A bacterium in a petri dish multiplies by a factor of 31 every day. On day O,
1 answer:
Answer:
Step-by-step explanation:
Given that,
The bacterial increase a factor of 31 every day ,
Day 0 = 1 bacterium = 31^0
Day 1 = 31 bacterial = 31¹
Day 2 = 31× 31 bacterial = 31²
Day 3 = 31 × 31 × 31 bacterial = 31³
And so on,
Then, nth day will have
Day n = 31ⁿ bacterial
So, when will the bacterial be 1,000,000,000
Then,
31ⁿ = 1,000,000,000
Take the Natural logarithmic to base 31 of both sides
Log_31 (31)ⁿ = Log_31 (1,000,000,000)
n•Log_31 (31) = Log_31 (1,000,000,000)
Then, log_a (a) = 1
n = Log_31 (1,000,000,000)
Check attachment for better view
Then,
The correct answer is A
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0.85(a + 60) ≤ 85
0.85a + 51 ≤ 85
-51 on both sides
0.85a ≤ 34
/0.85 on both sides
a ≤ 40
Answer:
a ≤ 40
Answer:
We know that for a line:
y = a*x + b
where a is the slope and b is the y-intercept.
Any line with a slope equal to -(1/a) will be perpendicular to the one above.
So here we start with the line:
3x + 4y + 5 = 0
let's rewrite this as:
4y = -3x - 5
y = -(3/4)*x - (5/4)
So a line perpendicular to this one, has a slope equal to:
- (-4/3) = (4/3)
So the perpendicular line will be something like:
y = (4/3)*x + c
We know that this line passes through the point (a, 3)
this means that, when x = a, y must be equal to 3.
Replacing these in the above line equation, we get:
3 = (4/3)*a + c
c = 3 - (4/3)*a
Then the equation for our line is:
y = (4/3)*x + 3 - (4/3)*a
We can rewrite this as:
y = (4/3)*(x -a) + 3
now we need to find the point where this line ( y = -(3/4)*x - (5/4)) and the original line intersect.
We can find this by solving:
(4/3)*(x -a) + 3 = y = -(3/4)*x - (5/4)
(4/3)*(x -a) + 3 = -(3/4)*x - (5/4)
(4/3)*x - (3/4)*x = -(4/3)*a - 3 - (5/4)
(16/12)*x - (9/12)*x = -(4/3)*a - 12/4 - 5/4
(7/12)*x = -(4/13)*a - 17/4
x = (-(4/13)*a - 17/4)*(12/7) = - (48/91)*a - 51/7
And the y-value is given by inputin this in any of the two lines, for example with the first one we get:
y = -(3/4)*(- (48/91)*a - 51/7) - (5/4)
= (36/91)*a + (153/28) - 5/4
Then the intersection point is:
( - (48/91)*a - 51/7, (36/91)*a + (153/28) - 5/4)
And we want that the distance between this point, and our original point (3, a) to be equal to 4.
Remember that the distance between two points (a, b) and (c, d) is:
distance = √( (a - c)^2 + (b - d)^2)
So here, the distance between (a, 3) and ( - (48/91)*a - 51/7, (36/91)*a + (153/28) - 5/4) is 4
4 = √( (a + (48/91)*a + 51/7)^2 + (3 - (36/91)*a + (153/28) - 5/4 )^2)
If we square both sides, we get:
4^2 = 16 = (a + (48/91)*a + 51/7)^2 + (3 - (36/91)*a - (153/28) + 5/4 )^2)
Now we need to solve this for a.
16 = (a*(1 + 48/91) + 51/7)^2 + ( -(36/91)*a + 3 - 5/4 + (153/28) )^2
16 = ( a*(139/91) + 51/7)^2 + ( -(36/91)*a - (43/28) )^2
16 = a^2*(139/91)^2 + 2*a*(139/91)*51/7 + (51/7)^2 + a^2*(36/91)^2 + 2*(36/91)*a*(43/28) + (43/28)^2
16 = a^2*( (139/91)^2 + (36/91)^2) + a*( 2*(139/91)*51/7 + 2*(36/91)*(43/28)) + (51/7)^2 + (43/28)^2
At this point we can see that this is really messy, so let's start solving these fractions.
16 = (2.49)*a^2 + a*(23.47) + 55.44
0 = (2.49)*a^2 + a*(23.47) + 55.44 - 16
0 = (2.49)*a^2 + a*(23.47) + 39.44
Now we can use the Bhaskara's formula for quadratic equations, the two solutions will be:
Then the two possible values of a are:
a = (-23.47 + 12.57)/4.98 = -2.19
a = (-23.47 - 12.57)/4.98 = -7.23