-3(10b+10)+5(b+2 help me solve this

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.

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:

1. 8+(30/(2+4)) = 8+(30/6) = 8+5 = 13

2. ((8+30)/2)+4 = (38/2)+4 = 19+4 = 23

Step-by-step explanation:

:)

Answer:

Step-by-step explanation:

23:

(8 + 30) ÷ 2 + 4

13:

8 + 30 ÷ (2 + 4)

please give me the brainliest if u can

Answer:

c

Step-by-step explanation:

use the distributive property to expand the expression:

-2(9 + 32n)

-2*9 + (-2*32n)

-18 - 64n

[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]

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:

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:  186.75 million which is the same as saying 186,750,000

Work Shown:

75% = 75/100 = 0.75

75% of 249 million = 0.75*249 million = 186.75 million

186.75 million = 186.75*10^6 = 186,750,000

Answer:

186,750,000

Step-by-step explanation:

Take 75% of 249 million

.75 * 249,000,000

186,750,000

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:

(-4, -3) y (3, -5)  

{-12 y, 15 y}

-12 y + 15 y = 3 y

(3 y)/2

3 sqrt(41) abs(y)

Answer:

answer is 10

explanation

when u add 7 with 10 u get 17 then double of 17 is 34

I hope It helps

It is found that the unknown number was 10.

An equation is an expression that shows the relationship between two or more numbers and variables. The addition is one of the mathematical operations. then the addition of two numbers results in the total amount of the combined value.

Given that "I add 7 to a certain number. I double the result. My final answer is 34".

Let consider the number be 10.

When we add 7 with 10 we get;

7 + 10 =  17

then double the result of 17 = 34

Hence, the unknown number was 10.

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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

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:

200 : 250 : 50

Step-by-step explanation:

Sum the parts of the ratio, 4 + 5 + 1 = 10 parts

Divide the amount by 10 to find the value of one part

500 ÷ 10 = 50 ← value of 1 part of ratio , then

4 parts = 4 × 50 = 200

5 parts = 5 × 50 = 250

500 = 200 : 250 : 50

Answer:

200, 250 and 50.

Step-by-step explanation:

First find the 'multiplier'.

4 + 5 + 1 = 10

500/10 = 50 = multiplier.

So the answer is

4*50 = 200

5 * 50 = 250

and 1 * 50 = 50.

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

65/100 * x or 0.65 * x

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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Answer:

15

Step-by-step explanation:

15×3=45

15×4=60

15×5=75

Answer:

15

Step-by-step explanation:

45 = 1 × 3^2 × 5

60 =  2^2 × 3 × 5

75 = 3 × 5^2

greatest number than divides 45, 60 and 75 without leaving remainder​ = GCF of 45,60,75 = 3 × 5 = 15

Answer: 20 ft³

Step-by-step explanation:

volume of triangular pyramid = [tex]\frac{1}{3} bh[/tex]

Therefore, the volume is:

[tex]\frac{1}{3} *10*6=\frac{1}{3}*60=\frac{60}{3}=20[/tex]

Answer: 10

Step-by-step explanation:

Sqrt (7- -5)^2+(-5-4)^2 =

Sqrt (12)^2+(-9)^2 =

Sqrt 225 = 15

2/3 * 15 = 30/3 = 10

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:

D.To teach a lesson

Step-by-step explanation:

Hope it helps you

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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Answer:

spread over a wide range

cannot be evaded

Everyone can contribute

convenient

Step-by-step explanation:

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:

[tex]0.04 = 4 \div 100 [/tex]

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:

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: Choice A

The number line graph is visually showing every number that is 19 or smaller; hence [tex]x \le 19[/tex]

Note the use of a closed or filled in circle at the endpoint (in contrast to an open circle). This indicates we are including the endpoint 19 as part of the solution set, and that's why we go for "or equal to" as part of the inequality sign.