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Gre4nikov [31]
4 years ago
7

The sum of squares of three consecutive numbers is 77 find the numbers​

Mathematics
1 answer:
ollegr [7]4 years ago
3 0

Answer:

4,5,6 are the three consecutive numbers. 16, 25 and 36 are their squares.

Step-by-step explanation:

Let the three consecutive numbers be x, (x+1), (x+2)

Now, the squares of these three numbers are  x^{2} ,(x+1)^{2} ,(x+2)^{2}

Sum = 77

∴by the problem ,

x^{2} +(x+1)^{2}+(x+2)^{2}  = 77\\x^{2} +(x^{2} +2x+1)+(x^{2}+4x +4) = 77\\x^{2} +x^{2} +2x+1+x^{2} +4x +4 = 77\\3x^{2} +6x+5 = 77\\3x^{2} +6x = 77-5\\3x^{2} + 6x = 72\\3x^{2} +6x-72= 0\\

{Taking 3 common }

x^{2} +2x- 24 = 0\\

{By factorization}

x^{2} +2x- 24 =0\\ x^{2} +6x-4x-24=0\\x(x+6)-4(x+6)=0\\(x+6)(x-4)=0\\

Therfore,

x= -6,4\\

<em>X can't be negetive </em>

∴ x =4\\x+1=5\\x+2=6

The squares of the three consecutive numbers are 16, 25, 36

The three consecutive numbers whose sum is 77 are 4, 5, 6

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Juan is working two summer jobs, making $12 per hour babysitting and making $16 per hour lifeguarding. In a given week, he can w
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\text{Let }b=

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\text{Let }l=

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Use a \le≤ symbol

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\text{Plug in }\color{green}{3}\text{ for }b\text{ and solve each inequality:}

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\begin{aligned}b+l\le 17\hspace{10px}\text{and}\hspace{10px}&12b+16l\ge 240 \\ \color{green}{3}+l\le 17\hspace{10px}\text{and}\hspace{10px}&12\left(\color{green}{3}\right)+16l\ge 240 \\ l\le 14\hspace{10px}\text{and}\hspace{10px}&36+16l\ge 240 \\ \hspace{10px}&16l\ge 204 \\ \hspace{10px}&l\ge 12.75 \\ \end{aligned}

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​  

 

12b+16l≥240

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16l≥204

l≥12.75

​  

 

\text{The values of }l\text{ that make BOTH inequalities true are:}

The values of l that make BOTH inequalities true are:

\{13,\ 14\}

{13, 14}

\text{(the final answer is this entire list)}

(the final answer is this entire list)

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