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Oxana [17]
3 years ago
6

Write 25*24*23*22 as a quotient of factorials and as a permutation

Mathematics
1 answer:
Kipish [7]3 years ago
5 0

Answer:

(a)

25\times 24\times 23\times 22=\frac{25!}{21!}

(b)

25\times 24\times 23\times 22=P(25,4)

Step-by-step explanation:

Quotient of factorials:

we are given

25\times 24\times 23\times 22

we can multiply top and bottom term by 21!

25\times 24\times 23\times 22=\frac{25\times 24\times 23\times 22\times 21!}{21!}

we can write as

25\times 24\times 23\times 22=\frac{25!}{21!}

As a permutation:

we know permutation formula

P(n,r)=\frac{n!}{(n-r)!}

now, we can compare and find 'n' and 'r'

n=25

n-r=21

we can plug back n=25

25-r=21

r=4

so, we can write

25\times 24\times 23\times 22=P(25,4)

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A right cylinder has a radius of 2 units and a height of 5
erica [24]

Answer:

62.8

Step-by-step explanation:

The volume of the cylinder is = base * height

base=Pi*radius*radius, where

base=3.1416*2*2 , height=5

Volume=3.1416*2*2*5

Volume =62.8 cubic units.

6 0
4 years ago
Read 2 more answers
Could somebody please help me with this?
BARSIC [14]

For this problem, we are going to use the Remainder Theorem. This says that for x - n to be a factor of a polynomial p(x), then p(n) = 0. Essentially, it says that x - n is a factor if when you substitute n into the polynomial you get a result of 0.


Thus, in our case, when we substitute x = 2 into the polynomial, we should get an answer of 0 if x - 2 is a factor of the polynomial. Given this information, we can solve for c:

p(2) = 2^3 - 4(2^2) + 2c + 2 = 0

8 - 16 + 2c + 2 = 0

2c - 6 = 0

2c = 6

c = 3


The solution is c = 3.

7 0
3 years ago
Read 2 more answers
Could someone please help me with this? Solution set of lx^2+mx+n=0 is...?
IgorLugansk [536]

Answer:

Simplifying

lx2 + mx + n = 0

Solving

lx2 + mx + n = 0

Solving for variable 'l'.

Move all terms containing l to the left, all other terms to the right.

Add '-1mx' to each side of the equation.

lx2 + mx + -1mx + n = 0 + -1mx

Combine like terms: mx + -1mx = 0

lx2 + 0 + n = 0 + -1mx

lx2 + n = 0 + -1mx

Remove the zero:

lx2 + n = -1mx

Add '-1n' to each side of the equation.

lx2 + n + -1n = -1mx + -1n

Combine like terms: n + -1n = 0

lx2 + 0 = -1mx + -1n

lx2 = -1mx + -1n

Divide each side by 'x2'.

l = -1mx-1 + -1nx-2

Simplifying

l = -1mx-1 + -1nx-2

Step-by-step explanation:

Hope this helped you!

6 0
4 years ago
Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)
maxonik [38]

Answer with Step-by-step explanation:

We are given that

r(t)=< t^2,\frac{2}{3}t^3,t >

We have to find T,N and B at the given point t > (1,2/3,1)

r'(t)=

\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1

T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}

Now, substitute t=1

T(1)=\frac{}{2+1}=\frac{1}{3}

T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}

T'(1)=-\frac{4}{9}+\frac{1}{3}

T'(1)=\frac{1}{9}=

\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}

N(1)=\frac{T'(1)}{\mid T'(1)\mid}

N(1)=\frac{}{\frac{2}{3}}=

N(1)=

B(1)=T(1)\times N(1)

B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}

B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})

B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k

B(1)=\frac{1}{3}

5 0
3 years ago
What is the value of f(x)= 3x + 7 x for 8?
Ira Lisetskai [31]

Answer:

3*8+7

31

hope this helps

5 0
3 years ago
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