Use completing the square:
Ex:
Now apply to your problem.
First factor out the '4' in front of the 'x^2' term
Apply completing the square:
![y = 4(x^2 + \frac{7}{4}x +(\frac{7/4}{2})^2 -(\frac{7/4}{2})^2) \\ \\ y = 4[(x^2 + \frac{7}{4}x+\frac{49}{64}) - \frac{49}{64} ] \\ \\ y = 4(x + \frac{7}{8})^2 - \frac{49}{16}](https://tex.z-dn.net/?f=y%20%3D%204%28x%5E2%20%2B%20%5Cfrac%7B7%7D%7B4%7Dx%20%2B%28%5Cfrac%7B7%2F4%7D%7B2%7D%29%5E2%20-%28%5Cfrac%7B7%2F4%7D%7B2%7D%29%5E2%29%20%5C%5C%20%20%5C%5C%20y%20%3D%204%5B%28x%5E2%20%2B%20%5Cfrac%7B7%7D%7B4%7Dx%2B%5Cfrac%7B49%7D%7B64%7D%29%20-%20%5Cfrac%7B49%7D%7B64%7D%20%5D%20%5C%5C%20%20%5C%5C%20y%20%3D%204%28x%20%2B%20%5Cfrac%7B7%7D%7B8%7D%29%5E2%20-%20%5Cfrac%7B49%7D%7B16%7D)
Answer:
The answer to your question is:
Step-by-step explanation: The third one is a factor
Factor
5x³ − 135
Find the prime factors of 135
135 3
45 3
15 3 Then 135 = 3³ x 5
5 5
1
5x³ - 5(3)³
Factor 5
5[ x³ - (3)³]
5 [ (x - 3)(x² + 3x + 9)]
So here you just add them as if they were regular numbers. -9v^5+4v^2, hope it helped.
Let h represent the height of the trapezoid, the perpendicular distance between AB and DC. Then the area of the trapezoid is
Area = (1/2)(AB + DC)·h
We are given a relationship between AB and DC, so we can write
Area = (1/2)(AB + AB/4)·h = (5/8)AB·h
The given dimensions let us determine the area of ∆BCE to be
Area ∆BCE = (1/2)(5 cm)(12 cm) = 30 cm²
The total area of the trapezoid is also the sum of the areas ...
Area = Area ∆BCE + Area ∆ABE + Area ∆DCE
Since AE = 1/3(AD), the perpendicular distance from E to AB will be h/3. The areas of the two smaller triangles can be computed as
Area ∆ABE = (1/2)(AB)·h/3 = (1/6)AB·h
Area ∆DCE = (1/2)(DC)·(2/3)h = (1/2)(AB/4)·(2/3)h = (1/12)AB·h
Putting all of the above into the equation for the total area of the trapezoid, we have
Area = (5/8)AB·h = 30 cm² + (1/6)AB·h + (1/12)AB·h
(5/8 -1/6 -1/12)AB·h = 30 cm²
AB·h = (30 cm²)/(3/8) = 80 cm²
Then the area of the trapezoid is
Area = (5/8)AB·h = (5/8)·80 cm² = 50 cm²
Answer:
no it is not 32.725 is 0.01 less than
