\right.\\\\\left\{\begin{array}{ccc}x^2&=&t^2+\dfrac{1}{t^2} +2\\\\y^2&=&t^2+\dfrac{1}{t^2} -2\\\end{array}\right.\\\\\\x^2-y^2=4: \ equilater\ hyperbola.\\[/tex]

Q2:

1)

[tex]\left\{\begin{array}{ccc}x&=&2t^2} \\\\y&=&4t \\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}t&=&\dfrac{y}{4} \\\\x&=&2*(\dfrac{y}{4})^2 \\\end{array}\right.\\\\\\\boxed{x=\dfrac{y^2}{8}} :\ parabola\ with\ x-axis\ as\ axis\ of\ symmetry[/tex]

2)

[tex]y=\dfrac{25}{x} \\[/tex]

equilater hyperbola (centre (0,0))

Answer:

First, let's find how far away the pier is.

Using the distance formula, we can see that the pier is [tex]\sqrt{58}[/tex] units away.

So, the radius is sqrt 58.

Area = pi (r)^2

So,  the area is 182.82 square units.

Let me know if this helps!

We have that The area that the light covers is  is mathematically given as

[tex]A=\pi x^2[/tex]

From the Question we are told that

Revolving beam of light as far as the pier

Let distance to pier be x

Generally the revolving beam turns a complete angle of 360

Therefore

Its goes in a circle

The area that the light covers is  is mathematically given as

[tex]A=\pi r^2[/tex]

[tex]A=\pi x^2[/tex]

In conclusion

The area that the light covers is  is mathematically given as

[tex]A=\pi x^2[/tex]

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The first, second, third and fourth order Maclaurin polynomials of f(x)=arctan(x) are:

So let's start by finding the first order maclaurin polynomial:

f(x)=f(0)+f'(0)x

so let's find each part of the function:

f(0)=arctan(0)

f(0)=0

now, let's find the first derivative of f(x)

f(x)=arctan(x)

This is a usual derivative so there is a rule we can use here:

[tex]f'(x)=\frac{1}{x^{2}+1}[/tex]

so now we can find f'(0)

[tex]f'(0)=\frac{1}{(0)^{2}+1}[/tex]

f'(0)=1

So we can now complete the first order Maclaurin Polynomial:

f(x)=0+1x

which simplifies to:

f(x)=x

Now let's find the second order polynomial, for which we will need to get the second derivative of the function:

[tex]f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}[/tex]

so:

[tex]f'(x)=\frac{1}{x^{2}+1}[/tex]

we can rewrite this derivative as:

[tex]f'(x)=(x^{2}+1)^{-1}[/tex]

and use the chain rule to get:

[tex]f''(x)=-1(x^{2}+1)^{-2}(2x)[/tex]

which simplifies to:

[tex]f''(x)=-\frac{2x}{(x^{2}+1)^{2}}[/tex]

now, we can find f''(0):

[tex]f''(0)=-\frac{2(0)}{((0)^{2}+1)^{2}}[/tex]

which yields:

f''(0)=0

so now we can complete the second order Maclaurin polynomial:

[tex]f(x)=0+1x+\frac{0}{2!}x^{2}[/tex]

which simplifies to:

f(x)=x

Now let's find the third order polynomial, for which we will need to get the third derivative of the function:

[tex]f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+\frac{f'''(0)}{3!}x^{3}[/tex]

so:

[tex]f''(x)=-\frac{2x}{(x^{2}+1)^{2}}[/tex]

In this case we can use the quotient rule to solve this:

Quotient rule: Whenever you have a function in the form , then it's derivative is:

[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]

in this case:

p=2x

p'=2

[tex]q=(x^{2}+1)^{2}[/tex]

[tex]q'=2(x^{2}+1)(2x)[/tex]

[tex]q'=4x(x^{2}+1)[/tex]

So when using the quotient rule we get:

[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]

[tex]f'''(x)=\frac{(2)(x^{2}+1)^{2}-(2x)(4x)(x^{2}+1)}{((x^{2}+1)^{2})^{2}}[/tex]

which simplifies to:

[tex]f'''(x)=\frac{-2x^{2}-2+8x^{2}}{(x^{2}+1)^{3}}[/tex]

[tex]f'''(x)=\frac{6x^{2}-2}{(x^{2}+1)^{3}}[/tex]

now, we can find f'''(0):

[tex]f'''(0)=\frac{6(0)^{2}-2}{((0)^{2}+1)^{3}}[/tex]

which yields:

f'''(0)=-2

so now we can complete the third order Maclaurin polynomial:

[tex]f(x)=0+1x+\frac{0}{2!}x^{2}-\frac{2}{3!}x^{3}[/tex]

which simplifies to:

[tex]f(x)=x-\frac{1}{3}x^{3}[/tex]

Now let's find the fourth order polynomial, for which we will need to get the fourth derivative of the function:

[tex]f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+\frac{f'''(0)}{3!}x^{3}+\frac{f^{(4)}(0)}{4!}x^{4}[/tex]

so:

[tex]f'''(x)=\frac{6x^{2}-2}{(x^{2}+1)^{3}}[/tex]

In this case we can use the quotient rule to solve this:

[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]

in this case:

[tex]p=6x^{2}-2[/tex]

p'=12x

[tex]q=(x^{2}+1)^{3}[/tex]

[tex]q'=3(x^{2}+1)^{2}(2x)[/tex]

[tex]q'=6x(x^{2}+1)^{2}[/tex]

So when using the quotient rule we get:

[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]

[tex]f^{4}(x)=\frac{(12x)(x^{2}+1)^{3}-(6x^{2}-2)(6x)(x^{2}+1)^{2}}{((x^{2}+1)^{3})^{2}}[/tex]

which simplifies to:

[tex]f^{4}(x)=\frac{12x^{3}+12x-6x^{3}+12x}{(x^{2}+1)^{4}}[/tex]

[tex]f^{4}(x)=\frac{6x^{3}+24x}{(x^{2}+1)^{4}}[/tex]

now, we can find f^{4}(0):

[tex]f^{4}(x)=\frac{6(0)^{3}+24(0)}{((0)^{2}+1)^{4}}[/tex]

which yields:

[tex]f^{4}(0)=0[/tex]

so now we can complete the fourth order Maclaurin polynomial:

[tex]f(x)=0+1x+\frac{0}{2!}x^{2}-\frac{2}{3!}x^{3}+\frac{0}{4!}x^{4}[/tex]

which simplifies to:

[tex]f(x)=x-\frac{1}{3}x^{3}[/tex]

you can find the graph of the four polynomials in the attached picture.

So the first, second, third and fourth order Maclaurin polynomials of f(x)=arctan(x) are:

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Answer:

B

Step-by-step explanation:

The function given is f(n) = n-3. Plug in n=0 and you will get an output - 3. Plug in n=2 and you will get an output - 1. Hence table B is the answer .

Answer:

D.To teach a lesson

Step-by-step explanation:

Hope it helps you

Answer:

[tex]10^{-3} \times10^{k} =10^{-3} =\frac{1}{10^{3} }[/tex]

[tex]10^{-3+1+k} =10^{-3}[/tex]

[tex]10^{-2+k} =10^{-3}[/tex]

Therefore, [tex]-2+k=-3[/tex]

[tex]k=-3+2[/tex]

[tex]k=-1[/tex]

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