Ask question

Ask question

All categories

3 years ago

7

How to find the answer for (5+4)3-2=1

1 answer:

suter [353]3 years ago

5 0

I thinks this equation is false it doesn’t have an answer

You might be interested in

Denise is building a fence around her 18ft by 34 foot rectangular garden. She put a post at each corner, put some posts along th

tigry1 [53]

Answer:

The greatest possible distance between each post and the next post is 17 ft

Step-by-step explanation:

Please find the attached document

3 0

3 years ago

If h(x)=x^4-2x^2-x+4 , then what is h(x)/(x+3)? You may use either long or synthetic division.

ozzi

Answer:

$\frac{x^4\:-\:2\:x^2\:-\:x\:+\:4}{x\:+\:3}=x^3-3x^2+7x-22+\frac{70}{x+3}$

Step-by-step explanation:

We are going to use long division to find the expression for $\frac{x^4\:-\:2\:x^2\:-\:x\:+\:4}{x\:+\:3}$

Step 1: Divide $\frac{x^4-2x^2-x+4}{x+3}$

Divide the leading coefficients of the numerator $x^4-2x^2-x+4$ and the divisor $x+3$

$\frac{x^4}{x}=x^3$

Quotient = $x^3$

Multiply $x+3$ by $x^3$

$x^4+3x^3$

Subtract $x^4+3x^3$ from $x^4-2x^2-x+4$ to get new remainder

Remainder = $-3x^3-2x^2-x+4$

Therefore,

$\frac{x^4-2x^2-x+4}{x+3}=x^3+\frac{-3x^3-2x^2-x+4}{x+3}$

Step 2: Divide $\frac{-3x^3-2x^2-x+4}{x+3}$

Quotient = $-3x^2$

Remainder = $7x^2-x+4$

Therefore,

$\frac{x^4-2x^2-x+4}{x+3}=x^3-3x^2+\frac{7x^2-x+4}{x+3}$

Step 3: Divide $\frac{7x^2-x+4}{x+3}$

Quotient = $7x$

Remainder = $-22x+4$

Therefore,

$\frac{x^4-2x^2-x+4}{x+3}=x^3-3x^2+7x+\frac{-22x+4}{x+3}$

Step 4:

Quotient = $-22$

Remainder = $70$

Therefore,

$\frac{x^4-2x^2-x+4}{x+3}=x^3-3x^2+7x-22+\frac{70}{x+3}$

3 0

3 years ago

Can someone answer me I don’t get it

valkas [14]

Answer:

B, C and F

Step-by-step explanation:

Negative numbers go farther and father to the left on a number line the higher in value they get.

5 0

3 years ago

yaroslaw [1]

Answer:

k=9

Step-by-step explanation:

K=9, because if you plug in 9 as k(2(9)=18)the equation is true.

5 0

2 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,

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

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

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

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

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

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

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

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

<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