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2 years ago

Choose the multiplication problem that correctly shows partial products. A) A B) B C) C

Mathematics

2 answers:

Kipish [7]2 years ago

6 0

Answer:

Step-by-step explanation:

5 x 2 = 0 (carry 1)

5 x 8 + 1 = 1 (carry 4 to bottom)

TiliK225 [7]2 years ago

3 0

Answer:

B) B

Step-by-step explanation:

In this case, the first partial product will be 2 * 5 which is 10. The only option that shows this is option B, so this is our answer.

We can also check the answers as a whole,

2 * 5 = 10

80 * 5= 400

400 + 1 = 410

Again, only option B gives us this. This is means that option B is our answer.

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Put the differential equation 9ty+ety′=yt2+81 into the form y′+p(t)y=g(t) and find p(t) and g(t). p(t)= help (formulas) g(t)= he

alexgriva [62]

Answer:

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And

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Step-by-step explanation:

Given that,

The differential equation is -

$9ty + e^{t}y' = \frac{y}{t^{2} + 81 }$

$e^{t}y' + (9t - \frac{1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t(t^{2} + 81 ) - 1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t^{3} + 729t - 1}{t^{2} + 81 } )y = 0\\y' + [\frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) } ]y = 0$

By comparing with y′+p(t)y=g(t), we get

$p(t) = \frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) }$

g(t) = 0

And

The differential equation $9ty + e^{t}y' = \frac{y}{t^{2} + 81 }$ is linear and homogeneous.

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