Answer:
This is a typical radioactive decay problem which uses the general form:
A = A0e^(-kt)
So, in the given equation, A0 = 192 and k = 0.015. We are to find the amount of substance left after t = 55 years. That would be represented by A. The solution is as follows:
A = 192e^(-0.015*55)
A = 84 mg
59 is a number
59 can be written as 59/1
Answer: d.h=−4
PLZ MARK BRAINLIEST!
Step-by-step explanation:
Let's solve your equation step-by-step.
−3(h+5)+2=4(h+6)−9
Step 1: Simplify both sides of the equation.
−3(h+5)+2=4(h+6)−9
(−3)(h)+(−3)(5)+2=(4)(h)+(4)(6)+−9(Distribute)
−3h+−15+2=4h+24+−9
(−3h)+(−15+2)=(4h)+(24+−9)(Combine Like Terms)
−3h+−13=4h+15
−3h−13=4h+15
Step 2: Subtract 4h from both sides.
−3h−13−4h=4h+15−4h
−7h−13=15
Step 3: Add 13 to both sides.
−7h−13+13=15+13
−7h=28
Step 4: Divide both sides by -7.
−7h
−7
=
28
−7
h=−4
Answer:
a.0.126
b.100.6356
c.5.010
Step-by-step explanation:
Answer:
Any value of x x makes the equation true. All real numbers Interval Notation: ( − ∞ , ∞ )
.