Direct variation is written in the form y = k * x or x = y * k. And, inverse variation is written in the form y = k/x or x = y/k.
For the first problem, if we isolated the y, we would get:
4y = 3x - 1
y = (3/4)x - 1/4
This is written as y = mx + b, which isn’t the form of direct or inverse variation, so the answer is NEITHER.
For the second equation, we could divide by 2 on both sides to isolate x and see if it is in the correct form:
x = 4/y
This is inverse form, so the second question is INVERSE with a constant of variation as 4.
Answer:
B. 3.
Step-by-step explanation:
OK lets try again.
The slope of the secant = slope of the tangent at a certain point ( The Mean Value Theorem).
Slope of the secant = f(5) - f(2) / (5 - 2)
= [(25-3) / (5-1) - (4-3) / (2-1)] / 3
= (22/4 - 1) / 3
= 9/2 / 3
= 9/6
= 3/2.
The derivative at c = the slope of the tangent at c.
Finding the derivative:
f'(x) = [2x(x - 1) - (x^2 - 3) ]/ (x - 1)^2 (where x = c).
= (x^2 - 2x + 3)/ (x - 1)^2 = the slope.
So equating the slopes:
(x^2 - 2x + 3) / (x - 1)^2 = 3/2
2x^2 - 4x + 6 = 3x^2 - 6x + 3
x^2 - 2x - 3 = 0
(x - 3)(x + 1) = 90
x = 3 , -1
x can't be -1 because we are working between the values 2 and 5 so
x = c = 3.
For this case we have that the weight of the bouquet of flowers will be:
4.3 * 10 ^ 4 - 2 * (7.5 * 10 ^ 2)
Rewriting we have:
4.1 * 10 ^ 4 miligrams
Thus, we have that the values of p and q are given by:
p = 4.1
q = 4
Answer:
4.1 * 10 ^ 4 miligrams
p = 4.1
q = 4
Answer:
Step-by-step explanation:
Algebraically, linear functions are polynomials with highest exponent equal to 1 or of the form y = c where c is constant. Nonlinear functions are all other functions. An example of a nonlinear function is y = x^2.
B. 11.33 is closer to the area of a circle with a radius of 2.1
