B, D, A
No problem make sure to heart my comment Helps
Answer:
7.5 L of 10% solution and 22.5 L of 30% solution
Step-by-step explanation:
Volume of 10% solution plus volume of 30% solution = total volume of 25% volume.
x + y = 30
Acid in 10% solution plus acid in 30% solution = total acid in 25% solution.
0.10 x + 0.30 y = 30 × 0.25
0.10 x + 0.30 y = 7.5
Solve the system of equations, using either substitution or elimination. I'll use substitution:
x = 30 − y
0.10 (30 − y) + 0.30 y = 7.5
3 − 0.10 y + 0.30 y = 7.5
0.20 y = 4.5
y = 22.5
x = 30 − y
x = 7.5
Sarah needs 7.5 L of 10% solution and 22.5 L of 30% solution.
In the future, please post the full problem with all included instructions. After doing a quick internet search, I found your problem listed somewhere else. It mentions two parts (a) and (b)
Part (a) asked for the equation of the line in y = mx+b form
That would be y = -2x+9
This is because each time y goes down by 2, x goes up by 1. We have slope = rise/run = -2/1 = -2. This indicates that the height of the candle decreases by 2 inches per hour. The slope represents the rate of change.
The initial height of the candle is the y intercept b value. So we have m = -2 and b = 9 lead us from y = mx+b to y = -2x+9
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Part (b) then asks you to graph the equation. Because this is a linear equation, it produces a straight line. We only need 2 points at minimum to graph any line. Let's plot (0,9) and (1,7) on the same xy grid. These two points are the first two rows of the table. Plot those two points and draw a straight line through them. The graph is below
Answer:
Two positive zeros, no negative zeros, two complex roots.
Step-by-step explanation:
The given function is 
According to the fundamental theorem of algebra, the function will have 4 roots.
The graph of the function intersects the positive axis at two points.
Hence the function has two positive zeros and no negative zeros.
The two remaining roots are imaginary. The function has two complex zeros.
See graph in attachment
Basically, what this asks you is to maximize the are A=ab where a and b are the sides of the recatangular area (b is the long side opposite to the river, a is the short side that also is the common fence of both corrals). Your maximization is constrained by the length of the fence, so you have to maximize subject to 3a+b=450 (drawing a sketch helps - again, b is the longer side opposite to the river, a are the three smaller parts restricting the corrals)
3a+b = 450
b = 450 - 3a
so the maximization max(ab) becomes
max(a(450-3a)=max(450a-3a^2)
Since this is in one variable, we can just take the derivative and set it equal to zero:
450-6a=0
6a=450
a=75
Plugging back into b=450-3a yields
b=450-3*75
b=450-225
b=215
Hope that helps!