Answer:
x = 4/3
y = 1/3
Step-by-step explanation:
System of equations! This is set up really well to make the second equation equal x then substitute.
x - y = 1
x = 1 + y
and then our substitution:
2 (1+y) + y = 3
and solve:
2 + 2y + y = 3
3y + 2 = 3
3y = 1
y = 1/3
And now we can substitute that value into one of our equations:
x - (1/3) = 1
x = 4/3
Next we should check by substituting these values into both of our equations:
2 (4/3) + (1/3) = 3
9 / 3 does equal 3 !
(4/3) - (1/3) does equal 1 !
Therefore, x = 4/3 , and y = 1/3
Answer:
339.42m
Step-by-step explanation:
Circumference of a circle=2x22/7x54
=339.42m
Where’s the answer choices ??
It should be 81 square feet.
3 yards x 9 yards = 27 yards squares.
It asks for your answer in square feet, so you multiply by 3. This is because there are 3 feet in ever yard.
27 x 3 feet = 81 square feet.
Answer:
This problem is incomplete, we do not know the fraction of the students that have a dog and also have a cat. Suppose we write the problem as:
"In Mrs.Hu's classroom, 4/5 of the students have a dog as a pet. X of the students who have a dog as a pet also have cat as a pet. If there are 45 students in her class, how many have both a dog and a cat as pets?"
Where X must be a positive number smaller than one, now we can solve it:
we know that in the class we have 45 students, and 4/5 of those students have dogs, so the number of students that have a dog as a pet is:
N = 45*(4/5) = 36
And we know that X of those 36 students also have a cat, so the number of students that have a dog and a cat is:
M = 36*X
now, we do not have, suppose that the value of X is 1/2 ("1/2 of the students who have a dog also have a cat")
M = 36*(1/2) = 18
So you can replace the value of X in the equation and find the number of students that have a dog and a cat as pets.
