First we need to find k ( rate of growth)
The formula is
A=p e^kt
A future bacteria 4800
P current bacteria 4000
E constant
K rate of growth?
T time 5 hours
Plug in the formula
4800=4000 e^5k
Solve for k
4800/4000=e^5k
Take the log for both sides
Log (4800/4000)=5k×log (e)
5k=log (4800/4000)÷log (e)
K=(log(4,800÷4,000)÷log(e))÷5
k=0.03646
Now use the formula again to find how bacteria will be present after 15 Hours
A=p e^kt
A ?
P 4000
K 0.03646
E constant
T 15 hours
Plug in the formula
A=4,000×e^(0.03646×15)
A=6,911.55 round your answer to get 6912 bacteria will be present after 15 Hours
Hope it helps!
I’d say 10 or 11 but I’m not 100%
We can write a proportion to resemble the problem;
AE/ED = AB/BC
AE = 9
ED = 6
AB = x
BC = 10
Substitute with the given values.
9/6 = x/10
9/6 * 10 = x/10 * 10
90/6 = x
15 = x
Therefore, the answer is 15.
Best of Luck!
f(x)= -2x+1 {-2,0,2,4,6}
If x=-2, then -2(-2)+1 = 5
If x=0, then -2(0)+1 = 1
If x=2, then -2(2)+1 = -3
If x=4, then -2(4)+1 = -7
If x=6, then -2(6)+1 = -11
Answer:
B {5,1,-3,-7,-11}
Bob has a garden
the area is 75 square feet
the legnth is 3 times the width
solve
75=lw
l=3w
subsitute
75=3w times w=3w^2
75=3w^2
sdivide by 3
25=w^2
square root
5=width
subsitute
3w=l
3 times 5=l
l=15
width=5
legnth=15
