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USPshnik [31]
3 years ago
8

There are three consecutive integers. The product of the first and the last is 35 more than 5 times the second number. Find thes

e integers.
Will give brainliest to best answer

will give thanks and 5 stars rating to good answers!


Please show all work (don't show 5+4 = 9 but you get the idea)
Mathematics
2 answers:
mamaluj [8]3 years ago
7 0

Answer:

8,9,10

Step-by-step explanation:

Let x-1, x and x+1 be the consecutive integers

(x-1)(x+1) = 5x + 35

x² - 1 - 5x - 35 = 0

x² - 5x - 36 = 0

x² - 9x + 4x - 36= 0

x(x - 9) + 4(x - 9) = 0

(x - 9)(x + 4) = 0

x = 9, -4

Rudik [331]3 years ago
7 0

Answer:

Step-by-step explanation:

hello :

let this three consecutive integers : x  , x+1  , x+2

x(x+2)=35+5(x+1)

x²+2x=35+5x+5

x²-3x-40=0

(x-8)(x+5)=0   .... x+5 ≠ 0 because : x integer

x-8=0   so : x=8

three consecutive integers are : 8 and 9 and 10

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\:  \:  \: \:  \:     \blue{\underline{\underline{ \large{ \tt{ \red{✧A \: N \: S \: W \: E \: R}}}}}}

\large{ \tt{✽  Step - By - Step \:Explanation}} :

\underline{ \underline{ \large{ \tt{G \: I\: V \: E \: N}}} }:

  • C.P of two mobile set = Rs 55555
  • Profit % on selling 1 mobile = 10%
  • Loss % on selling another mobile = 10%

\underline{ \underline{ \large{ \tt{S\: O \: L \:V \:I \: N \: G}} }}....

\begin{array}{  |c| c | } \tt{Case \: I}& \text{Case \: II}  \\  \hline \hline  \tt{CP = x} &  \tt{ CP = 55555 - x} \\  \tt{Profit\% =10   }  & \tt{Loss\% = 10\%} \\\   \tt{SP_{1} =   \frac{cp(100 + P\%)}{100} }\\    &   \tt{SP_{2} =  \frac{cp(100 - L\%)}{100}   }\\   = \tt{\frac{x(100 + 10)}{10} }  &  =   \tt{\frac{9 \cancel{0}(55555 - x)}{1 \cancel{00}}}  \\  \\ =  \frac{11}{10}  x  &  =   \tt{\frac{499995 - 9x}{10}} \end{array}

\underline{ \large{ \tt{According \: To \:Question \: (ATQ)}} }:

\large{ \tt{ SP_{1} =  SP_{2}}}

⤏ \large{ \tt{ \frac{11}{10} x =  \frac{499995 - 9x}{10}}}

⤏ \large{ \tt{110x = 4999950 - 90x}}

⤏ \tt{110x + 90x = 4999950}

⤏ \large{ \tt{200x = 4999950}}

⤏ \large{ \tt{x =  \frac{4999950}{200}}}

⤏ \large{ \tt{x = 24999.97}}

\large{ \tt{Replacing \: Value}} :

⤏ \large{ \tt{ SP = 2( \frac{11}{10 } \times 24999.97) = \underline{Rs \: 54999.934}}}

\large{ \tt{CP = Rs \: 55555}}

Since SP < CP , he / she made a loss

\large{ \tt{Loss\% =  \frac{CP- SP}{CP}  \times 100\% }}

⟶ \large \tt \:  \frac{55555 - 54999.934}{55555}  \times 100\%

⟶ \large{ \tt{0.99\%}}

⟶\boxed{ \large{ \tt{1\%}}}

Hence , Loss percent = 1%

Hope I helped ! ♕

Have a wonderful day / night ! ツ

✏\underbrace{ \overbrace{  \mathfrak{Carry \: On \: Learning}}} !!

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

3 0
3 years ago
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