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1 year ago

What number should go in the empty boxes to complete the calculation for finding the product of

Mathematics

2 answers:

LekaFEV [45]1 year ago

7 0

Answer: Box 1: 2 Box 2: 2

Step-by-step explanation:

So, for box one it would be 2 because 3 x 4 = 12, put the 2 on the bottom and the 1 on the 6, then 6 x 4 + 1 = 25, then put the 5 on the bottom and the 2 on the top, then 4 x 0 + 2 = 2, so the answer would be 2.

For box 2, 252 + 1?60 = 0.1512, so we have to find the missing number. So, 2 + 0 = 2, so put the two on the bottom, then 5 + 6 = 11, so put the 1 on the bottom and the other 1 on the 2 from the first box, so 2 + 1 = 3, plus 2 more would equal 5, so the second box would be 2.

Hope this helps!

larisa86 [58]1 year ago

4 0

To do the multiplication, we multiply 0.63 by each of the digits & decimal places of 0.24 (and align the products that we get) then add them together.

0.63 x 4 = 2.52, therefore the first missing spot is a 2.

0.63 x 2 = 1.26, therefore the second missing spot is a 2 (the 0 is so that the decimal point can be aligned for the result.)

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Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

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Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

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The auxiliary equation (A.E) f(m) = 0

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$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

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<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

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$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

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$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

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The general solution is $y = y_{C} +y_{P}$

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What number should go in the empty boxes to complete the calculation for finding the product of ​

What number should go in the empty boxes to complete the calculation for finding the product of