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laiz [17]
3 years ago
10

Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.

Mathematics
1 answer:
Mice21 [21]3 years ago
5 0

Answer:

<em>LCM</em> = 8y^{4}(y+ 10)^{2}(y + 8)

Step-by-step explanation:

Making factors of 8y^{6}+ 144y^{5}+ 640y^{4}

Taking 8y^{4} common:

\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)

Using <em>factorization</em> method:

\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times  4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)

Now, Making factors of 2y^{4} + 40y^{3} + 200y^{2}

Taking 2y^{2} common:

\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)

Using <em>factorization</em> method:

\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10)        ............ (2)

The underlined parts show the Highest Common Factor(HCF).

i.e. <em>HCF</em> is 2y^{2} (y+ 10).

We know the relation between <em>LCM, HCF</em> of the two numbers <em>'p' , 'q'</em> and the <em>numbers</em> themselves as:

HCF \times LCM = p \times q

Using equations <em>(1)</em> and <em>(2)</em>: \Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)

Hence, <em>LCM</em> = 8y^{4}(y+ 10)^{2}(y + 8)

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Answer:

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Step-by-step explanation:

12 - x^2 = 0

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3 years ago
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In the right triangle shown, m∠A = 30 degrees, and BC = 6√2, how long is AC?
mart [117]

They're trying to trick you with √2.  Remember in the 30/60/90 triangle the sides are in ratio 1:√3:2, with the "1" opposite the 30 degrees.

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Three monkey's walk into a motel on the Curious George Hotel and ask for a room. The desk clerk says a room costs 30 bananas, so
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Step 1

Given;

\begin{gathered} Cost\text{ for a room= \$30} \\ Each\text{ monkey pays 30 bananas} \\ Clerk\text{ ask bell boy to refund 5 bananas} \\ Bell\text{ boy refunds 3 bananas} \\ Bell\text{ boy keeps 2 bananas} \end{gathered}

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Step 2

Here is the break down; the balance should be like this;

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Hence, there is no extra 1 banana, The monkeys spent 27 bananas of which the bellboy took 2 bananas and 25 bananas went to the room, and 3 bananas where returned to them. A total of 30 bananas

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1 year ago
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