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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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Step-by-step explanation:

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3 years ago
Simplify :<br>(a+b)²<br>Thanks ​
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(a+b)(a+b)

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6 0
3 years ago
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Can you please help me with 23 and 25 I do not understand!!! Please help
adell [148]

when you multiply a number by a value that's greater than 1, that number increases, for example 2 and then we multiply it like 2 * 3, clearly 2*3 is greater than 2.

now when you multiply a number  by a value that's lesser than 1, that number decreases, for example 2 and then we multiply it by something like 1/4, so 2 * 1/4 = 1/2, well, clearly 1/2 is smaller than 2, so 1/2 is a decreased value of 2.

23)

we have here 4/7, and then a product, well, the product is 4/7 and 9/10, however 9/10 is less than 1, recall 1 = 10/10, 9/10 is just a tiny bit less than 1.

because we're multiplying 4/7 by some amount less than 1, the product is a smaller value than 4/7, thus

 \bf \cfrac{4}{7}~~>~\left( \cfrac{9}{10}\times \cfrac{4}{7} \right)

25)

we have here 5/6 and we multiply it by 7/7, wait a minute!!! 7/7 = 1, we're really just multiplying 5/6 by 1 then, what does that give us?  well, 5/6 * 1  = 5/6, just the number itself,

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3 years ago
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3 years ago
Sherman goes golfing every 6^\text{th}6 th 6, start superscript, start text, t, h, end text, end superscript day and Brad goes g
nydimaria [60]

Answer:

In every 42 days,  Sherman and Brad will go golfing on the same day.

Step-by-step explanation:

Given:

Sherman goes golfing every 6th day.

Brad goes golfing every 7th day.

If they both went golfing today, we need to determine how many days unit they will go golfing on the same day again.

Solution:

In order to determine how many days unit Sherman and Brad will go golfing on the same day again, we will take least common multiple of 6 and 7.

To find least common multiple of 6 and 7, we will list out their multiples.

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We find out that the least common multiple of 6 and 7 =42.

Thus, we can conclude that Sherman and Brad will go golfing on same days on every 42nd day after they meet..

7 0
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