Answer:
8ft x 6ft
Step-by-step explanation:
For this problem, we want to create measurements of length and width that double our current area. First we need to find the current area.
C_Length = 6 ft
C_Width = 4 ft
C_Area = C_Length * C_Width
C_Area = 6 ft * 4 ft = 24 ft^2
Now we want our new area to be double the old area.
N_Area = 2C_Area
N_Area = 2 * 24 ft^2
N_Area = 48 ft^2
Now we simply need to find the increase in width and length, distributed equally, that will create the new area.
N_Area = (C_Length + x) * (C_Width + x)
Where x is the amount that each variable is increased by to create the new area.
Now we simply solve for x.
N_Area = (C_Length + x) * (C_Width + x)
48 = (6 + x) * (4 + x)
48 = 24 + 10x + x^2
0 = -24 + 10x + x^2
Now we can either apply the quadratic formula or attempt to factor using the constant break down method.
(x + 12) (x - 2) = 0
x = -12; x = 2
Note that we will choose x = 2 since the other option would require us to have negative material, which is impossible.
Now we can check to validate our answer by plugging in the value of x.
48 ?= (6 + 2) * (4 + 2)
48 ?= (8) * (6)
48 == 48
Thus, we have found the dimensions of the new aquarium, 8ft x 6ft.
Cheers.
<h3>Steps</h3>
3 60/79
•Finding the top part of the fraction
#Firstt times 79 with 3
79× 3 = 237
#Then, u plus it with 60
237 + 60 = 297
•Bottom part
#The bottom part of the fraction stay the same so,
297/79
<h3>
Answer : 297/79</h3>
<em>#</em><em>H</em><em>o</em><em>p</em><em>e</em><em> </em><em>my</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>corre</em><em>ct</em><em> </em><em>cuz</em><em> </em><em>i </em><em>dont use</em><em> </em><em>calc</em><em> </em><em>to</em><em> </em><em>do</em><em> </em><em>this</em>
Well if the equation is -16t^2 + vt + h0,
and we know v = 30, because that's the velocity,
and we also know h0 = 6 because that's the height,
then the equation will read: -16t^2 + 30t + 6
The length of the missing side is 80
because if you add 82+18 = 100 and you need it to add up to 180
When you divide a whole number by something, you're asking:
"How many times can I put this thing
into the whole number ?"
If the thing is ' 1 ', then each time you put it into the whole number,
it takes up the space of ' 1 ', and you can do that exactly the same
whole number of times.
If the thing is more than ' 1 ', then each time you stuff it into the
whole number, it takes up the space of more than ' 1 ', so you
can't do that as many times as the whole number.
If the thing is less than ' 1 ', then each time you stuff it into the
whole number, it only takes up the space of less than ' 1 ', so
there'll be enough space in there to let you do that more than
the whole number of times.
______________________________________
Another way to look at it:
When you divide a whole number by something, you're asking:
"How many times can I take this thing away from
the whole number?"
If the thing is ' 1 ', then each time you take it away from the
whole number, you take away exactly ' 1 ', and you can do that
exactly the same number of times as the whole number.
If the thing is more than ' 1 ', then each time you take it away from
the whole number, you take away more than ' 1 ', so you can't do
that as many times as the whole number.
If the thing is less than ' 1 ', then each time you you take it away
from the whole number, you only take away a piece of ' 1 ', so
you can do that more times than the whole number.
