Answer:
7.4r+0.3s+12q
Step-by-step explanation:
I am in 8th grade and we do these problems for honors algebra and if you add (8.5q+1.6r)+(3.5q-4.1s)+(4.4s+5.8r) you will get the answer above
Answer:
The sum of all the sweets is 54 sweets
Step-by-step explanation:
Here, we want to calculate the number of sweets the three have altogether.
Let the number of sweet Faith has be x
Mathematically;
Faith has 4 sweets fewer than the average number of sweets
The average is the sum of all the sweets divided by 3
The average will be (10 + 30 + x)/3
Thus;
(10 + 30 + x)/3 - 4 = x
(40 + x)/3 = x + 4
40 + x = 3(x + 4)
40 + x = 3x + 12
40-12 = 3x -x
28 = 2x
x = 28/2
x = 14 sweets
So the sum of all the sweets would be 10 + 30 + 14 = 54 sweets
Answer:
x must equal 5 and y must equal 2
therefore,
5a + 2b = 23 and
5a - 2b = 7.
I'm gonna take 5a + 2b = 23 and add 7 to each side:
5a + 2b + 7 = 30
and substitute 5a -2b in for 7 on the left side:
5a + 2b + (5a-2b) = 30
10a = 30
a = 3
Now we know a. Now take
5a -2b = 7
and substitute in 3 for a
5*(3) - 2b = 7
15 -2b = 7
8 = 2b
b = 4
a = 3 and b = 4
Step-by-step explanation:
1 and one - twelfth
because 12gies 3 times in 37
