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Paul [167]
3 years ago
15

Find two consecutive odd integers such that their product is 23 more than 5 times their sum.

Mathematics
1 answer:
Zina [86]3 years ago
7 0

If x is the smaller one, and is odd, the next consecutive odd integer is x+2.

x(x+2) = 23 + 5(x+ (x+2))

x^2 + 2x = 23 +5(2x+2)

x^2 + 2x = 23 + 10x + 10

x^2 -8x - 33 = 0

(x-11)(x+3) = 0

x = 11\textrm{ or }x=-3

Check:

11(13) = 143

23 + 5(11 + 13) = 23 + 120 = 143, check

-3(-1) = 3

23 + 5(-3 + -1) = 3, check

Answer: 11,13 or -3,-1



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Answer:

<h3>x=60</h3>

Step-by-step explanation:

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6 0
2 years ago
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How many solution does this have and how do you solve it
Vadim26 [7]

Answer:

Infinite solutions

Step-by-step explanation:

To determine the number of solutions a system has, you need its slope. If the slopes are the same, check if the equations are equivalent or not.

To find the slope, rearrange the equation to isolate "y", which makes it slope-intercept form y = mx + b.

"x" and "y" represent points on the line.

"m" represents the slope, which is how steep a line is.

"b" represents the y-intercept (where the graph hits the y-axis).

Isolate "y" in both equations:

x + 3y = 0

3y = -x        Subtract 'x' from both sides

y = -x/3        Divide both sides by 3.

y = -\frac{1}{3}x

The slope is -1/3.

9y = -3x

y = -3x/9        Divide both sides by 9

y = -\frac{3}{9}x        Reduce the fraction to lowest terms

y = -\frac{1}{3}x

The slope is -1/3, but the two equations are the same.

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5 0
3 years ago
Point B has coordinates ​(1​,2​). The​ x-coordinate of point A is negative 8. The distance between point A and point B is 15 uni
olchik [2.2K]

Answer:

The point A will be  (-8,14) or (-8,-10).

Step-by-step explanation:

Point B has coordinates (1,2) and the x-coordinate of point A is - 8.

Let us assume that the coordinates of point A are (-8,k).

Now, given that the point A is 15 units apart from point B.

Therefore, from the distance formula, we can write that  

\sqrt{(1 - ( -8))^{2} + (2 - k)^{2}} = 15

Now,squaring both sides, we get  

(1 - ( -8))^{2} + (2 - k)^{2} = 225

⇒ (2 - k)^{2} = 225 - 9^{2} = 144

⇒ 2 - k = ± 12

⇒ k = 14 or -10.

Therefore, the point A will be  (-8,14) or (-8,-10). (Answer)

We know that the distance between two points on the coordinate plane (x_{1}, y_{1}) and (x_{2}, y_{2}) is given by  

\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}.

7 0
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Sorry again... but i need help on this one too:(
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The answer is A
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