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swat32
3 years ago
5

Determine the solution set of (2x - 5)2 = 11.

Mathematics
2 answers:
zhenek [66]3 years ago
8 0
I think what you meant was

(2x - 5)² = 11 -- (1)

Square root both sides of (1), i.e.

√(2x - 5)² = ± √11 -- (2)

From (2), we have

2x - 5 = ± √11 -- (3)

By adding 5 to both sides in (3), we have

2x = 5 ± √11 -- (4)

Divide both sides of (4) by 2, and we obtain

x = (5 ± √11)/2 -- (5)

From (5), the solution set of (1) is

x = (5 + √11)/2, (5 - √11)/2 ...Ans.
Fynjy0 [20]3 years ago
3 0

Taking square root on both sides, that becomes 2x - 5 = (+/-) √11

Then:

2x = (+/-)√11 + 5

x = [5 +/- √11 ] / 2

x = 5/2 +/- (√11)/2

That is, x = 5/2 +(√11)/2 and x = 5/2 - (√11)/2

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Step-by-step explanation:

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4.166666.... which is equal to 25/6

Step-by-step explanation:

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X^2+16x+64 can anyone show the work for this and find the factor.
Alexandra [31]

Answer:

should i simplify it? or factorise it?

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x^2+16+64

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8 0
3 years ago
Which equations represent the line that is perpendicular to the line 5x − 2y = −6 and passes through the point (5, −4)? Select t
nignag [31]

For this case we have that by definition, the equation of a line in the slope-intersection form is given by:

y = mx + b

Where:

m: It's the slope

b: It is the cut-off point with the y axis

On the other hand we have that if two lines are perpendicular, then the product of their slopes is -1. So:

m_ {1} * m_ {2} = - 1

The given line is:

5x-2y = -6\\-2y = -6-5x\\2y = 5x + 6\\y = \frac {5} {2} x + \frac {6} {2}\\y = \frac {5} {2} x + 3

So we have:

m_ {1} = \frac {5} {2}

We find m_ {2}:m_ {2} = \frac {-1} {\frac {5} {2}}\\m = - \frac {2} {5}

So, a line perpendicular to the one given is of the form:

y = - \frac {2} {5} x + b

We substitute the given point to find "b":

-4 = - \frac {2} {5} (5) + b\\-4 = -2 + b\\-4 + 2 = b\\b = -2

Finally we have:

y = - \frac {2} {5} x-2

In point-slope form we have:

y - (- 4) = - \frac {2} {5} (x-5)\\y + 4 = - \frac {2} {5} (x-5)

ANswer:

y = - \frac {2} {5} x-2\\y + 4 = - \frac {2} {5} (x-5)

3 0
3 years ago
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