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Dafna11 [192]
2 years ago
15

The sum of two consecutive even integers is 122. What are the two numbers?

Mathematics
2 answers:
irakobra [83]2 years ago
7 0

Answer:

  1. Let the first number = x
  2. Therefore, the other number = x + 2 (x+2 is so as in two consecutive even numbers there is 1 odd number)
  3. Now, According to question
  4. x + x + 2 = 122
  5. 2x + 2 = 122
  6. 2x = 122-2
  7. 2x = 120
  8. x = 120/2
  9. x = 60
  10. Now, first no. = x = 60
  11. Therefore, second no. = x + 2 = 60 + 2 = 62
  12. Equation 60 + 62 = 122

Step-by-step explanation:

If you like my answer than please mark me brainliest

leonid [27]2 years ago
3 0

Answer:

The numbers are : 60 , 62

Step-by-step explanation:

Let the two consecutive even numbers be 2x , 2x + 2

2x + 2x + 2 = 122

Combine like terms

4x + 2 = 122

Subtract 2 from both sides

4x = 122 - 2

4x = 120

x = 120/4

x = 30

2x = 2*30 = 60

2x + 2 = 2*30 + 2 = 60 + 2 = 62

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abruzzese [7]

Answer:

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Step-by-step explanation:

For h hours, Brandee's pay will be ...

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  pay = 40r +1.5hr -60r . . . . . eliminate parentheses

  pay = 1.5hr -20r . . . . . . . . . . collect terms

Solving for h, we have ...

  pay +20r = 1.5hr . . . . . . . . . . add 20r

  h = (pay +20r)/(1.5r) . . . . . . . . divide by 1.5r

For the given pay of 800 +240 = 1040, we have ...

  h = (1040 +20r)/(1.5r)

  h = 13.333 +693.333/r . . . . . simplify to 2 terms

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<em>Additional comments</em>

You need to know r to find the number of hours Brandee worked. She got paid $800 for a presumed 40-hours of regular time, so made r = $20 per hour. The above formula will tell you she worked 48 hours in the pay period.

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3 years ago
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Daniel [21]

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3 years ago
Show work please<br> \sqrt(x+12)-\sqrt(2x+1)=1
Nesterboy [21]

Answer:

x=4

Step-by-step explanation:

Given \displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1, start by squaring both sides to work towards isolating x:

\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2

Recall (a-b)^2=a^2-2ab+b^2 and \sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}:

\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1

Isolate the radical:

\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}

Square both sides:

\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2

Expand using FOIL and (a+b)^2=a^2+2ab+b^2:

\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36

Move everything to one side to get a quadratic:

\displaystyle-\frac{1}{4}x^2+7x-24=0

Solving using the quadratic formula:

A quadratic in ax^2+bx+c has real solutions \displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}. In \displaystyle-\frac{1}{4}x^2+7x-24, assign values:

\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24

Solving yields:

\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}

Only x=4 works when plugged in the original equation. Therefore, x=24 is extraneous and the only solution is \boxed{x=4}

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Alexxandr [17]

Answer:

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Step-by-step explanation:

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lilavasa [31]
H + 732 = -194

* combine like terms

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