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masya89 [10]
3 years ago
7

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

Mathematics
1 answer:
suter [353]3 years ago
5 0
I thinks this equation is false it doesn’t have an answer
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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
2k=18
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
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