The answer to this question is c
Because the discriminant is less than zero, there are no real solutions in the equation.
Answer:
the equation should be corrected to fit the data of the problem. With the corrected equation a mass of 0.5 grams remains after 150 years
Step-by-step explanation:
for the mass y( in grams)
y=23* (1/2)^(t/45), t ≥ 0.
the initial mass is at t=0 , then
y= 23 grams → should be 16 grams
half-life from the equation = 45 years → should be 30 years
the correct equation should be
y=16*(1/2)^(t/30), t ≥ 0
then after 150 years → t= 150
y=16*(1/2)^(150/30)= 16*(1/2)^5 = 16/32 = 0.5 grams
then a mass of 0.5 grams remains after 150 years
Solve the following system using substitution:
{y + 2.3 = 0.45 x
{-2 y = -3.6
In the second equation, look to solve for y:
{y + 2.3 = 0.45 x
{-2 y = -3.6
-3.6 = -18/5:
-2 y = -18/5
Divide both sides by -2:
{y + 2.3 = 0.45 x
{y = 9/5
Substitute y = 9/5 into the first equation:
{4.1 = 0.45 x
{y = 9/5
In the first equation, look to solve for x:
{4.1 = 0.45 x
{y = 9/5
4.1 = 41/10 and 0.45 x = (9 x)/20:
41/10 = (9 x)/20
41/10 = (9 x)/20 is equivalent to (9 x)/20 = 41/10:
{(9 x)/20 = 41/10
{y = 9/5
Multiply both sides by 20/9:
Answer: {x = 82/9
{y = 9/5
