A plot of land is a pentagon with vertices Q(-4, 4), R(2, 4), S(4, 1), T(2, -4), and U(-4, -4).

1 answer:

8 0

The measure of QR is 6 units, which is 6 meters, since every unit on the coordinate grid is 1 meter.
The measure of UT is also 6 meters.
The measure of QU is 8 meters.
To find the measure of RS, we use the distance formula:
$\sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}$
We substitute the coordinates of R and S into the formula and have:
$\sqrt{(2-4)^2+(4-1)^2} \\ =\sqrt{(-2)^2 + (3)^2} \\ =\sqrt{4+9} \\=\sqrt{13} \\=3.61$
To find the measure of ST we use the distance formula again, substituting the coordinates for S and T into the formula:
$\sqrt{(4-2)^2+(1--4)^2} \\=\sqrt{(2)^2 + (1+4)^2} \\=\sqrt{4+5^2} \\=\sqrt{4+25} \\=\sqrt{29} = 5.39$
Now we find our perimeter by adding the sides together: 5.39 + 3.61 + 6 + 6 + 8 = 29 meters
Since fencing costs $24.75 per meter, we multiply this by our perimeter:
29(24.75) = $717.75

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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

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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

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

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