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3 years ago

Evaluate 3x(x-2)/2x-1 when x=0

Mathematics

1 answer:

Tpy6a [65]3 years ago

6 0

Answer:

The answer is zero, because the entire equation is being multiplied by zero. The only way that will give you non-zero value is if the denominator in the fraction came to zero as well, but in this case it does not.

Step-by-step explanation:

$3x\frac{x - 2}{2x - 1} \\3(0)\frac{0-2}{2(0) - 1}\\0$

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Order the following values from least to greatest.

ch4aika [34]

Would be -91,-24,12,50.8

6 0

3 years ago

Read 2 more answers

How can i prove this property to be true for all values of n, using mathematical induction.

chubhunter [2.5K]

Proof -

So, in the first part we'll verify by taking n = 1.

$\implies \: 1 = {1}^{2} = \frac{1(1 + 1)(2 + 1)}{6}$

$\implies{ \frac{1(2)(3)}{6} }$

$\implies{ 1}$

Therefore, it is true for the first part.

In the second part we will assume that,

$\: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} = \frac{k(k + 1)(2k + 1)}{6} }$

and we will prove that,

$\sf{ \: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 1 + 1) \{2(k + 1) + 1\}}{6}}}$

$\: {{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 2) (2k + 3)}{6}}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k (k + 1) (2k + 1) }{6} + \frac{(k + 1) ^{2} }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k(k+1)(2k+1)+6(k+1)^ 2 }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)\{k(2k+1)+6(k+1)\} }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2 +k+6k+6) }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2+7k+6) }{6}$

${1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(k+2)(2k+3) }{6}$

Henceforth, by using the principle of mathematical induction 1²+2² +3²+....+n² = n(n+1)(2n+1)/ 6 for all positive integers n.

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8 0

2 years ago

I really need help with this and I have so many assignments to do pls help me so I don’t get my phone taken

solong [7]

Step-by-step explanation:

Ithink a

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3 years ago

What is the product of the polynomial 2x^3 + 8x^2 + x + 3 times the monomial x?

Elis [28]

Answer:

$\huge\boxed{2x^4 + 8x^3 + x^2 + 3x}$

Step-by-step explanation:

When we multiply a polynomial by a specific monomial, all terms in the polynomial get multiplied by the monomial.

So we can do this by term.

$2x^3 \cdot x$

Multiplying x by x to the something power will just increase the exponent on the x term by one (since powers are just x multiplied by itself). This means the exponent here will increase by 1, so this becomes $2x^4$ .

Same logic applies to the second term: $8x^2$ , raise the exponent by one, $8x^3$ .

For the third term, if there is no exponent on a term you can assume it's exponent is 1 (so $x^1$ in this case - as $x^1=x$ ). Again, using the same logic, this turns into $x^2$ .

For the last term, when we multiply a constant by a variable, we get a term with a variable and a coefficient - the constant becomes the coefficient and the variable stays the variable. Therefore, $3 \cdot x = 3x$ .

Adding these terms all together gets us $2x^4 + 8x^3 + x^2 + 3x$ .

Hope this helped!

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4 years ago

Please help me with number 9

Mrrafil [7]

The answer is 8 pints

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3 years ago