Let find the least of common multiple = LCM it’s for the denominators.
Multiple of the numerator then the denominator to get the denominators
Don’t forget to to add the numerator but leave the denominators the same
Answer:
xy = 1
k = 79
Step-by-step explanation:
Question One
The first and third frames look to me to be the same. I'll treat them that way.
y = x^2 Equate y = x^2 to the result of 2y + 6 = 2x + 6
2y + 6 = 2(x + 3) Remove the brackets
2y + 6 = 2x + 6 Subtract 6 from both sides
2y = 2x Divide by 2
y = x
Now solve these two equations.
so x^2 = x
x > 0
1 solution is x = 0 from which y = 0. This won't work. x must be greater than 0. So the other is
x(x) = x Divide both sides by x
x = 1
y = x^2 Put x = 1 into x^2
y = 1^2 Solve
y = 1
The second solution is
(1,1)
xy = 1*1
xy = 1
Answer: A
Question Two
square root(k + 2) - x = 0
Subtract x from both sides
sqrt(k + 2) = x Square both sides
k + 2 = x^2 Let x = 9
k + 2 = 9^2 Square 9
k + 2 = 81
k = 81 - 2
k = 79
Answer:
Step-by-step explanation:
You should do 76 + 48 to get 124 and since the triangle is 180 you would do 180 minus 124 and you would get 56
Answer:
y = 2
Step-by-step explanation:
The quantity of the radioactive material which would remain 2 years from now is; 4 grams.
<h3>What is the quantity remaining in 2 years?</h3>
Since, it follows from the task content that after two year, the quantity of the radioactive substance decrease by 1 gram. Consequently, it follows from proportion that the quantity remaining in 2 years from now is; 5-1 = 4 grams.
Read more on radioactive material;
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