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ioda
3 years ago
5

Find the product of all real values of r for which 1/2x=r-x/7

Mathematics
1 answer:
Dahasolnce [82]3 years ago
8 0

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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Answer:

Option 1: 0.32

Step-by-step explanation:

Let

P(A) be the experimental probability of getting two

And

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And

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3 years ago
Find an equation for the line with the given properties. Perpendicular to the line 7x - 3y = 68; containing the point (8, -8)
Sever21 [200]

Answer:

y=\dfrac{-3}{7}x-\dfrac{32}{7}

Step-by-step explanation:

Given that,

A line 7x - 3y = 68 and containing the point (8, -8).

The equation can be written as :

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The slope is :7/3

Line is perpendicular so use m = –3/7

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So, required equation is :

y=\dfrac{-3}{7}x-\dfrac{32}{7}

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