If one-third of a number, x, is greater than five less than twice the number, which of the following is true?

A decorator adds vases to a mantle to decorate. She wants to use 2 matching green vases, 3 matching blue vases, and 4 matching w

hjlf

There are 2 green, 3 blue, and 4 white vases.
The green vases can be arranged in 2! = 2*1 = 2 ways.
The blue vases can be arranged in 3! = 3*21 = 6 ways.
The white vases can be arranged in 4! = 4*3*2*1 = 24 ways.

The total number of arrangements is
2*6*24 = 288

Answer: 288

3 0

3 years ago

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Find the center of the circle that can be circumscribed about EFG with E(4, 4), F(4, 2), G(8, 2). (3, 6) (4, 2) (4, 4) (6, 3)

Lesechka [4]

Answer:

(6,3)

Step-by-step explanation:

8 0

3 years ago

What is 3=7(4 - 2u)-6u

alisha [4.7K]

Answer:

u = 5/4

Step-by-step explanation:

to evaluate for the value of u we would simply open the bracket and then evaluate for the value of u by collecting the like terms together.

solution

3=7(4 - 2u)-6u

3 = 28- 14u - 6u

collect the like terms

3 + 14u + 6u = 28

20u = 28 - 3

20u = 25

divide both sides by the coefficient of u which is 20

20u/20 = 25/20

u = 5/4

5 0

3 years ago

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7) PG & E have 12 linemen working Tuesdays in Placer County. They work in groups of 8. How many

BabaBlast [244]

Part A

Since order matters, we use the nPr permutation formula

We use n = 12 and r = 8

$_{n}P_{r} = \frac{n!}{(n-r)!}\\\\_{12}P_{8} = \frac{12!}{(12-8)!}\\\\_{12}P_{8} = \frac{12!}{4!}\\\\_{12}P_{8} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1}\\\\_{12}P_{8} = \frac{479,001,600}{24}\\\\_{12}P_{8} = 19,958,400\\\\$

There are a little under 20 million different permutations.

<h3>Answer: 19,958,400</h3>

Side note: your teacher may not want you to type in the commas

============================================================

Part B

In this case, order doesn't matter. We could use the nCr combination formula like so.

$_{n}C_{r} = \frac{n!}{r!(n-r)!}\\\\_{12}C_{8} = \frac{12!}{8!(12-8)!}\\\\_{12}C_{8} = \frac{12!}{4!}\\\\_{12}C_{8} = \frac{12*11*10*9*8!}{8!*4!}\\\\_{12}C_{8} = \frac{12*11*10*9}{4!} \ \text{ ... pair of 8! terms cancel}\\\\_{12}C_{8} = \frac{12*11*10*9}{4*3*2*1}\\\\_{12}C_{8} = \frac{11880}{24}\\\\_{12}C_{8} = 495\\\\$

We have a much smaller number compared to last time because order isn't important. Consider a group of 3 people {A,B,C} and this group is identical to {C,B,A}. This idea applies to groups of any number.

-----------------

Another way we can compute the answer is to use the result from part A.

Recall that:

nCr = (nPr)/(r!)

If we know the permutation value, we simply divide by r! to get the combination value. In this case, we divide by r! = 8! = 8*7*6*5*4*3*2*1 = 40,320

So,

$_{n}C_{r} = \frac{_{n}P_{r}}{r!}\\\\_{12}C_{8} = \frac{_{12}P_{8}}{8!}\\\\_{12}C_{8} = \frac{19,958,400}{40,320}\\\\_{12}C_{8} = 495\\\\$

Not only is this shortcut fairly handy, but it's also interesting to see how the concepts of combinations and permutations connect to one another.

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<h3>Answer: 495</h3>

5 0

2 years ago

If m (x) = StartFraction x + 5 Over x minus 1 EndFraction and n(x) = x – 3, which function has the same domain as (m circle n) (

Andrew [12]

Answer:

The function h (x) = StartFraction 11 Over x minus 4 EndFraction

has the domain of h (x) = ( - ∞, 4) U (4, ∞ ) which is same as the

domain of (m ∘ n)(x) = ( - ∞, 4) U (4, ∞ ).

Step-by-step explanation:

m (x) = StartFraction x + 5 Over x minus 1 and n(x) = x – 3

As m (x) could be written as:

$m(x) = \frac{x+5}{x-1}$

n(x) = x – 3

(m ∘ n)(x) = m(n(x)) = m(x – 3)

= $\frac{(x-3) + 5}{(x-3) - 1}$

= $\frac{x+2}{x-4}$

In order to find the domain of (m ∘ n)(x) = $\frac{x+2}{x-4}$ , we need to

make sure that denominator can not be zero,

So,

x - 4 = 0

x = 4

So, our domain can be anything except for 4.

Hence, Domain = D = ( - ∞, 4) U (4, ∞ )

Now, compare the domain of (m ∘ n)(x) i.e D = ( - ∞, 4) U (4, ∞ ) with all the options:

Option A: h (x) = x + 5 / 11

Option A has the Domain of h (x) = Set of all real numbers

So, Option A is false.

Option B: h (x) = 11 / x - 1

x - 1 = 0

x = 1

Option B has the domain of h (x) = ( - ∞, 1) U (1, ∞ )

So, Option B is false.

Option C: h (x) = 11 / x - 4

x - 4 = 0

x = 4

Domain C has the domain of h (x) = ( - ∞, 4) U (4, ∞ )

So, Option C is true.

Option D: h (x) = 11 / x - 3

x - 3 = 0

x = 3

Option D has the domain of h (x) = ( - ∞, 3) U (3, ∞ )

So, option D is also false.

Hence, only option C is true as the option C i.e. h (x) = StartFraction 11 Over x minus 4 EndFraction has the domain of h (x) = ( - ∞, 4) U (4, ∞ ) which is same as the domain of (m ∘ n)(x) = ( - ∞, 4) U (4, ∞ ).

keywords: Domain, composite function

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3 years ago

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