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3 years ago

X and y table for y=x+2

Mathematics

2 answers:

LUCKY_DIMON [66]3 years ago

8 0

X+2=y
1+2=3
    so x=1, y=3
2+2=4
   so x=2, y=4
3+2=5
   so x=3, y=5
4+2=6
   so x=4, y=6

Andreyy893 years ago

5 0

Subsiutte
x,y
-10,-8
-9,-7
-8,-6
-7,-5
-6,-4
-5,-3
-4,-2
-3,-1
-2,0
-1,1
0,2
1,3
2,4
3,5
4,6
5,7
6,8
7,9
8,10
9,11
10,12

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Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

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2 years ago

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Answer:

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Step-by-step explanation:

Common denominator is already presented which is 4.

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you can also convert to decimal form -3/4 + 2 3/4 is the same as:

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1 year ago