Answer: D
Step-by-step explanation:
(5z + 15) - (11z + 2)
= (5z-11z) +(15-2)
= -6z+13
Answer:
Step-by-step explanation:
Hello!
A line that is perpendicular to another has an opposite-reciprocal slope.
Meaning:
- Flip the sign (+/-)
- Flip the fraction
The slope perpendicular to 3/5 would be -5/3.
We can solve for the y-intercept by plugging in the x and y values given from the coordinate into the equation with out new slope.
<h3 /><h3>Solve for B</h3>
The y-intercept of the new line is 20.
The equation is
Recall the definition of the cross product:
i x i = j x j = k x k = 0
i x j = k
j x k = i
k x i = j
The cross product is antisymmetric, or anticommutative, meaning that for any vectors u and v, we have u x v = - (v x u).
It's also distributive, so for any vectors u, v, and w, we have (u + v) x w = (u x w) + (v x w).
Taking all of these properties together, we get
b x a = (6i - j + 2k) x (2i + 2j - 5k)
b x a = 12 (i x i) - 2 (j x i) + 4 (k x i)
............. + 12 (i x j) - 2 (j x j) + 4 (k x j)
............. - 30 (i x k) + 5 (j x k) - 10 (k x k)
b x a = 1 (j x k) + 34 (k x i) + 14 (i x j)
b x a = i + 34j + 14k
X^2/(x- 9 = 81/(x - 9)
This is the equation for which you want the solution.
Multiplying both sides of the equation by (x - 9) we get
x^2(x - 9)/(x - 9) = 81(x - 9)/(x - 9)
So the (x - 9) goes out from both the denominator and the numerator and then the simplified equation becomes
x^2 = 81
x ^2 = (9)^2
x = 9
So the value of the unknown variable x comes out to be 9.
Answer:
The average rate of change of the cost of a pack is 22 cents per year.
Another name for the average <em>rate of change</em> is <em>slope</em>.
Step-by-step explanation:
The average rate of change of the cost of a pack (
), in monetary units per year, is equal to the change in the average cost of a pack (
), in monetary units, divided by the change in time (
), in years. Then, the average rate of change is:
The average rate of change of the cost of a pack is 22 cents per year.
Another name for the average <em>rate of change</em> is <em>slope</em>.
