The given expression is:
(x^3 * y^4) / (3y^4)
We can notice that the term y^4 is a common term found in both the numerator and the denominator of the given expression, therefore, we can cancel this term from the both numerator and denominator (as if you divided both numerator and denominator by y^4).
Doing this, we will have the simplified form of the expression as follows:
(x^3) / (3)
Let us assume last year the price of the item = $x.
After increase in price by 15% the new price = x + 15% of x = x+ 0.15x = 1.15x.
There is a 25% discount for employee.
25% of 1.15x = 0.25 × 1.15x = 0.2875 x.
Price after discount = 1.15x - 0.2875 x = 0.8625x.
The employee pays $172.50.
Therefore,
0.8625x = 172.50.
Dividing both sides by 0.08625.
<h3>x=200.</h3><h3>Therefore, last year the price was $200.</h3>
Answer:
x+20⩾50
Step-by-step explanation:
in inequality, the word at least means that is equal or more than. so that means that Allie might equal or more than 50 hours in the model but if you're to solve it, Allie will need 30 or more hours. and since she already has 20, add the 20.
Answer: p = 30 q = 30 (if not touching edge of hexagon but if it is touching edge one of the pairs) then p or q = 120 and p and q = 30
Step 1) Sum of an interior angle of a polygon = 720 degree where n = 6 as 6 exterior sides = (n-2) * 180 = (6-2) * 180 = 4 * 180 = 720 degree Where the measure of each angle of a hexagon = 720/ 6 = 120 degree Step 2) Then show 3 angle letter names = 180 degree Step 3) Angle name letters + 2 angle (other two names letters inside within triangle) add up to 180 degree Step 3) State since triangle name (of all letters within one triangle) is Isosceles Step 4) <u>Triangle (state same triangle letters with number 2 in front) = 180 - 120 = 60 then state triangle (letter name) / 2 = 30 degrees </u>obviously the 120 and 60 differentiates if not a hexagon to a pentagon if not a hexagon = (3 *180) / 5 = 108 then due to isosceles (180 - 108)/ 2 = 72/2 = 36 degree for a p<u>entagon </u>or 30 degree each for p + q for a <u>hexagon</u> etc
