Answer:
There will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days
Step-by-step explanation:
The given parameters are;
The half life of the radioactive substance = 45 days
The mass of the substance initially present = 6.2 grams
The expression for evaluating the half life is given as follows;
Where;
N(t) = The amount of the substance left after a given time period = 1 gram
N₀ = The initial amount of the radioactive substance = 6.2 grams
= The half life of the radioactive substance = 45 days
Substituting the values gives;
The time that it takes for the mass of the radioactive substance to remain 1 g ≈ 118.45 days
Therefore, there will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days.
Answer:
x+-(1)/(2),-3
Step-by-step explanation:
Thank you:)
Answer:
Attachment 1 : 5x + 6y = 5, Attachment 2 : 4cotθcscθ
Step-by-step explanation:
Remember that we have three key points in solving these types of problems,
• x = r cos(θ)
• y = r sin(θ)
• x² + y² = r²
a ) For this first problem we need not apply the third equation.
( Multiply either side by 5 cos(θ) + 6 sin(θ) )
r 
( 5 cos(θ) + 6 sin(θ) ) = 5,
( Distribute r )
5r cos(θ) + 6r sin(θ) = 5
( Substitute )
5x + 6y = 5 - the correct solution is option c
b ) We know that y² = 4x ⇒
r²sin²(θ) = 4r cos(θ),
r = 4cos(θ) / sin²(θ) = 4 cot(θ) csc(θ) = 4cotθcscθ - again the correct solution is option c
Answer:
15 1/4 pages im thinking please correct me if im wrong
Step-by-step explanation:
Answer:
There are no enough information to determine the length of the fence, assuming we were given the perimeter of the fence, and say, the dimension of the fence, then we can easily find the length.
Perimeter of the fence, P = 2(L + B).. If the fence is a rectangular.
L = (P/2) - B
If the fence is square, P = 4L
L = P/4
