One way is to factor and group and get every 3
729=3 times 3 times 3 times 3 times 3 times 3
so we group the ones that happen 3 times
729=(3*3*3) times (3*3*3)
we know that we can take the cube root of each group and multiply the result
729=
![( \sqrt[3]{3*3*3})( \sqrt[3]{3*3*3})](https://tex.z-dn.net/?f=%28%20%5Csqrt%5B3%5D%7B3%2A3%2A3%7D%29%28%20%5Csqrt%5B3%5D%7B3%2A3%2A3%7D%29)
=(3)(3)=9
the answer is 9
A
So, first we look at the equations
y= (x+3)^2+4 changes to
y= (x+1)^2+ 6
So, the first one is saying that we start at (-3,4)
(since when in the parentheses, it's opposite) and the second one is saying start at (-1,6), so it moved on the x-axis 2 units to the right and on the y-axis, it moved 2 units up
The formula for circumference of a circle
=
2
π
r
The radius
=
3
c
m
(assuming units to be in cm)
2
π
r
=
2
×
22
7
×
3
=
132
7
=
18.86
c
m
That will depend, you have to ask yourself first how many kilobytes one picture is. Let's just say that the size of one picture is 100 kb (which is the average size of a picture) Then first you multiply 100 kb with the number of pictures which is 43. Now you have a total used up memory of 4300 kb. After that, you minus the used up memory which is 4300 kb, to the total available space which is 32,834.5 and you will get an available space of 28534.5 kb. After that, you divide the remaining available space with the size of each picture. So this will be 28534.5 divided by 100. You will get 285. You can still take 285 pictures.
Answer:
The answer is (3x^2-2x+5)
