Answer:
The price for each kilogram of strawberries is $7.50
Step-by-step explanation:
<u><em>The question is</em></u>
Blues berry farm charges Percy a total of $24.75 for entrance and 2.5 kilograms of strawberries. The entrance fee is $6 and the price for each kilogram of strawberries is constant
Determine the price for each kilogram of strawberries
Let
x ----> the price for each kilogram of strawberries
we know that
The entrance fee plus the price of each kilogram of strawberries multiplied by the number of kilogram of strawberries must be equal to $24.75
so
The linear equation that represent this situation is

solve for x
subtract 6 both sides

divide by 2.5 both sides

therefore
The price for each kilogram of strawberries is $7.50
Answer:
Step-by-step explanation:
Given a triangle has points:
(-5,1),(2,1), (2,-1)
Let us label the points:
A(2,1),
B(-5,1) and
C(2,-1)
To find:
Distance between (−5, 1) and (2, −1) i.e. BC.
Horizontal leg AB and
Vertical leg, AC.
Solution:
Please refer to the attached diagram for the labeling of the points on xy coordinate plane.
We can simply use Distance formula here, to find the distance between two coordinates.
Distance formula
:

For BC:


Horizontal leg, AC:


Vertical Leg, AB:


You would be 3ft above the sea level, if you are -6ft under the sea level and if you are standing 3ft “above” you would have to clime back up 9ft to get back to the positive’s which is “positive 3” so yes, your elevation will be opposite of the “plain” since you went to a -6 to a positive 3, with different numbers.
Answer:
a) The continuous rate of growth of this bacterium population is 45%
b) Initial population of culture at t = 0 is 950 bacteria
c) number of bacterial culture contain at t = 5 is 9013 bacteria
Step-by-step explanation:
The number of bacteria in a culture is given by
n(t) = 950
a) rate of growth is
Bacteria growth model N(t) = no
Where r is the growth rate
hence r = 0.45
= 45%
b) Initial population of culture at t = 0
n(0) = 950
= 950
= 950 bacteria
c) number of bacterial culture contain at t = 5
n(5) = 950
= 950
= 9013.35
= 9013 bacteria