Answer:
0.08
Step-by-step explanation:
1.6 liters of the drink was distributed in 20 cups
The litres of drink in each cup can be calculated as follows
= 1.6/20
= 0.08
Hence 0.08 liters of drink would be in each cup
Answer:
<u>Point-slope form</u>: y - 12 = 10 (x - 5)
<u>Slope-Intercept form</u>: y= 10x - 38
Step-by-step explanation:
im not fully sure which equation ur looking for but i think thats right lol. hopefully this helps ! :)
Answer:
convert 9.5
n-9=27
Step-by-step explanation:
I converted it for you.
Convert each mixed number to an improper fraction
--------------------------------------------
Convert 6&1/2 to an improper fraction
6&1/2 = (6*2+1)/2
6&1/2 = (12+1)/2
6&1/2 = 13/2
--------------------------------------------
Do the same for 9&1/4
9&1/4 = (9*4+1)/4
9&1/4 = (36+1)/4
9&1/4 = 37/4
--------------------------------------------
Now multiply the improper fractions 13/2 and 37/4. Multiply straight across
(13/2)*(37/4) = (13*37)/(2*4)
(13/2)*(37/4) = 481/8
--------------------------------------------
The last step is to use long division to find the mixed number equivalent of 481/8
Or we an say
60*8 = 480
So 60*8+1 = 480+1
which means
481/8 = (480+1)/8 = 480/8+1/8 = 60&1/8
Making the final answer to be choice C
Another way to get the answer is to convert 481/8 into a decimal to get 60.125 and then convert that back to a mixed number. You should get 60&1/8
If

is odd, then

while if

is even, then the sum would be

The latter case is easier to solve:

which means

.
In the odd case, instead of considering the above equation we can consider the partial sums. If

is odd, then the sum of the even integers between 1 and

would be

Now consider the partial sum up to the second-to-last term,

Subtracting this from the previous partial sum, we have

We're given that the sums must add to

, which means


But taking the differences now yields

and there is only one

for which

; namely,

. However, the sum of the even integers between 1 and 5 is

, whereas

. So there are no solutions to this over the odd integers.