Factor by GCF
1 answer:
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Answer:
Step-by-step explanation:
we are given half-life of PO-210 and the initial mass
we want to figure out the remaining mass <u>after</u><u> </u><u>4</u><u>2</u><u>0</u><u> </u><u>days</u><u> </u>
in order to solve so we can consider the half-life formula given by
where:
- f(t) is the remaining quantity of a substance after time t has elapsed.
- a is the initial quantity of this substance.
- T is the half-life
since it halves every 140 days our T is 140 and t is 420. as the initial mass of the sample is 5 our a is 5
thus substitute:
reduce fraction:
By using calculator we acquire:
hence, the remaining sample after 420 days is 0.625 kg
Let by the smallest of the 6 integers. Then
But 189 is odd, so it cannot be divisible by 6, which means is not an integer...
Answer:
(-2.2, -1.6), (3, 1)
Step-by-step explanation:
You don't have to go far to find the equations. They are right there in your problem statement. Perhaps you want to find the solutions to the equations.
Use the first equation to write an expression for x, then substitute that into the second equation:
x = 2y +1
y^2 -3(2y+1)(y) +8 = 0
-5y^2 -3y +8 = 0
-(5y +8)(y -1) = 0
y = -8/5 or y = 1
The corresponding values of x are ...
x = 2(-8/5)+1 = -11/5
x = 2(1) +1 = 3
The solutions are (x, y) = (-2.2, -1.6) and (3, 1).
