find the first 3 terms of the sequence below.​

Parameterize the surface (I'll call it S) by

r(u, v) = (1 - u) (1 - v) i + u (1 - v) j + v k

with 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1.

Take the normal vector to this surface to be

n = ∂r/∂u × ∂r/∂v = ((v - 1) i + (1 - v) j) × ((u - 1) i - u j + k) = (1 - v) (i + j + k)

with magnitude

||n|| = √3 (1 - v)

Then in the integral, we have

[tex]\displaystyle\iint_SG(x,y,z)\,\mathrm ds = \int_0^1\int_0^1 G((1-u)(1-v),u(1-v),v) \|\mathbf n\| \,\mathrm du\,\mathrm dv \\\\= \sqrt3 \int_0^1\int_0^1uv(1-v)^2\,\mathrm du\,\mathrm dv \\\\= \boxed{\frac1{8\sqrt3}}[/tex]

Alternatively, if you're not familiar with parameterizing surfaces, you can use the "projection" formula:

[tex]\displaystyle\iint_S G(x,y,z)\,\mathrm ds = \int_{S_{xy}}G(x,y,z)\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}\,\mathrm dx\,\mathrm dy[/tex]

where I write [tex]S_{xy}[/tex] to mean the projection of the surface onto the (x, y)-plane, and z = f(x, y). We would then use

x + y + z = 1   ==>   z = f(x, y) = 1 - x - y

and [tex]S_{xy}[/tex] is the triangle,

{(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 - x}

Then the integral becomes

[tex]\displaystyle\int_0^1\int_0^{1-x}y(1-x-y)\sqrt{1+(-1)^2+(-1)^2}\,\mathrm dy\,\mathrm dx \\\\= \sqrt3\int_0^1\int_0^{1-x} y(1-x-y)\,\mathrm dy\,\mathrm dx \\\\= \frac{\sqrt3}{24} \\\\= \boxed{\frac1{8\sqrt3}}[/tex]

Answer:

t>-10

31>k

-4>h

f≥6.8

this is the answer

Answer:

12

Step-by-step explanation:

PEMDAS is the order

P = parenthesis

E = exponent

M = *

D = division

A = +

S = -

so first 8*2 = 16

and then -8/2 = -4

and then -4 + 16

= 12

Answer:

[tex]CI=189.5,194.5[/tex]

Step-by-step explanation:

From the question we are told that:

Sample size [tex]n=40[/tex]

Mean [tex]\=x =192[/tex]

Standard deviation[tex]\sigma=8[/tex]

Significance Level [tex]\alpha=0.05[/tex]

From table

Critical Value of [tex]Z=1.96[/tex]

Generally the equation for momentum is mathematically given by

 [tex]CI =\=x \pm z_(a/2) \frac{\sigma}{\sqrt{n}}[/tex]

 [tex]CI =192 \pm 1.96 \frac{8}{\sqrt{40}}[/tex]

 [tex]CI=192 \pm 2.479[/tex]

 [tex]CI=189.5,194.5[/tex]

Answer:

40 degrees un edge

Step-by-step explanation:

Answer:

The person above me got this correct, so the answer to this is 40! I just did the Unit Test and got a 100%!

Answer:

10 Ghuups I believe. I am sorry if this is wrong

Answer:

x = 16, or if you didn't want the value for x,

2x + 3 = 35

Step-by-step explanation:

Three more: +3

Twice a number: 2x

Combined:

2x + 3 = 35.

Get rid of the 3 by subtracting it from both sides:

2x = 32

Get rid of the 2 by dividing it from both sides:

x = 16

Answer:

The number is 16.

Step-by-step explanation:

Let the unknown number be x.

Now we translate the sentence into an equation piece by piece.

Three more than twice a number is 35.

2x

Three more than twice a number is 35.

2x + 3

Three more than twice a number is 35.

2x + 3 = 35

Now we solve the equation.

Subtract 3 from both sides.

2x = 32

Divide both sides by 2.

x = 16

Answer: The number is 16.

P.S. Notice that x was a variable that was introduced solely to solve the problem. The original problem is a word problem, not an equation, and has no x in it. The correct answer makes no reference to x since x was used to solve the equation but is not part of the given problem. The person asking the question has no idea what x is. He just wants a number as an answer.

Answer:

just keep writing down outcome on a sheet of paper then count total

Step-by-step explanation:

Answer: The answer is b

Answer:

It should be a 2 sample t-test for the sample mean mu 1 - mu 2 at alpha = 0.05.

Step-by-step explanation:

To solve, you can just insert the data values into a graphing calculator and it should work. Remember to check the conditions and write out the null and alternate hypotheses.

Null: mu 1 - mu 2 = 0 There is no difference

Alternate: mu 1 - mu 2 =/= 0 There is a difference

Conditions:

-Random sample? Yes b/c assume that it is from a simple random sample.

-10%? Assume that there are more than 120 men and 80 women in the population"

-Normal Distribution? If the data for each respective sample is approx normal, assume they come from a normally distributed population. Large counts and the central limit theorem do not work here.

After this, insert ur data values into the graphing calculator n solve for p.

Once you get p, make a conclusion based on alpha = 0.05. If p is less than alpha, you can reject the null and conclude that you have significant evidence that the alternate is true. If p is greater than alpha, you cannot reject the null and conclude that you do not have significant evidence that the alternate is true.

Answer:

l don't know

Step-by-step explanation:

Answer:

(1/x + 1/y)k is the answer :)

1*2^3+0*2^2+1*2^1+1*2^0

8+0+2+1

=11

Answer:

We know that every number can be written as a product of prime numbers.

The method to find the factorized form of a number depends on the number, we just try to find the different factors by dividing by them, for example for the number 1000 we have:

1000 is an even number, then we can divide it by 2 (2 is a prime number)

1000 = 2*500   (so we already found a prime factor)

500 is also an even number, so we can divide it by 2

1000 = 2*500 = 2*2*250  (we found another prime factor)

dividing by 2 again we get:

1000 = 2*2*250 = 2*2*2*125

1000 = (2*2*2)*125

now we just need to factorize 125

we know that 125 is a multiple of 5, such that:

125 = 5*25 = 5*5*5

(5 is a prime number, so it is completely factorized).

Then the factorization of 1000 is:

1000 = (2*2*2)*(5*5*5) = 2^3*5^3

Now with another example, 1422

1422 is an even number, so we again start using the factor 2:

1422 = 2 = 711

then:

1422 = 2*711

we already found a factor.

711 is a multiple of 3 (the sum of its digits is a multiple of 3), then:

711/3 = 237

We can write our number as:

1422 = 2*3*237

237 is also a multiple of 3

237/3 = 79

then:

1422 = 2*3*3*79

and 79 is a prime number, so we already have 1422 completely factorized.

Answer:

[tex]A"B" = \frac{AB}{2}[/tex]

Step-by-step explanation:

Given

[tex]k = \frac{1}{2}[/tex] --- scale factor

Required

Relationship between AB and A"B"

[tex]k = \frac{1}{2}[/tex] implies that the sides of A"B"C" are smaller than ABC

i.e.

[tex]A"B" = k * AB[/tex]

[tex]A"B" = \frac{1}{2} * AB[/tex]

This gives:

[tex]A"B" = \frac{AB}{2}[/tex]

Answer:

- Surveyors

- developers of the GPS system

Step-by-step explanation:

Both the question and options given doesn't seem to be properly formatted. A well formatted form of the question is written in the comment section below.

Answer:

10a^4

Step-by-step explanation:

Given the expression :

7a^4+3a^4

The sum of the expression given above could be taken directly Since the power of each individual value is the same.

7a^4+3a^4

Adding the coefficients

(7+3)a^4

10a^4

Answer:

a) {3,5}{3,10}{5,10}

b) [tex]P(A)=\frac{1}{3}[/tex]

c) [tex]P(B)=\frac{2}{3}[/tex]

d) [tex]P(C)=\frac{1}{3}[/tex]

e) [tex]P(A and C)=0[/tex]

f) [tex]P(A or B)=1[/tex]

g) [tex]P(B and C)=\frac{1}{3}[/tex]

h) [tex]P(A or C)=\frac{2}{3}[/tex]

i) [tex]P(C given B)=\frac{1}{2}[/tex]

j) [tex]P(C given A)=0[/tex]

k) [tex]P(not B)=\frac{1}{3}[/tex]

l) [tex]P(not C)=\frac{2}{3}[/tex]

Yes, events A and B are mutually exclusive. Because the results can either be even or odd, not both. No, events B and C are not mutually exclusive because the result can be both, odd and prime.

Step-by-step explanation:

a)

In order to solve part a of the problem, we need to find the possible outcomes, in this case, the possible outcomes are:

{3,5}{3,10} and {5,10}

We could think of the oppsite order, for example {5,3}{10,3}{10,5} but these are basically the same as the previous outcomes, so we will just take three outcomes in our sample space. We can think of it as drawing the two chips at the same time.

b)

Now the probability of the sum of the chips to be even. There is only one outcome where the sum of the chips is even, {3,5} since 3+5=8 the other outcomes will give us an odd number, so:

[tex]P=\frac{#desired}{#possible}[/tex]

[tex]P(A)=\frac{1}{3}[/tex]

c) For the probability of the sum of the chips to be odd, there are two outcomes where the sum of the chips is odd, {3,10} since 3+10=13 and {5,10} since 5+10=15 the other outcomes will give us an even number, so:

[tex]P(B)=\frac{2}{3}[/tex]

d) The probability of the sum of the chips is prime. There is only one outcome where the sum of the chips is prime, {3,10} since 3+10=13 the other outcomes will give us non prime results, so:

[tex]P(C)=\frac{1}{3}[/tex]

e) The probability of the sum of the chips to be even and prime. There are no results where we can get an even and prime number, since the only even and prime number there is is number 2 and no outcome will give us that number, so:

P(A and C)=0

f) The probability of the sum of the chips is even or odd. We can either get even or odd results, so no matter what outcome we get, we will get an odd or even result so:

[tex]P(A or B)=1[/tex]

g) The probability of the sum of the chips is odd and prime. There is only one outcome where the sum of the chips is odd and prime, {3,10} since 3+10=13 the other outcomes will give us non prime results, so:

[tex]P(B and C)=\frac{1}{3}[/tex]

h) The probability of the sum of the chips is even or prime. There are two outcomes where the sum of the chips is even or prime, {3,10} since 3+10=13 and {3,5} since 3+5=8 so:

[tex]P(A or C)=\frac{2}{3}[/tex]

i) The probability of the sum of the chips is prime given that the sum of the chips is odd. There are two possible results where the sum of the chips is odd {3,10} and {5,10} and only one of those results is even, {3,10}, so

[tex]P(C given B)=\frac{1}{2}[/tex]

j) The probability of the sum of the chips is prime given that the sum of the chips is even. There is only one possible even result: {3,5} but that result isn't prime, so

[tex]P(C given A)=0[/tex]

k) The probability of the sum of the chips is not odd. There is only one outcome where the sum of the chips is not odd (even), {3,5} so:

[tex]P(not B)=\frac{1}{3}[/tex]

l) The probability of the sum of the chips is not prime. There are two outcomes where the sum of the chips is not prime, {3,5} and {5,10} so:

[tex]P(not C)=\frac{2}{3}[/tex]

Are events A and B mutually exclusive?

Yes, events A and B are mutually exclusive.

Why or why not?

Because the results can either be even or odd, not both.

Are events B and C mutually exclusive?

No, events B and C are not mutually exclusive.

Why or Why not?

Because the result can be both, odd and prime.

Answer:

a = 19

Step-by-step explanation:

Form a equation using y = mx + c form

(27) = (-8)(10) + c

27 = -80 + c

107 = c

y = -8x + 107

Substitute the x value to find a

y = -8(11) + 107

y = 19

Answer: 62

Step-by-step explanation:

Given

p = 7, q = 2, r = 4

Solve

q ( 5p - r )

Substitute

(2) (5(7) - (4))

Simplify

(2) (35 - 4)

(2) (31)

62

Hope this helps!! :)

Please let me know if you have any questions

Answer:

xlog(64)−xlog(2)−2

Step-by-step explanation:

Simplify 6log(2) by moving 6 inside the logarithm.

log(2^6)x − log(2)x − 2

Raise 2 to the power of 6.

log(64)x − log(2)x − 2

Reorder factors in log(64)x − log(2)x −2.

Answer:

71, 71, 75, 85, 98, 98, 104, 111, 114, 122, 164

The middle quartile is 98.

The lower quartile is 80

The upper quartile is 112.5

Answer: Hello attached below is the well written complete question

answer:

T = 40 + 30e^-0.0288t

Step-by-step explanation:

To( outside temperature ) = 40°F

T  = ?

k( thermal conductivity ) = constant

A( area of heat transfer) = constant

Hence Differential equation for the temperature of the house

T = 40 + 30e^-0.0288t

Attached below is the detailed solution of the problem

Answer: [tex]24\ cm^2/s[/tex]

Step-by-step explanation:

Given

Each side of square is increasing at a rate of 4 cm/s

Side of a square when area is [tex]9\ cm^2[/tex]

Suppose a is side of the square

[tex]\Rightarrow a=\sqrt{9}\\\Rightarrow a=3\ cm[/tex]

Area is given by

[tex]\Rightarrow A=a^2\\\text{Differentiate area w.r.t time}\\\\\Rightarrow \dfrac{dA}{dt}=2a\dfrac{da}{dt}\\\\\Rightarrow \dfrac{dA}{dt}=2\times 3\times 4\\\\\Rightarrow \dfrac{dA}{dt}=24\ cm^2/s[/tex]

Area is increasing at a rate of [tex]24\ cm^2/s[/tex]

Answer:

24cm^2/ s

Step-by-step explanation:

9514 1404 393

Answer:

  (c)  3x^4

Step-by-step explanation:

The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

  (a^b)^c = a^(bc)

___

  [tex]\displaystyle\sqrt[3]{9x^4}\cdot\sqrt[3]{3x^8}=\sqrt[3]{9\cdot3x^4x^8}=\sqrt[3]{27x^{12}}=\sqrt[3]{(3x^4)^3}=\boxed{3x^4}[/tex]

Answer:

916.67 (corrected to 2 decimal places)

Hope I'm right, fingers crossed

Step-by-step explanation:

[tex]55 \times 50 = 2750 \div \frac{3}{4 } = 916.6666667[/tex]

Answer:

65 students.

Step-by-step explanation:

Given that :

Every student planted as many plant as their number ;

Then let the number of student = x

Then the number of plant planted by each student will also = x

The total number of plants planted by all the students = 4225

The Number of students can be obtained thus ;

Total number of plants = Number of plants * number of plants per student

4225 = x * x

4225 = x²

√4225 = x

65 = x

Hence, there are 65 students

Answer:

14) 4x+10=8x-26 (corresponding angles are equal)

4x-8x=-26-10

-4x=-36

x= -36/-4= 9

x=9

15) perimeter of rectangle= 2(l+b)

2( l+ [tex]\frac{2l}{3}[/tex]) = 40m

2l+ [tex]\frac{4l}{3}[/tex] =40

Take LCM as 3

[tex]\frac{2l}{1}[/tex] * [tex]\frac{3}{3}[/tex] + [tex]\frac{4l}{3}[/tex] =40

[tex]\frac{6l+4l}{3}[/tex] = 40

[tex]\frac{10l}{3}[/tex] = 40

10l=40*3

10l = 120

l= 120/10 =12 cm

l=12cm

b= 2/3 *12 = 8cm

16) 2:3:4

It can be written as 2x+3x+4x

sum of angles of a triangle =180 degree

so 2x+3x+4x=180

9x=180

x=180/9=20 degree

1st angle=2x=2*20= 40 degree

2nd angle= 3x=3*20 =60 degree

3rd angle= 4x=4*20= 80 degree

17) sum of interior angles of a pentagon is 540 degree

so, 125+88+128+60+x=540 degree

401 +x= 540 degree

x=540-401= 139 degree

Hope this helps

Please mark me as brainliest

Answer:The 95% Rule states that approximately 95% of observations fall within two ... about 95% will be within two standard deviations of the mean, and about 99.7% will be ... Suppose the pulse rates of 200 college men are bell-shaped with a mean of 72 ... 1.2 - Samples & Populations ... 3.5 - Relations between Multiple Variables.

Step-by-step explanation: