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denpristay [2]
3 years ago
12

How much must be subtracted from the first term of the sequence 12, 16, 64 so that the sequence becomes geometric?

Mathematics
1 answer:
insens350 [35]3 years ago
7 0

Answer: 8 should be subtracted to 4,16,64


Step-by-step explanation:


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$600 invested with compound interest at a rate of 4% per year for 10 years <br> formula:M= P(1+i)^n
makkiz [27]
Answer: M = 2100$

Explanation:

M = P(1 + I)^n

P = 600$, I = 4% = 0.04, n = 10

=> M = 600(1 + 0.04)^10
M = 600(1.04)^10
M = 600(3.5)
M = 2100$
8 0
2 years ago
Is the function even,odd or neither and how do you know?<br><br> ASAP
wlad13 [49]

Answer:

Odd

Step-by-step explanation:

6 0
2 years ago
Write the slope-intercept form of the equation of the line described.
Ad libitum [116K]

                                                  7)

Given the line

y=-x+2

We know that the slope-intercept form of the line equation is

y=mx+b

here

m is the slope and b is the intercept

Thus, the slope = -1

We know that the parallel lines have the same. Thus, the slope of the parallel line is: -1

Using point-slope of the line equation

y-y_1=m\left(x-x_1\right)

substituting the values m = -1 and the point (4, 0)

y-0 = -1 (x-4)

Writing in the slope-intercept form

y = -x+4

Thus, the slope-intercept form of the equation of the line equation parallel to y=-x+ 2 will be:

  • y = -x+4

                                                      8)

<em>Note: </em><em>Your line is a little bit unclear. But, I am assuming the</em>

<em>ine is: </em>y = -x-1

<em />

Given the assumed line

y = -x-1

We know that the slope-intercept form of the line equation is

y=mx+b

here

m is the slope and b is the intercept

Thus, the slope = m = -1

We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:

slope = m = -1

perpendicular slope = – 1/m = -1/-1 = 1

Using point-slope of the line equation

y-y_1=m\left(x-x_1\right)

substituting the values m = 1 and the point (4, 3)

y - 3 = 1 (x-4)

Writing in the slope-intercept form

y-3 = x-4

y = x-4+3

y = x - 1

Thus, the slope-intercept form of the equation of the line equation perpendicular to y = -x-1 will be:

  • y = x - 1

5 0
2 years ago
Compute show all of your work: 8X4-4/2+1
Pepsi [2]
8*4-4/2+1    
32-4/2+1
32-2+1
=31 answer 
So, the answer to this question is 31

5 0
3 years ago
(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:

<u><em>Explanation</em></u>:-

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

<em>Step(i)</em>:-

Given differential problem

                           y′+3 y=9 t

<em>Take the Laplace transform of both sides of the differential equation</em>

                L( y′+3 y) = L(9 t)

 <em>Using Formula Transform of derivatives</em>

<em>                 L(y¹(t)) = s y⁻(s)-y(0)</em>

  <em>  By using Laplace transform formula</em>

<em>               </em>L(t) = \frac{1}{S^{2} }<em> </em>

<em>Step(ii):-</em>

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}

<em>Step(iii</em>):-

<em>By using partial fractions , we get</em>

\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)

 <em>  B = 9/3 = 3</em>

  Put s = -3 in equation(i)

  9 = C(-3)²

  <em>C = 1</em>

 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)

 <em> 0 = A + C</em>

<em>put C=1 , becomes A = -1</em>

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

<u><em>Step(iv):-</em></u>

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})

<em>By using inverse Laplace transform</em>

<em></em>L^{-1} (\frac{1}{s} ) =1<em></em>

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

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

<u><em>Final answer</em></u>:-

<em>Now the solution , we get</em>

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

           

           

5 0
3 years ago
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