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3 years ago

12

Solve 2x2 + 5x + 5 = 0. Round solutions to the nearest hundredth.

2 answers:

kotykmax [81]3 years ago

7 0

Aight, so I just did all of that work to find There Are No Real Solutions.

Hope this helps y'all & that it saves ya from the trouble of doing work :D

borishaifa [10]3 years ago

3 0

I hope this helps you

2x^2+5x+5=0

disctirminant =5^2-4.2.5

disctirminant = -15

x1= -5+i .square root of 15/2.2

x2= -5- i .square root of 15/2.2

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3 years ago

Maria throws a softball straight up into the air with an initial velocity of 50 ft/sec. The ball leaves her hand when it is exac

Nataly_w [17]

Answer:

The equation that models the situation is $s_{f} = 4+\frac{v_{f}^{2}-2500}{64.348}$ .

Step-by-step explanation:

Let suppose that effects from air friction and Earth's rotation can be neglected, so that the softball can be modelled experiment a free fall, that is, an uniform accelerated motion due to gravity. From we know the initial velocity and position of the position and we can determine the final position of the ball as a function of the final velocity:

$v_{f}^{2} = v_{o}^{2}+2\cdot a \cdot (s_{f}-s_{o})$ (1)

Where:

$s_{o}$ , $s_{f}$ - Initial and final position of the softball, measured in feet.

$a$ - Acceleration, measured in meters per square second.

$v_{o}$ , $v_{f}$ - Initial and final velocities of the softball, measured in feet per second.

If we know that $v_{o} = 50\,\frac{ft}{s}$ , $a = -32.174\,\frac{m}{s^{2}}$ and $s_{o} = 4\,ft$ , then the equation that models the situation is:

$v_{f}^{2} = 2500-64.348\cdot (s_{f}-4)$

Then, we clear the final position of the softball:

$-\frac{v_{f}^{2}-2500}{-64.348} = s_{f}-4$

$s_{f} = 4+\frac{v_{f}^{2}-2500}{64.348}$

The equation that models the situation is $s_{f} = 4+\frac{v_{f}^{2}-2500}{64.348}$ .

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2 years ago