Answer:
2z+6
Step-by-step explanation:
You are close, but it is not the correct answer.
1 + 2k = 5k + 7
First, since we are trying to get the variable by itself, we need to move all things to one side. Let's subtract 1 from each side.

Second, we can now subtract (7 - 1) to get 6. Let's add that into our problem and then subtract 5 from each side, still trying to get the variable by itself.

Third, subtract (2k - 5k) to get -3k. Hint the negative symbol now. If the problem were switched the other way, it would still be a positive.

Fourth, we can now divide each side by -3. This results in a fraction and also getting the variable by itself on one side of the problem. Remember the negative sign in front of the fraction since (-3) carried it.
Fifth, this is our last step. We need to simplify the fraction. Since out fraction is 6/3, which number multiples by 3 to get the answer of 6? The answer is 2, since (3 × 2 = 6). Don't forget the negative!

Answer:
You need 864.52 sq. in. of blue tile.
We need to find the area of the large circle (including blue and yellow) and the area of the yellow circle.
We have the circumference of the blue circle, 113.097.
C = πd
113.097 = 3.14d
Divide both sides by 3.14:
113.097/3.14 = 3.14d/3.14
36.018 = d
The diameter is twice as much as the radius, so r = 36.018/2 = 18.009.
The area of a circle is given by the formula
A = πr²
A = 3.14(18.009)² = 1018.38 sq. in.
The area of the small yellow circle is given by
A = 3.14(7²) = 3.14(49) = 153.86
Subtracting these two, we have
1018.38 - 153.86 = 864.52 sq. in.
Answer:
54
Step-by-step explanation:4545
Answer:
The weight of the parcel is approximately 6 kilograms.
Step-by-step explanation:
The weight of the parcel is 12 lbs and 8 oz.
1 lb = 16 oz
So, 8 oz = lb.
That means, 12 lbs and 8 oz = (12 + 0.5) lbs = 12.5 lbs
Now, 1 lb = 454 g
So, 12.5 lbs = (12.5 × 454) g = 5675 g
As, 1 kilogram = 1000 g
That means, 1 g = 0.001 kilogram
So, 5675 g kilograms. (Rounded to the nearest whole kilogram)
Thus, the weight of the parcel is approximately 6 kilograms.