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2 years ago

9 + (3x + 4) = 9 + (4 + 3x) shows and example of what property?

Mathematics

2 answers:

Blababa [14]2 years ago

6 0

Answer:

A. Communative Property of Addition

Step-by-step explanation:

Ann [662]2 years ago

5 0

A.communicative property

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I need help on this question!

agasfer [191]

Answer:

I think its 1234.

sorry if im wrong

4 0

2 years ago

Read 2 more answers

Calculus 3 help please.

Reptile [31]

I assume each path $C$ is oriented positively/counterclockwise.

(a) Parameterize $C$ by

$\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}$

with $-\frac\pi2\le t\le\frac\pi2$ . Then the line element is

$ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt$

and the integral reduces to

$\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt$

The integrand is symmetric about $t=0$ , so

$\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt$

Substitute $u=\sin(t)$ and $du=\cos(t)\,dt$ . Then we get

$\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}$

(b) Parameterize $C$ by

$\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{3^2+4^2} \, dt = 5\,dt$

and

$\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt$

Integrate by parts with

$u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}$

$\displaystyle \int u\,dv = uv - \int v\,du$

$\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}$

$\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt$

and

$\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}$

8 0

1 year ago

What is the solution of the matrix equation?

katrin [286]

Hello from MrBillDoesMath!

Answer:

The first choice ( -12, 20)

Discussion:

Using matrix multiplication we find the equations:

1x + 1y = 8 (A)

2x + 3y = 36 (B)

Next, compute

2(A) - B =>

(2x + 2y) - (2x + 3y) = 2*8 - 36 =>

(2x - 2x) + ( 2y-3y) = 16 - 36 =>

-y = -20 =>

y = 20

Substituting this y value in (A) gives

x + 20 = 8 =>

x = 8 - 20 =>

x = -12

The answer is the first choice ( -12, 20)

Thank you,

MrB

6 0

2 years ago

What is |9|? <br> -18<br> -9<br> 9<br> 18

Veseljchak [2.6K]

Answer:

the true answer is 9

Step-by-step explanation:

Because the product of the absolute value is the same number and positive

6 0

2 years ago

Read 2 more answers

The percentage of cotton in material used to manufacture men's shirts follows.

velikii [3]

Answer:

Median = 34.55

Quartiles:

Q1 = 33.1

Q2 = 34.55

Q3 = 35.6

Inter Quartile range = 2.5

Step-by-step explanation:

Step 1

We arrange the given the following data from lowest to the highest

32.0, 32.1, 32.1, 32.2, 32.4, 32.5, 32.5, 32.5, 32.8, 32.8, 32.8, 32.8, 32.9, 32.9, 33.0, 33.0, 33.2, 33.2, 33.2, 33.2, 33.5, 33.6, 33.8, 33.8, 33.9, 34.0, 34.0, 34.1, 34.1, 34.3, 34.4, 34.5, 34.6, 34.6, 34.6, 34.7, 34.7, 35.0, 35.0, 35.1, 35.1, 35.2, 35.4, 35.5, 35.5, 35.5, 35.5, 35.6, 35.6, 36.1, 36.3, 36.4, 36.4, 36.4, 36.5, 36.6, 36.7, 36.7, 37.0, 37.1, 37.4, 37.4, 37.6, 37.8

a. Construct a stem-and-leaf display for the data.

Stem and leaf Display

Stem | Leaf

32 | 0,1,1,2,4,5,5,5,8,8,8,8,9,9

33 | 0,0,2,2,2,2,5,6,8,8,9

34 | 0,0,1,1,3,4,5,6,6,6,7,7

35 | 0,0,1,1,2,4,5,5,5,5,6,6,

36 | 1,3,4,4,4,5,6,7,7,

37 | 0,1,4,4,6,8

b. Calculate the median and quartiles of these data.

Number of terms = 64

1) Median = 1/2(n + 1)th value

n = 64

= 1/2(64 + 1)th

= 1/2(65)th

= 32.5 th value

This means it is between the 32nd and 33rd value

32nd = 34.5

33rd = 34.6

= 34.5 + 34.6/2

= 69.1/2

= 34.55

2) First Quartile

1/4(n + 1)th value

n = 64

= 1/4(64 + 1)th

= 1/4(65)th

= 16.5th value

This means it is between the 16th and 17th value

16th value = 33.0

17tj value = 33.2

= 33.0 + 33.2/2

= 66.2/2

= 34.55

Q1 --> 33.1

3)Second Quartile = Median

1/2(n + 1)th value

n = 64

= 1/2(64 + 1)th

= 1/2(65)th

= 32.5 th value

This means it is between the 32nd and 33rd value

32nd = 34.5

33rd = 34.6

= 34.5 + 34.6/2

= 69.1/2

= 34.55

Q2 --> 34.55

Third Quartile

3/4(n + 1)th value

n = 64

= 3/4(64 + 1)th

= 3/4(65)th

= 48.75 th value

This means it is towards the 49th value

32nd = 34.5

Hence,

Q3 --> 35.6

Inter Quartile range

Q3 - Q1

= 35.6 - 33.1

= 2.5

8 0

3 years ago