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4 years ago

4. Simplify 1000^2/3 using a calculator

Mathematics

1 answer:

____ [38]4 years ago

4 0

1000² ÷ 3
333 333.33
the 3 after the decimal point goes on forever

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Please help me with this question

adoni [48]

Answer:

$y = e^{x^2 - 3x}\\\\\\Let \ u = x^2 - 3x\\\\y = e^u\\\\\frac{dy}{du} = e^u\\\\\frac{du}{dx} = 2x - 3\\\\y' = \frac{dy}{dx} = \frac{dy}{du} *\frac{du}{dx} = e^u * (2x -3)\\\\Substitute \ u = x^2 -3\\\\y' = (2x -3)e^{x^2-3x}\\\\option A$

5 0

3 years ago

A rectangle has an area of 48 square units and a perimeter of 32 units. What are its dimensions?

baherus [9]

Area is length x width, and perimeter is adding up all the sides together
so if it’s 32 for the perimeter, then the dimensions should be 12 and 4
12x4 is 48
12+12+4+4 is 32

5 0

3 years ago

Helppp!!!!laura is playing a card game with her friend.she will draw two cards from those pictured.

puteri [66]

Answer:

30%

Step-by-step explanation:

In this case it would be the combination of 2 taken from 5.

That is, when n = 5 and r = 2

nCr = n! / r! * (n-r)!

replacing

5C2 = 5! / (2! * 3!)

5C2 = 10

Which means that the sample space is 10 possible combinations.

the probability that the card you draw is even is 3/5, since there are 3 even cards that exist and 5 is the total number of cards.

Now, so that both are for it would be the multiplication of the previous probability and 2/4 because an even card and a card in general have already been drawn:

(3/5) * (2/4) = 0.3

So the probability that both are even is 30%

5 0

4 years ago

3/4 x3/4 an exponential form

Kobotan [32]

3/4 times 3/4 is the same as 3/4 squared or (3/4)^2

3 0

4 years ago

Solve the following congruence equations for X a) 8x = 1(mod 13) b) 8x = 4(mod 13) c) 99x = 5(mod 13)

xxMikexx [17]

Answer:

a) 5+13k where k is integer

b) 20+13k where k is integer

c)12+13k where k is integer

Step-by-step explanation:

(a)

$8x \equiv 1 (mod 13) \text{ means } 8x-1=13k$ .

8x-1=13k

Subtract 13k on both sides:

8x-13k-1=0

Add 1 on both sides:

8x-13k=1

I'm going to use Euclidean Algorithm.

13=8(1)+5

8=5(1)+3

5=3(1)+2

3=2(1)+1

Now backwards through the equations:

3-2=1

3-(5-3)=1

3-5+3=1

(8-5)-5+(8-5)=1

2(8)-3(5)=1

2(8)-3(13-8)=1

5(8)-3(13)=1

So compare this to:

8x-13k=1

We see that x is 5 while k is 3.

Anyways 5 is a solution or 5+13k is a solution where k is an integer.

$8x \equiv 4 (mod 13)$

8x-4=13k

Subtract 13k on both sides:

8x-13k-4=0

Add 4 on both sides:

8x-13k=4

We got this from above:

5(8)-3(13)=1

If we multiply both sides by 4 we get:

8(20)-13(12)=4

So x=20 and 20+13k is also a solution where k is an integer.

[tex]99x \equiv 5 (mod 13)[/tex

99x-5=13k

Subtract 13k on both sides:

99x-13k-5=0

Add 5 on both sides:

99x-13k=5

Using Euclidean Algorithm:

99=13(7)+8

13=8(1)+5

Go back through the equations:

13-8=5

13-(99-13(7))=5

8(13)-99=5

99(-1)+8(13)=5

Compare this to 99x-13k=5 and see that x=-1 or -1+13=12 or 12+13k is a solution where k is an integer.

8 0

3 years ago

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