A. (2,6)(5,0)
slope = (0 - 6) / (5 - 2) = -6/3 = -2
y = mx + b
slope(m) = -2
(2,6)..x = 2 and y = 6
sub and find b, the y int
6 = -2(2) + b
6 = -4 + b
6 + 4 = b
10 = b
so line A , equation is : y = -2x + 10
============================
B. (0,2)(6,6)
slope = (6 - 2) / (6 - 0) = 4/6 which reduces to 2/3
y = mx + b
slope(m) = 2/3
(0,2)...x = 0 and y = 2
sub and find b, the y int
2 = 2/3(0) + b
2 = b
LIne B, equation is : y = 2/3x + 2
=========================
so we have :
y = 2/3x + 2
y = -2x + 10
2/3x + 2 = -2x + 10 ..multiply by 3
2x + 6 = -6x + 30
2x + 6x = 30 - 6
8x = 24
x = 24/8
x = 3
y = -2x + 10
y = -2(3) + 10
y = -6 + 10
y = 4
solution is (3,4).....3/4 is the solution to both lines <===
Step-by-step explanation:
-3x + 4 > 25
-3x > 21 (subtract 4 on both sides)
x < -7. (divide -3 on both sides, flip the sign)
Answer: -4 is the answer
Step-by-step explanation: google caculator
Answer:
Step-by-step explanation:
The top one is the third one the bottom one is the first one
<span>Assuming the reaction is of 1st order, we can
start using the formula for rate of 1st order reaction:</span>
dN / dt = k * N
Rearranging,
dN / N = k dt
Where N = amount of sample, k = rate constant, t = time
Integrating the equation from N = Ni to Nf and t = ti to
tf will result in:
ln (Nf / Ni) = k (tf – ti)
Since k is constant, we can equate to situations.
Situation 1 is triple in size every days, situation 2 is after 20 days.
ln (Nf / Ni) / (tf – ti) = k
ln (3Ni / Ni) / 4 = ln (Nf / 40) / 20
Calculating for Nf,
<span>Nf =
9,720 bacteria </span>
