Answer:
all work is shown and pictured
To solve this we use the
equation,
<span> M1V1 = M2V2</span>
<span> where M1 is the
concentration of the stock solution, V1 is the volume of the stock solution, M2
is the concentration of the new solution and V2 is its volume.</span>
<span>
</span>
<span>M1 = .80 g H2SO4 / g solution (1018 g solution / L solution ) (1 mol / 98.08) = 8.30 M</span>
<span>
</span>
<span>8.30 (V1) = (5.0 M) (1.5 L)</span>
<span>V1 = 0.9036 L or 903.6 mL of the stock needed</span>
a) The domain of a function refers to the set of values of x that satisfy the function. On the graph, it is between the minimum value of x on the left and the maximum value of x on the right. Since the line continues to the left and right, it tends towards infinity on both sides. Hence,
Domain = (- infinity, infinity)
b) The range of a function refers to the set of values of y that satisfy the function. On the graph, it is between the minimum value of at the bottom and the maximum value of y at the top. The maximimum value of y is - 1 and the minimum is negative infinity. Hence,
Range = (- infinity, - 1}
c) The x intercept is the value of x when y is zero. It occurs at the point where the line cuts the x axis. Since the line doesn't cut the x axis, there is no x intercept.
d) The y intercept is the value of y when x is zero. It occurs at the point where the line cuts the y axis. It cuts the y axis at x = - 4
Thus, y intercept = - 4
e) the function value, f(- 4) is the value of y at x = - 4. From the graph,
f(- 4) = - 2
the function value, f(0) is the value of y at x = 0. From the graph,
f(0) = - 4
Answer:
Step-by-step explanation:
Suppose the dimensions of the playground are x and y.
The total amount of the fence used is given and it is 780 ft. In terms of x and y this would be 3x+2y=780 (we add 3x because we want it to be cut in the middle). Therefore, y= 780/2-3/2x. Now, the total area (A )to be fenced is
A=x*y= x*(390-3/2x)=-3/2 x^2+390x
Calculating the derivative of A and setting it equals to 0 to find the maximum
A'= -3x+390=0
This yields x=130.
Therefore y=780/2-3/2*130=195
Thus, the maximum area is 130*195=25,350ft^2
