F(x) + k - Moves the graph k units up.
k f(x) stretches the graph parallel to y-axis by a facor k
f (kx) stretches the graph by a factor 1/k parallel to x-axis
f(x + k) moves the graph 3 units to the left.
For k negative the first one moves it k units down
for second transform negative does same transfoormation but also reflects the graph in the x axis
For the third transform negative k :- same as above but also reflects in y axis
4th transform - negative k moves graph k units to the right
Answer:
225.78 grams
Step-by-step explanation:
To solve this question, we would be using the formula
P(t) = Po × 2^t/n
Where P(t) = Remaining amount after r hours
Po = Initial amount
t = Time
In the question,
Where P(t) = Remaining amount after r hours = unknown
Po = Initial amount = 537
t = Time = 10 days
P(t) = 537 × 2^(10/)
P(t) = 225.78 grams
Therefore, the amount of iodine-131 left after 10 days = 225.78 grams
Answer:
11:1
Step-by-step explanation:
(3x+8)(3x+8) is the answer :))
Answer:
already in standard form
Step-by-step explanation:
the equation of a quadratic in standard form is
y = ax² + bx + c ( a ≠ 0 )
y = x² + 4x + 3 ← is in standard form
with a = 1 , b = 4 , c = 3