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Bezzdna [24]
2 years ago
5

Using the equation √4x+15=3√x How many potential solutions are there?

Mathematics
1 answer:
geniusboy [140]2 years ago
5 0

Answer:

Using the equation √4x+15=3√x

How many potential solutions are there?

3

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Factor out the GCF from the terms of the polynomial –4y5 + 6y3 + 8y2 – 2y. A. –2y4 + 3y2 + 4y – 1 B. y(–4y4 + 6y2 + 8y – 2) C. 2
andreyandreev [35.5K]

Answer:

Option D) -2y(2y^4-3y^2-4y+1) is correct

Therefore -4y^5+6y^3+8y^2-2y=-2y(2y^4-3y^2-4y+1)

Step-by-step explanation:

Given polynomial is -4y^5+6y^3+8y^2-2y

To factorise the given polynomial by taking out the common terms of the given polynomial :

  • -4y^5+6y^3+8y^2-2y
  • =2y(-2y^4+3y^2+4y-1) ( here 2y is GCF common term so taking outside the terms of the polynomial )
  • =-2y(2y^4-3y^2-4y+1) ( now taking (-) outside )

Therefore -4y^5+6y^3+8y^2-2y=-2y(2y^4-3y^2-4y+1)

Option D) -2y(2y^4-3y^2-4y+1) is correct

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3 years ago
Pre college need help
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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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2 years ago
WHat is the greatest common factor (GCF) of 9 and 27
xxTIMURxx [149]

Answer:

9

Step-by-step explanation:

9/9=1

27/9=3

5 0
3 years ago
Determine whether the given function is a solution to the given differential equation.
lesantik [10]

Given :

A function , x = 2cos t -3sin t               .....equation 1.

A differential equation , x'' + x = 0      .....equation 2.

To Find :

Whether the given function is a solution to the given differential equation.

Solution :

First derivative of x :

x'=\dfrac{d(2cos t - 3sin t)}{dt}\\\\x'=\dfrac{d(2cost)}{dt}-\dfrac{(3sint)}{dt}\\\\x'=-2sint-3cost

Now , second derivative :

x''=\dfrac{d(-2sint-3cost)}{dt}\\\\x''=-\dfrac{d(2sint)}{dt}-\dfrac{d(3cost)}{dt}\\\\x''=-2cost+3sint

( Note : derivative of sin t is cos t and cos t is -sin t )

Putting value of x'' and x in equation 2 , we get :

=(-2cos t + 3sin t ) + ( 2cos t -3sin t )

= 0

So , x'' and x satisfy equation 2.

Therefore , x function is a solution of given differential equation .

Hence , this is the required solution .

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