Answer:
11 tshirts would cost £106.15
Step-by-step explanation:
67.55 divided by 7 = 9.65
9.65 x 11 = 106.15
Therefore the total would be £106.15.
Answer:
n=-3/4
Step-by-step explanation:
First, let us restate the given conditions
c is 5 more than variable a ( c = a + 5)
c is also three less than variable a (c = a - 3)
Now, lets look at the answer choices and or given
c = a − 5
c = a + 3
Here, c is 5 less than "a"...so automatically disqualified
a = c + 5
a = 3c − 3
Here, we have to get "C" by itself in both top and bottom equation.
So,
Simplified version :
c = a - 5
Here, c is 5 less than "a"...so automatically disqualified
a = c − 5
a = 3c + 3
Here also, we have to get "C" by itself in both top and bottom equation.
So,
simplified version:
c = a + 5
Here, c is 5 more than "a"...so we continue
c = (a - 3) / 3
Here, c is 3 less than "a" <u>divided by 3</u><u /> . So, this is not correct
c = a + 5
c = a − 3
Here, c is 5 more than "A"
Also, c is 3 less than "a"
Which satisfies the given.
So, our answer is going to be the last one:
c = a + 5
c = a - 3
Answer:
-0.9090... can be written as
.
Explanation:
Any <em>repeating </em>decimal can be written as a fraction by dividing the section of the pattern to be repeated <em>by </em>9's.
We can start by listing out
0.909090... = 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...
Now. we let this series be equal to x, that is
= 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...
Now, we'll multiply both sides by 100
.
= 90 + 0 + 9/10 + 0/100 + 9/1000 + 0/10000 + ...
Then, subtract the 1st equation from the second like so:
= 90 + 0 + 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...
= - 9/10 - 0/100 - 9/1000 - 0/10000 - 9/100000 - 0/1000000 - ...
And we end up with this:

Finally, we divide both sides by 99 in order to isolate x and get the fraction we're looking for.

Which can be reduced and simplified to

Hope this helps!
Answer:
C & D because vertical angles have the same measures and thoses angles intersects each other
Step-by-step explanation: