Round 823 to the nearest ten

Mathematics

2 answers:

kirza4 [7]2 years ago

7 0

Answer:

820

Step-by-step explanation:

Find the number in the ten place 2 and look one place to the right for the rounding digit 3. Round up if this number is greater than or equal to

5 and round down if it is less than 5.

bulgar [2K]2 years ago

4 0

820

bc it is not 825 or higher it stays the same

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What is the quotient: (6x4 + 15x3 + 10x2 + 10x + 4) ÷ (3x2 + 2)?

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Answer: Option 'C' is correct.

Step-by-step explanation:

Since we have given that

$\frac{\left(6x^4+15x^3+10x^2+10x+4\right)}{\left(3x^2+2\right)}$

Now, we will find the quotient by factoring the numerator:

$\mathrm{Use\:the\:rational\:root\:theorem}\\a_0=4,\:\quad a_n=6\\\\\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:4,\:\quad \\\mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2,\:3,\:6\\\\\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:4}{1,\:2,\:3,\:6}\\\\-\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+2\\\\=\left(x+2\right)\frac{6x^4+15x^3+10x^2+10x+4}{x+2}\\\\=\frac{6x^4+15x^3+10x^2+10x+4}{x+2}=6x^3+3x^2+4x+2\\\\$

Now, we will factor it again:

$=\left(6x^3+3x^2\right)+\left(4x+2\right)\\\\=2\left(2x+1\right)+3x^2\left(2x+1\right)\\\\=\left(2x+1\right)\left(3x^2+2\right)$

At last we get our factorised form :

$=\left(x+2\right)\left(2x+1\right)\left(3x^2+2\right)\\\\=\frac{\left(x+2\right)\left(2x+1\right)\left(3x^2+2\right)}{3x^2+2}\\\\=\left(x+2\right)\left(2x+1\right)\\\\=2x^2+5x+2$