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3 years ago

A tile is selected from seven tiles, each labeled with a different letter from the first seven letters of the alphabet.

Mathematics

1 answer:

MArishka [77]3 years ago

7 0

Answer:

Step-by-step explanation:

U= {A,B,C,D,E,F,G}

X = {A, B, C, D}

Y = {A, C, E, F}

a) X or Y = {A, B, C, D} and {A, C, E, F} = {A, B, C, D, E, F}

B) X and Y = {A, B, C, D} or {A, C, E, F} = {A, C}

c) Complement of X = {A,B,C,D,E,F,G} - {A, B, C, D} = {B,D,G}

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How to add it because I don not know how to do it

koban [17]

1 3   4
1 -- + 1 -- =    2 --
5 5    5

Explanation:

1 + 1 = 2
1/5 + 3/5 = 4/5<span />

7 0

3 years ago

Read 2 more answers

"find the reduction formula for the integral" sin^n(18x)

dexar [7]

Let

$I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx$
For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

$\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx$
$\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$

For the remaining integral, we can integrate by parts, taking

$u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx$ $\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax$

$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
$\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)$

So putting everything together, we found

$I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$
$I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)$
$I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$
$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

7 0

3 years ago

What is the area of a book that has a length of 5 inches and a width of 10 inches? how do i find the answer?

wlad13 [49]

I hope this helps you

Area=length.width

Area=5.10

Area=50

4 0

3 years ago

Kendra used these steps to solve the equation 2x + 5 = 3 + 2(x + 1). Which choice describes the meaning of her result, 5 = 5?

11Alexandr11 [23.1K]

Hello!

We can solve this algebraically

2x + 5 = 3 + 2(x+1)

Distribute the 2

2x + 5 = 3 + 2x + 2

combine like terms

2x + 5 = 2x + 5

Since both sides of the equation is the same all values of x make the equation true

The answer is A

Hope this helps!

7 0

3 years ago

Evaluate the expression. r = <5, 9, -4>, v = <3, 3, -8>, w = <9, -6, -1> v ⋅ w (5 points)

oee [108]

Answer:

Step-by-step explanation:

7 0

3 years ago