0.5 as a decimal
hope this partially helped:)
sorry couldn't answer completely
let me know if it helped a bit:)
Question is unclear to me ,, can you rewrite it in a nicer way so i can answer?
Answer:
28. 120 degrees
29. 30 degrees
30. 56 degrees & 124 degrees
31. 72 degrees, 108 degrees, and 18 degrees
Step-by-step explanation:
We assign variable x for the answer we are looking for (28-29).
28.
Supplement means x + y = 180 degrees. We also know x = 2y. Substitution gives us 3y = 180 degrees, so y = 60 degrees and x = 120 degrees.
29.
Complement means x + y = 90 degrees. We are given 2x = y. Substitution brings us 3x = 90 degrees, x = 30 degrees.
30.
Supplement means x + y = 180 degrees. We are told that y = 2x + 12, so we substitute. This gives 3x + 12 = 180 degrees, x = 56 degrees. Substituting that back into the equation for y, we get 124 degrees.
31.
Supplement means x + y = 180 degrees. Complement means x + z = 90 degrees. Using our given info, we know y = 6z. We can substitute that in to get x + 6z = 180. Subtracting our second and third equations, we get 5z = 90, z = 18 degrees. Therefore, x = 72 degrees, y = 108 degrees.
True they are equivalent fractions both are divided by -3 to get 14/15
You can just 1) multiply the binomial by itself, or you can use 2) the square of a binomial pattern. I'll show it to you both ways.
1) Multiply the binomial by itself.
(3x - 2)^2 = (3x - 2)(3x - 2) =
Multiply every term of the first binomial by every term of the second binomial, then collect like terms. (This is often called using FOIL.)
= 9x^2 - 6x - 6x + 4
= 9x^2 - 12x + 4
2) Use the square of a binomial pattern
The square of a binomial is
(a - b)^2 = a^2 - 2ab - b^2
a^2 is the square of the first term.
b^2 is the square of the second term.
-2ab is the product of the two terms and 2.
You have
(3x - 2)^2,
where the first term is 3x, and the second term is -2
square the first term: 9x^2
square the last term: 4
the product of the terms and 2 is: -12x
Put it all together, and you get
9x^2 - 12x + 4
just like we got above with the other method.
