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cricket20 [7]
3 years ago
15

How many different ways can you arrange 10 letters?

Mathematics
1 answer:
Sloan [31]3 years ago
8 0
Multiply all of them together.10*9*8 and so on...
3,628,800 ways !
You might be interested in
53w + 13 < 56w + 16 solve for w
Sergio039 [100]

Answer:

w>-1

Step-by-step explanation:

1 Move terms

53w-56w<16-13

2 Collect like terms then subtract

-3w< 3

3 Divide both sides by -3

Answer: w>-1

4 0
3 years ago
Read 2 more answers
What is the value of x in the equation one-fifthx – two-thirdsy = 30, when y = 15? 4 8 80 200
vova2212 [387]

Answer:

x = 200

Step-by-step explanation:

Given

\frac{1}{5} x - \frac{2}{3} y = 30 ← substitute y = 15 into the equation

\frac{1}{5} x - \frac{2}{3} × 15 = 30 , that is

\frac{1}{5} x - 10 = 30 ( add 10 to both sides )

\frac{1}{5} x = 40 ( multiply both sides by 5 to clear the fraction )

x = 200

5 0
3 years ago
What is 0.54 converted to a simplified fraction?
nadya68 [22]
0.54=0.54/1=5.4/10=54/100=27/50
4 0
3 years ago
Please help!
erik [133]
For the answer to the question above, 
1 + nx + [n(n-1)/(2-factorial)](x)^2 + [n(n-1)(n-2)/3-factorial] (x)^3 

<span>1 + nx + [n(n-1)/(2 x 1)](x)^2 + [n(n-1)(n-2)/3 x 2 x 1] (x)^3 </span>

<span>1 + nx + [n(n-1)/2](x)^2 + [n(n-1)(n-2)/6] (x)^3 </span>

<span>1 + 9x + 36x^2 + 84x^3 </span>

<span>In my experience, up to the x^3 is often adequate to approximate a route. </span>

<span>(1+x) = 0.98 </span>

<span>x = 0.98 - 1 = -0.02 </span>

<span>Substituting: </span>

<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 </span>

<span>approximation = 0.834 </span>

<span>Checking the real value in your calculator: </span>

<span>(0.98)^9 = 0.834 </span>

<span>So you have approximated correctly. </span>

<span>If you want to know how accurate your approximation is, write out the result of each in full: </span>

<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 = 0.833728 </span>

<span> (0.98)^9 = 0.8337477621 </span>

<span>So it is correct to 4</span>
5 0
3 years ago
Read 2 more answers
Now, he wants to wrap it with wrapping paper. If the length of the shoebox measures 10 in, the width measures 5 in, and the heig
julia-pushkina [17]

The amount of wrapping paper he need to cover the shoebox is 190 square inches

<h3>Surface area of a box</h3>

The formula for calculating the surface area of a box is expressed as:

SA = 2(lw + wh + lh)

where

l is the length

w is the width

h is the height

Given the following parameters

l = 10in

w = 5in

h =3in

Substitute

S = 2(10*5 + 5*3 + 10*3)

S = 2(50+15+30)

S = 2(95)

S = 190 square inches

Hence the amount of wrapping paper he need to cover the shoebox is 190 square inches

Learn more on surface area of box here: brainly.com/question/26161002

#SPJ1

6 0
2 years ago
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