3 years ago

Did I do anything wrong? how would I solve it if I did do something wrong??

Mathematics

2 answers:

s344n2d4d5 [400]3 years ago

5 0

Thats right!well done

levacccp [35]3 years ago

3 0

Answer:

No its correct x=25

Step-by-step explanation:

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3x(2-x)+2x(x-1)=5x(x+3)

mylen [45]

Step-by-step explanation:

6x - 3x^2 + 2x^2 -2x = 5x^2 + 15x

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-11x - 6x^2 = 0

-6x^2 - 11x = 0

use the quadratic formula

(-b±√(b²-4ac))/(2a)

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3 years ago

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2 years ago

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3 years ago

Select the two values of x that are roots of this equation.

charle [14.2K]

Answer:

$x1 = \frac{ - 3 - \sqrt{33} }{2} \\ x2 = \frac{ - 3 + \sqrt{33} }{2}$

Step-by-step explanation:

$x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac} }{2a} \\ = \frac{ - 3 + - \sqrt{9 + 24} }{2} \\ x1 = \frac{ - 3 - \sqrt{33} }{2} \\ x2 = \frac{ - 3 + \sqrt{33} }{2}$

7 0

3 years ago

I’m needing some help with this

Vikentia [17]

Answer: place points at (0,3) and (5,5)

Step-by-step explanation: the formula for a line is y = mx + b, where m is the slope and b is the y-intercept (the point where the line meets the y-axis). Since b is positive 3, place a point at (0,3).

m = 2/5, and since slope is rise over run (distance traveled in y axis over distance traveled in x axis) and it is positive, your next point is 2 up and 5 right, which is (5,5). You could also place a point at (-2,-2).

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1 year ago

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Did I do anything wrong? how would I solve it if I did do something wrong??​

Did I do anything wrong? how would I solve it if I did do something wrong??