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4 years ago

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The sum of the first and third of three consecutive even integers is 136. find the three even integers.

1 answer:

tatyana61 [14]4 years ago

5 0

Based on your problem, you only have one given. So, you can't make an equation for this because there are not limits to the equation. The only thing that you know is that the three numbers are consecutive even integers. My way of solution for this is trial-and-error. However, it's really quite easy.

For example: 42 + 46 = 88. I have to increase the numbers more to reach 136. Suppose: 82 + 86 = 168. That exceeded 136. So, it must be between 46 and 82. Suppose again: 66 + 70 = 136. Therefore, the sequence of the consecutive even integers are 66, 68, and 70.

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Pick up rational number from the following numbers : -6/7,0,100/0,5 3/4,0/2,24/1

Dominik [7]

Answer:

All of them

Step-by-step explanation:

A rational number is a number that can be in the form in a/b like 5/4, 8/9

100 can be write as: 1000/10

So all of the number that you list is all RATIONAL NUMBERS

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4 years ago

. Use the quadratic formula to solve each quadratic real equation. Round

Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

$\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}$ , with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}$

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}$

$\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5$

So B has two solutions of 5 and -1.5.

Now to C!

$\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}$

$\frac{44}{8} = 5.5$

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

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