Answer:
<em>a</em><em>)</em><em>=</em><em> </em><em>5x</em><em>+</em><em>10</em>
<em>b</em><em>)</em><em>=</em><em> </em><em>7y-6</em>
<em>c</em><em>)</em><em>=</em><em> </em><em>6x-6</em>
<em>d</em><em>)</em><em>=</em><em> </em><em>2x</em><em>^</em><em>2</em><em>+</em><em>8x-5</em>
<em>e</em><em>)</em><em>=</em><em> </em><em>6x</em><em>+</em><em>9</em>
<em>f</em><em>)</em><em>=</em><em> </em><em>4xy-4y</em><em>^</em><em>2</em><em>+</em><em>8y</em>
<em>g</em><em>)</em><em>=</em><em> </em><em>-12y</em><em>^</em><em>2</em><em>+</em><em>10y</em>
<em>h</em><em>)</em><em>=</em><em> </em><em>6x</em><em>+</em><em>154</em>
<em>i</em><em>)</em><em>=</em><em> </em><em>6x</em><em>+</em><em>21</em>
<em>h</em><em>)</em><em>=</em><em> </em><em>2x</em><em>^</em><em>2</em><em>+</em><em>16</em>
<span>The maxima of a differential equation can be obtained by
getting the 1st derivate dx/dy and equating it to 0.</span>
<span>Given the equation h = - 2 t^2 + 12 t , taking the 1st derivative
result in:</span>
dh = - 4 t dt + 12 dt
<span>dh / dt = 0 = - 4 t + 12 calculating
for t:</span>
t = -12 / - 4
t = 3
s
Therefore the maximum height obtained is calculated by
plugging in the value of t in the given equation.
h = -2 (3)^2 + 12 (3)
h =
18 m
This problem can also be solved graphically by plotting t
(x-axis) against h (y-axis). Then assigning values to t and calculate for h and
plot it in the graph to see the point in which the peak is obtained. Therefore
the answer to this is:
<span>The ball reaches a maximum height of 18
meters. The maximum of h(t) can be found both graphically or algebraically, and
lies at (3,18). The x-coordinate, 3, is the time in seconds it takes the ball
to reach maximum height, and the y-coordinate, 18, is the max height in meters.</span>
Answer:
Option C
Step-by-step explanation:
Got it on the unit test in edge
600 * .34 = 204 are in 8th grade
204 * 1/4 = 51 51 8th graders walk to school
D