Equation A simplifies to 0 = 0. It is always true.
Equation B simplifies to 1 = -1 for a ≠ 0. It is never true.
Equation C simplifies to 2a = 0. It is true only for a = 0.
Equation D simplifies to 2a = 0. It is only true for a = 0.
The equation that is true for all values of "a" is ...
A. Equation A
We have the following expression:
y = logbx
We clear x of the expression.
We have then:
b ^ y = b ^ (logbx)
Rewriting:
x = b ^ y
Substituting we have:
x = b ^ 0
x = 1
Answer:
If (x, 0) lies on the graph of y = logbx, then:
x = 1
Orio has 9 more than Ted. Using the equation y=mx+b to find Orios number of comic books, we can plug 2 months to substitute for the variable x.
2(2)+15 becomes 4+15, which is 19.
We know Ted has 10 comics after 2 months from the graph.
19-10= 9. :)
Answer:c
Step-by-step explanation:I just put it
Answer:
The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h
Step-by-step explanation:
At noon the location of Lan = 300 km north of Makenna
Lan's direction = South
Lan's speed = 60 km/h
Makenna's direction and speed = West at 75 km/h
The distance Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km
The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km
The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km
Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.
By Pythagoras' theorem, we have;
s² = x² + y²
The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61
ds²/dt = dx²/dt + dy²/dt
2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt
2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900
ds/dt = 900/(2×30·√61) ≈ 1.92
The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h