<img src="https://tex.z-dn.net/?f=f%5Cleft%28x%5Cright%29%3D%5Cfrac%7B%5Cleft%28x%5E%7B3%7D%2B8x%5E%7B2%7D%2B21x%2B18%5Cright%29

Ask question

Ask question

All categories

3 years ago

10

%7D%7B%5Cleft%28x%2B3%5Cright%29%7D" id="TexFormula1" title="f\left(x\right)=\frac{\left(x^{3}+8x^{2}+21x+18\right)}{\left(x+3\right)}" alt="f\left(x\right)=\frac{\left(x^{3}+8x^{2}+21x+18\right)}{\left(x+3\right)}" align="absmiddle" class="latex-formula">

1 answer:

scoundrel [369]3 years ago

8 0

Answer:

f = $\frac{x^{3}+8x^{2}+39}{x^{2}+3x }$

Step-by-step explanation:

Solved for f!

You might be interested in

Can anyone help me on this last one

andreev551 [17]

Answer: 7 cm

The length looks around the same size as the base.

8 0

2 years ago

Read 2 more answers

A water park charges a rental fee plus $1.50 per hour to rent inflatable rafts. The total cost to rent a raft for 6 hours is $15

lawyer [7]

Answer:

6*1.50= 9 is the answer

Step-by-step explanation:

7 0

3 years ago

Which number sentence is true?

UNO [17]

Answer:

B

Step-by-step explanation:

|-82| < |-83|

If the absolute values are replaced by an integer, 82 < 83

7 0

2 years ago

Read 2 more answers

The net of the right circular cylinder shown is made of

max2010maxim [7]

That may be any kinda right circular cylinder it always contain a net like attachment.

So

There is a rectangle which is not a square for LSA
And two bases which are circles

Option D

3 0

2 years ago

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,

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

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

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

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

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

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

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

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

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

_______________________________

<em>Please scroll left - right to view the full solution.</em>

8 0

2 years ago