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3 years ago

Do even numbers have more factors in odd numbers or do i numbers have more factors than even numbers

1 answer:

4 0

It is because even numbers always have a factor of two, and therefore, larger composite even numbers will have factors of two and other even numbers based around two, such as 4, 8, 16, 32, and so on. On the other hand, numbers which are odd can have factors of 3, 5, and 7 for example, and their numbers based around them(3, 9, 27; 5, 10, 15; 7, 49, 343; and so on). If we look into it, notice how for odd numbers the space between the numbers based around 3, 5, and 7 are increasingly further apart. This is the reason why less large odd integers to have numerous factors. It is because odd numbers cannot have the prime factor 2, this will reduce their factor number. And is is also because even numbers are already divided by 2, this will give them more factors over the odd numbers.

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Two number have a sum of 71 and a difference of 37

larisa86 [58]

The right answer for the question that is being asked and shown above is that:

Two numbers have a sum of 71 and a difference of 37
x + y = 71
x - y = 37

So in order to get the two numbers, here it is:
x + y = 71
x - y = 37
------------
2x = 108
x = 54
y = 71 - 54
y = 17

So the two numbers are 54 and 17

6 0

3 years ago

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Mars2501 [29]

Answer:

Answer is $110.25. You figure out how.

Step-by-step explanation:

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2 years ago

A positive integer is 4 more than 12 times another. Their product is 5896. Find the two integers.

maksim [4K]

X = integer one
y = integer 2

x = 12y + 4

x * y = 5896

Solve the system of equations

8 0

3 years ago

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(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti

Rashid [163]

Answer:

The solution

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}$

Step-by-step explanation:

Explanation:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

Step(i):-

Given differential problem

y′+3 y=9 t

Take the Laplace transform of both sides of the differential equation

L( y′+3 y) = L(9 t)

Using Formula Transform of derivatives

L(y¹(t)) = s y⁻(s)-y(0)

By using Laplace transform formula

$L(t) = \frac{1}{S^{2} }$

Step(ii):-

Given

L( y′(t)) + 3 L (y(t)) = 9 L( t)

$s y^{-} (s) - y(0) + 3y^{-}(s) = \frac{9}{s^{2} }$

$s y^{-} (s) - 7 + 3y^{-}(s) = \frac{9}{s^{2} }$

Taking common y⁻(s) and simplification, we get

$( s + 3)y^{-}(s) = \frac{9}{s^{2} }+7$

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

Step(iii):-

By using partial fractions , we get

$\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}$

$\frac{9}{s^{2} (s+3} = \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}$

On simplification we get

9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Put s =0 in equation(i)

9 = B(0+3)

B = 9/3 = 3

Put s = -3 in equation(i)

9 = C(-3)²

C = 1

Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

9 = A s²+3 A s +B(s)+3 B +C(s²)

0 = A + C

put C=1 , becomes A = -1

$\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}$

Step(iv):-

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

$y^{-}(s) =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}$

Applying inverse Laplace transform on both sides

$L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})$

By using inverse Laplace transform

$L^{-1} (\frac{1}{s} ) =1$

$L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}$

$L^{-1} (\frac{1}{s+a} ) =e^{-at}$

Final answer:-

Now the solution , we get

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}$

5 0

3 years ago

Please help!!!!!!!!!

vovikov84 [41]

Answer:

hopes it's helps

Step-by-step explanation:

I got 20 blocks for my answer

3 0

3 years ago

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