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3 years ago

What are the solution(s) to the quadratic equation x2 – 25 = 0?

2 answers:

8 0

X^2 - 25 = 0
Add both sides by 25
x^2 = 25
Take the positive and negative square root.

x = 5
OR
x = -5

Have an awesome day! :)

Amiraneli [1.4K]3 years ago

4 0

Answer:

there are two solutions are x = (+5) and x = (-5)

Step-by-step explanation:

The given quadratic equation is x² - 25 = 0

Now we can get the solutions of this equation by zero factors.

[ since (a² - b²) = ( a-b ) ( a+b ) ]

so x² - 25 = (x-5) (x+5)

( x-5 ) = 0 ⇒ x = 5

or x + 5 = 0 ⇒ x = -5

Therefore, there are two solutions are x = (+5) and x = (-5)

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The percentage in fraction in its simplest form is 5/4%

<h3>How to find percentage</h3>

Given:

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3 1/3 = x% of 800

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Plug it into the equation above:

3^2 + b^2 = 8^2

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3 years ago

Line segment BD divides trapezoid ABCE into a rectangle and a right

grin007 [14]

<h3>Answer: 96 units</h3>

===========================================================

Explanation:

Since ABDE is a rectangle, the opposite sides AE and BD are the same length. This makes BD = 24. Similarly, ED = 20 since AB = 20.

Focus on triangle BCD. Let x be the length of segment CD. We can use the pythagorean theorem to find x

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(BD)^2 + (CD)^2 = (BC)^2

24^2 + x^2 = 25^2

576 + x^2 = 625

x^2 = 625 - 576

x^2 = 49

x = sqrt(49)

x = 7

Segment CD is 7 units long.

Since ED = 20, this means,

EC = ED + DC

EC = 20 + 7

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P = AB+BC+CE+EA

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3 years ago

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4-15 all this is overdue and I cant understand anything

attashe74 [19]

I'm not 100% sure how to do this, but it would help if you had any notes/online notes/textbook that shows how to do it (also to see if what I did is the same as the notes)

4. B<u>reak-even point</u> (It's basically the intersection point. Or the point when the two lines meet/the point that is the same for both equations)

I plugged in numbers for x into both equations, until C and R were the same number.

I had x = 1, so I plugged in 1 for x into both equations:

C = 15x + 150 C = 15(1) + 150 = 15 + 150 = 165

R = 45x R = 45(1) = 45

Then I had x = 2, then x = 3, and so on until C and R were the same number (meaning they had the same point), which was (5, 225)

Idk how you are suppose to check the solution because there are multiple ways to do this. I doubt this, but one option is plugging (5, 225) back into the equations:

C = 15x + 150 R = 45x

225 = 15(5) + 150 225 = 45(5)

225 = 75 + 150 225 = 225

225 = 225

5. (I made another sort of table? because I didn't know what kind you were suppose to use) Do the same as #4, and plug in numbers for x until C and R were the same number

6. Do the same as #4 (If you have any questions or information I should know then let me know)

7. y = mx + b

"m" is the slope, "b" is the y-intercept [the y value when x = 0 (0, y)]

The y-intercepts are -2 and 13, so the graph is B. Now you need to estimate the solution(point of intersection) by looking at the graph. It looks around (6,7).

Now you need to check the solution by plugging the point into both equations.

y = 1.5x - 2

7 = 1.5(6) - 2

7 = 9 - 2

7 = 7

y = -x + 13

7 = -6 + 13

7 = 7 The point is the solution

8. The graph is A. Do the same as #7 and estimate a point. I will say around (2.5,6.5) And check this by plugging the point into both equations

y = x + 4

6.5 = 2.5 + 4

6.5 = 6.5

y = 3x - 1

6.5 = 3(2.5) - 1

6.5 = 7.5 - 1

6.5 = 6.5 The point is the solution

9. Graph C, do the same as #7 and #8

10. y = mx + b

"b" is (0,b) or (0,y)

"m" is the slope

( $slope=\frac{rise}{run}$

Rise is the number of units you go up(+) or down(-)

Run is the number of units you go to the right)

You need to graph the lines of each equation, and find the point of intersection.

11-15. Do the same as #10

If "y" is not by itself, get it by itself.

#13:

x + y = 27 Subtract x on both sides to get "y" by itself

y = 27 - x

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3 years ago