Answer:
B; (2,-4)
Step-by-step explanation:
The general equation of a straight line is;
y = mx + b
where m is the slope and b is the y-intercept
So from the question, when we look at the given equation, its slope is -1/2
mathematically, when two lines are perpendicular, the product of their slopes is -1
thus;
m1 * m2 = -1
m2 * -1/2 = -1
-m2 = -2
m2 = 2
The general equation form is;
y-y1 = m(x-x1)
where (x1,y1) = (4,0)
y-0 = 2(x-4)
y = 2x - 8
So, now we look at the point that will work for this equation
For the line that will work, if we substitute its x-value, we get the y-value
Let us take a look at option B
y = 2(2) -8 = 4-8 = -4
we can see that (2,-4) works
The given equations are

(1)

(2)
When t=0, obtain

Obtain derivatives of (1) and find x'(0).
x' (t+1) + x - 4√x - 4t*[(1/2)*1/√x = 0
x' (t+1) + x - 4√x -27/√x = 0
When t=0, obtain
x'(0) + x(0) - 4√x(0) = 0
x'(0) + 9 - 4*3 = 0
x'(0) = 3
Here, x' means

.
Obtain the derivative of (2) and find y'(0).
2y' + 4*(3/2)*(√y)*(y') = 3t² + 1
When t=0, obtain
2y'(0) +6√y(0) * y'(0) = 1
2y'(0) = 1
y'(0) = 1/2.
Here, y' means

.
Because

, obtain

Answer:
The slope of the curve at t=0 is 1/6.
Replace every x you see in the function with 1 and simplify.
Let x be 1.
f(1) = 4(1)^2 -(1) + 3
f(1) = 4(1) - 1 + 3
f(1) = 4 - 1 + 3
f(1) = 3 + 3
f(1) = 6
Done!
We use P = i•e^rt for exponential population growth, where P = end population, i = initial population, r = rate, and t = time
P = 2•i = 2•15 = 30, so 30 = 15 [e^(r•1)],
or 30/15 = 2 = e^(r)
ln 2 = ln (e^r)
.693 = r•(ln e), ln e = 1, so r = .693
Now that we have our doubling rate of .693, we can use that r and our t as the 12th hour is t=11, because there are 11 more hours at the end of that first hour
So our initial population is again 15, and P = i•e^rt
P = 15•e^(.693×11) = 15•e^(7.624)
P = 15•2046.94 = 30,704