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Luden [163]
3 years ago
6

What is the least common multiple of x^2 + 11

Mathematics
1 answer:
artcher [175]3 years ago
4 0
These 2 terms have nothing whatsoever in common, other than the multiplier (+1), so there really is no LCM.

If you had 5x^2 + 5(11), then the LCM would be 5.

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There is no table!!!! Please put a table the next time!!!!
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What is 6.7 x (-2.3) round to the nearest hundredth
Vaselesa [24]
Your answer would be -15.41. If you ever need any other kind of help, just ask me.
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Stephanie is hosting an art show and she would like to raise at least $800. She plans on
Masteriza [31]

Answer: 10, 15

Step-by-step explanation:

Given

Stephanie wants to raise at least $800

she plans on selling VIP tickets for $20 each and general admission for $40 each

suppose there are x and y tickets for VIP and general admission

\therefore x+y\leq 25\\\\\Rightarrow 20x+40y\geq 800

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7 0
3 years ago
Help me guys <br> thx so much
navik [9.2K]

Answer:

Option D, 10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}

Step-by-step explanation:

<u>Step 1:  Multiply</u>

<u />(\sqrt{10x^4} -x\sqrt{5x^2} )*(2\sqrt{15x^4} + \sqrt{3x^3})\\ (\sqrt{10 * x^2 * x^2} -x\sqrt{5 * x^2} ) * (2\sqrt{15 * x^2 * x^2} +\sqrt{3 * x^2 * x})\\(x^2\sqrt{10} -x^2\sqrt{5} )*(2x^2\sqrt{15} +x\sqrt{3x}) \\\\

(x^2\sqrt{10}*2x^2\sqrt{15} )+(x^2\sqrt{10}*x\sqrt{3x} ) + (-x^2\sqrt{5} *2x^2\sqrt{15}) + (-x^2\sqrt{5} *x\sqrt{3x}

(2x^4\sqrt{150} ) + (x^3\sqrt{30x}) + (-2x^4\sqrt{75}) + (-x^3\sqrt{15x}  )

2x^4\sqrt{5^2*6} + x^3\sqrt{30x} -2x^4\sqrt{5^2*3} -x^3\sqrt{15x}

10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}

Answer:  Option D, 10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}

6 0
3 years ago
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Ok whoever answers this correct will get 100 brainly points
Snezhnost [94]

Answer:

We are given an area and three different widths and we need to determine the corresponding length and perimeter.  

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The second width that is given is 5 yards and in order to get an area of 100 yards we need to multiply by 20 yards.  This would mean that our length is 20 yards and our perimeter would be 2(l + w) which is 2(20 + 5) = 50 yards.

The final width that is given is 10 yards and in order to get an area of 100 yards we need to multiply by 10.  This would mean that our length is 10 yards and our perimeter would be 2(l + w) which is 2(10 + 10) = 40 yards.

Therefore the field that would require the least amount of fencing (the smallest perimeter) is option C, field #3.

<u><em>Hope this helps!</em></u>

3 0
2 years ago
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