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joja [24]
3 years ago
14

The product of a + 3b and –a2 + 4ab + 2b is –a3 + xa2b + 2ab + 12ab2 + yb2. What are the values of x and y in the expression?

Mathematics
2 answers:
NeX [460]3 years ago
8 0

Answer:

X=1   y=6

hope this helps i just took the test!

Anuta_ua [19.1K]3 years ago
6 0
First, we multiply the given expressions:
(a + 3b) (-a² + 4ab + 2b)
= -a³ + 4a²b + 2ab - 3a²b + 12ab² + 6b²
Simplifying:

-a³ + a²b + 2ab + 12ab² + 6b²

By comparison, we can determine the values of x and y, which work out to be:
x = 1
b = 6
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Evaluate the integral. 3 2 t3i t t − 2 j t sin(πt)k dt
sveta [45]

∫(t = 2 to 3) t^3 dt

= (1/4)t^4 {for t = 2 to 3}

= 65/4.

----

∫(t = 2 to 3) t √(t - 2) dt

= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2

= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du

= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}

= 26/15.

----

For the k-entry, use integration by parts with

u = t, dv = sin(πt) dt

du = 1 dt, v = (-1/π) cos(πt).


So, ∫(t = 2 to 3) t sin(πt) dt

= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt

= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]

= 5/π + 0

= 5/π.

Therefore,

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