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4 years ago

Find the value of x. (5x+15) (6x-10)

Mathematics

2 answers:

Annette [7]4 years ago

7 0

Answer:

We get x = -5/3 and x = -3

Step-by-step explanation:

$(5x+15) (6x-10)$

Multiplying this and equating to zero, to find x.

$30x^{2} -50x+90x-150=0$

$30x^{2} +40x-150=0$

Factoring out 10.

$3x^{2} +4x-15=0$

$3x^{2} +9x-5x-15=0$

3x(x+3)-5(x+3)=0 $[tex](3x-5)(x+3)=0$ [/tex]

We get x = -5/3 and x = -3

Blizzard [7]4 years ago

4 0

(5x+15)(6x-10)
= 30x^2 + 30x - 50x - 150
= 30x^2 - 20x - 150

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The value of y varies directly with x, and y = 48 when x = 16. Find y when x = 176.

emmainna [20.7K]

Try to find what y is when x is equal to one.

48/16= 3

So, when x is 1, y is 3.

(1,3)

All you need to do is multiply 176 by 3 for the correct value of y.

176*3= 528

When x= 176, y= 528. (176,528)

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3 years ago

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Nataly [62]

Answer:

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Step-by-step explanation:

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2 years ago

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3 years ago

X²+Y²=250 and XY=117<br> What are the values of X and Y?

yan [13]

Answer:

Step-by-step explanation:

$x^2+y^2=250\\\\x^2-2xy+y^2+2xy=250\\\\(x-y)^2=250-2xy\\\\(x-y)^2=250-2\cdot117\\\\ (x-y)^2=16\\\\x-y=4\qquad\qquad\vee\qquad \qquad x-y=-4\\\\x=4+y \qquad\qquad \vee\qquad\qquad x=-4+y\\\\(y+4)y=117\qquad\vee\qquad\quad (y-4)y=117\\\\y^2+4y-117=0\qquad\vee\qquad y^2-4y-117=0\\\\y=\dfrac{-4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\qquad\vee\qquad y=\dfrac{4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\\\\y=\dfrac{-4\pm\sqrt{16+468}}{2}\qquad\ \ \vee\qquad y=\dfrac{4\pm\sqrt{16+468}}{2}$

$y_1=\dfrac{-4-22}{2}\ ,\quad y_2=\dfrac{-4+22}{2}\ ,\quad y_3=\dfrac{4-22}{2}\ ,\quad y_4=\dfrac{4+22}{2}\\\\y_1=-13\ ,\qquad y_2=9\ ,\qquad\quad\qquad\ y_3=-9\ ,\qquad y_4=13\\\\x_{1,2}=4+y_{1,2}\qquad\qquad\qquad\qquad\qquad x_{3,4}=-4+y_{3,4}\\\\x_1=-9\ ,\qquad x_2=13\ ,\qquad\quad\qquad x_3=-13\ ,\qquad x_4=9$

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4 years ago