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3 years ago

Given the equation 2 Square root of x minus 5 = 2, solve for x and identify if it is an extraneous solution.

2 answers:

8 0

2 square root (x - 5) = 2. 2^2 (x -5) = 2^2. 4(x - 5) = 4. 4x - 20 = 4. 4x = 24. x = 6. This solution is not extraneous, because extraneous solutions emerge from solving the problem but are not actually valid solutions for the initial problem. With rounding, this solution is valid for the initial problem.

mylen [45]3 years ago

8 0

Answer:

x = 6, solution is not extraneous

Step-by-step explanation:

Given the equation 2 Square root of x minus 5 = 2

$2\sqrt{x-5} =2$ . solve for x

Divide by 2 on both sides

$\sqrt{x-5} =1$

Take square on both sides to remove square root

x-5= 1

add 5 on both sides

x= 6

Now we plug in x=6 in our original equation

$2\sqrt{6-5} =2$

2 = 2 is true

So x= 6 satisfies our given equation

X=6 is not an extraneous solution

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3 years ago

A right angle triangle is shown with hypotenuse equal to 13 centimeters. An acute angle of the triangle is labeled as x degrees.

erastova [34]

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3 years ago

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A shop has determined that the number of bottles of water they sell on a summer day is a function of the day’s high temperature,

Alik [6]

Answer:

$T=t+94$

Where T is the hight temperature and t is the day

Explanation:

The table shows that every day the High temperature, T (degrees F) increases 1 unit.

Then, this is a linear function with slope = 1ºF / day.

You can use the point-slope equation to determine the function of the high temperature:

Point-slope equation:

$y-y_1=m(x-x_1)$

Choose any point from the table and m = 1. I will use the first point (1,95)

m = 1

x₁ = 1

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Substitute:

$y-95=1(x-1)\\\\y=x-1+95\\\\y=x+94$

Make the variables y = T, and x = t:

$T=t+94$

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3 years ago

If the point (a, b) is equidistant from the points (-a, 2) and (2, -b), then prove that 3(a+b)+4=0

erastova [34]

Answer: see proof below

Step-by-step explanation:

Since (a, b) is equidistant from (-a, 2) and (2, -b), then it is the midpoint of the those two points. Use Midpoint formula to find (a, b).

$M_x=\dfrac{x_1+x_2}{2}\qquad \qquad \qquad M_y=\dfrac{y_1+y_2}{2}\\\\\\a=\dfrac{-a+2}{2}\qquad \qquad \qquad \quad b=\dfrac{2-b}{2}\\\\\\2a=-a+2\qquad \qquad \qquad \quad 2b=2-b\\\\\\3a=2\qquad \qquad \qquad \qquad \qquad 3b=2\\\\\\a=\dfrac{2}{3}\qquad \qquad \qquad \qquad \qquad b=\dfrac{2}{3}$