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3 years ago

How many different ways can the first twelve letters of the alphabet be arranged?

2 answers:

4 0

So there are 12 spaces right. In first blank u have a choice to fill any of the alphabet .(first twelve alphabets) <span>and it goes down to just 1 letter in the last space. but do i just multiply them.

hope dis helps</span>

photoshop1234 [79]3 years ago

4 0

12!
12!=12*11*10*9*8*7*6*5*4*3*2*1
=479001600

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Write an expression that can be used to multiply 6x198 mentally

VashaNatasha [74]

6×100
6×90
6×8
hope this helps

7 0

3 years ago

What is the equation of the graph below?On a coordinate plane, a curve crosses the y-axis at (0, negative 1). It has a minimum o

zzz [600]

ANSWER:

4th option: y = cosine (x + pi)

STEP-BY-STEP EXPLANATION:

For the function y = cosine (x), when x = 0, y = 1 and when x = , y = -1. The graph shows the opposite, when x = 0, y = -1 and when x = , y = 1, so we must add the same amount x = , for y to be positive 1.

So the equation of the graph would be:

$y=\cos(x+)$

The correct answer is then the 4th option y = cosine (x + pi)

3 0

1 year ago

Convert the following decimal to a common fraction. Reduce to lowest terms 0.2 = 2/1 0/2 1/2 1/5

Phoenix [80]

0.2 = $\frac{2}{10} = \frac{1}{5}$
So, $\frac{1}{5}$

7 0

3 years ago

I’m lost again can somebody help asap ??

serious [3.7K]

Fully detailed and equipped information process right below. The answer for this question is "25" (simplified form).

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(5^{- 5} \times 2^8 \times 1 \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(2^8 \times 1 \times \dfrac{1}{5^5} \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(1 \times \dfrac{2^8 \times 1}{5^5} \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(1 \times \dfrac{256}{3125} \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(\dfrac{3125}{256} \Bigg)^2}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{1}{2^3 \times 5^2} \Bigg)^4 \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{1^4}{(2^3 \times 5^2)^4} \Bigg) \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{1^4}{(5^2)^4 \times (2^3)^4} \Bigg) \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \dfrac{1}{2^{12} \times 5^8} \times \dfrac{3125^2}{256^2}}$

$\mathbf{\dfrac{3125^2 \times 1 \times 2^{28}}{256^2 \times 5^8 \times 2^{12}}}$

$\mathbf{\dfrac{2^{(28 - 12)} \times 3125^2}{256^2 \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times 3125^2}{256^2 \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times (5^5)^2}{(2^8)^2 \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times 5^{10}}{2^{16} \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times 5^{10 - 8}}{2^{16}}}$

$\mathbf{\dfrac{5^{10 - 8}}{1}}$

$\mathbf{\therefore \quad 5^2}$

$\mathbf{\therefore \quad 25}$

$\boxed{\mathbf{\underline{\therefore \quad Final \: \: Answer \: \: is: \: 25}}}$

Hope it helps.

7 0

3 years ago

Pliz ans this question

andrey2020 [161]

Answer:

Hello,

- 6 (x - 2)² - 1

Step-by-step explanation:

- 6 x² + 24 x - 25 = - 6 (x² - 4 x) - 25

= - 6 ( (x - 2)² - 4) - 25

= - 6 (x - 2)² + 24 - 25

= - 6 (x - 2)² - 1

5 0

3 years ago

Read 2 more answers