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3 years ago

the area of a rectangular pool is given by the trinomial to 2y^2+y - 3 what are the possible dimensions of a pool. use factoring

1 answer:

4 0

2y² + y - 3

(2y+3)(y-1)

2y(y-1) + 3(y-1)
2y² - 2y + 3y - 3
2y² + y - 3

2y + 3 = 0
2y = -3
y = -3/2
y = -1 1/2

y - 1 = 0
y = 1

2y² + y - 3 = 0
2(-3/2)² + (-3/2) - 3 = 0
2(9/4) - 3/2 - 3 = 0
18/4 - 3/2 - 3 = 0
18/4 - (3/2 * 2/2) - 3(4/4) = 0
18/4 - 6/4 - 12/4 = 0
(18 - 6 - 12)/4 = 0
(18 -18)/4 = 0
0/4 = 0
0 = 0

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2(x/8 + 3) = 7 + 1/4x what is x

malfutka [58]

Answer:

x1 = 2 - $\sqrt{5}$

x2= 2 + $\sqrt{5}$

Step-by-step explanation:

x/4 + 6 = 7 + 1/4x

(x^2 - 1)/4x = 1

x^2 - 1 = 4x

x^2 - 4x -1 = 0

x= (4+ $\sqrt{20}$ )/2 = 2 - $\sqrt{5}$

x= 2 + $\sqrt{5}$

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2 years ago

Help please!! :)))) Thanks!

baherus [9]

The answer is C. 16!!

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antoniya [11.8K]

The similar circles P and Q can be made equal by dilation and translation

The horizontal distance between the center of circles P and Q is 11.70 units
The scale factor of dilation from circle P to Q is 2.5

<h3>The horizontal distance between their centers?</h3>

From the figure, we have the centers to be:

P = (-5,4)

Q = (6,8)

The distance is then calculated using:

d = √(x2 - x1)^2 + (y2 - y1)^2

So, we have:

d = √(6 + 5)^2 + (8 - 4)^2

Evaluate the sum

d = √137

Evaluate the root

d = 11.70

Hence, the horizontal distance between the center of circles P and Q is 11.70 units

<h3>The scale factor of dilation from circle P to Q</h3>

We have their radius to be:

P = 2

Q = 5

Divide the radius of Q by P to determine the scale factor (k)

k = Q/P

k = 5/2

k = 2.5

Hence, the scale factor of dilation from circle P to Q is 2.5