${2}x^{2} + 8 {y}^{2} = 16 \\ \\ 1. \: {2x}^{2} = - 8 {y}^{2} + 16 \\ 2. \: {x}^{2} = - 4y^{2} + 8 \\ 3. \: x = \sqrt{ - 4y^{2} + 8} \\ x = - \sqrt{ - 4y^{2} + 8 } \\ \\ answer \\ x = \sqrt{ - 4y^{2} + 8 } \\ x = - \sqrt{ - 4y^{2} + 8}$

4 0

3 years ago

An object is dropped from a small plane. As the object falls, its distance, d, above the ground after t seconds, is given by the

DedPeter [7]

The inequality can be used to find the interval of time taken by the object to reach the height greater than 300 feet above the ground is:

$d = -16t^2 + 1000$

Solution:

The object falls, its distance, d, above the ground after t seconds, is given by the formula:

$d = -16t^2 + 1000$

To find the time interval in which the object is at a height greater than 300 ft

Frame a inequality,

$-16t^2 + 1000 > 300$

Solve the inequality

Subtract 1000 from both sides

$-16t^2 + 1000 - 1000 > 300 - 1000\\\\-16t^2 > -700$

$16t^2 < 700\\\\Divide\ both\ sides\ by\ 16\\\\t^2 < \frac{700}{16}\\\\Take\ square\ root\ on\ both\ sides\\\\t < \sqrt{\frac{700}{16}}\\\\t < \pm 6.61$

Time cannot be negative

Therefore,

t < 6.61

And the inequality used is: $-16t^2 + 1000>300$

6 0

3 years ago