Answer:
x = 1/2 or 2
Step-by-step explanation:
You can factor this as ...
2x^2 -4x -x +2 = 0 . . . rewrite the middle term to enable factoring
2x(x -2) -1(x -2) = 0 . . . . factor by grouping
(2x -1)(x -2) = 0 . . . . factor out (x -2)
x = 1/2, 2 . . . . . . values of x that make the factors zero
Answer:
1) ![2y^{2}+4y-9](https://tex.z-dn.net/?f=2y%5E%7B2%7D%2B4y-9)
2) ![18x^{5}y^{4}z](https://tex.z-dn.net/?f=18x%5E%7B5%7Dy%5E%7B4%7Dz)
3) ![6x^{5}-12x^{4}+9x^{3}](https://tex.z-dn.net/?f=6x%5E%7B5%7D-12x%5E%7B4%7D%2B9x%5E%7B3%7D)
4) 40
5) ![x^{3}-7x^{2}+3x+36](https://tex.z-dn.net/?f=x%5E%7B3%7D-7x%5E%7B2%7D%2B3x%2B36)
Step-by-step explanation:
1) Distribute the negative sign that is outside the parentheses and then you must add like terms, as following:
![(y^{2}-3y-5)-(-y^{2}-7y+4)=y^{2}-3y-5+y^{2}+7y-4=2y^{2}+4y-9](https://tex.z-dn.net/?f=%28y%5E%7B2%7D-3y-5%29-%28-y%5E%7B2%7D-7y%2B4%29%3Dy%5E%7B2%7D-3y-5%2By%5E%7B2%7D%2B7y-4%3D2y%5E%7B2%7D%2B4y-9)
2) According to the Product property of exponents, when you multiply powers with the same base, you must add the exponents. Then:
![(6xy^{3}z)(3x^{2}yx^{2})=18x^{5}y^{4}z](https://tex.z-dn.net/?f=%286xy%5E%7B3%7Dz%29%283x%5E%7B2%7Dyx%5E%7B2%7D%29%3D18x%5E%7B5%7Dy%5E%7B4%7Dz)
3) Apply the Distributive property and the Product property of exponents. Then, you obtain:
![-3x^{3}(-2x^{2}+4x-3)=6x^{5}-12x^{4}+9x^{3}](https://tex.z-dn.net/?f=-3x%5E%7B3%7D%28-2x%5E%7B2%7D%2B4x-3%29%3D6x%5E%7B5%7D-12x%5E%7B4%7D%2B9x%5E%7B3%7D)
4)
is a square of a sum, then, by definition you have:
![(a+b)^{2}=a^{2}+2ab+b^{2}](https://tex.z-dn.net/?f=%28a%2Bb%29%5E%7B2%7D%3Da%5E%7B2%7D%2B2ab%2Bb%5E%7B2%7D)
Then:
![(4a+5)^{2}=(4a)^{2}+2(4a)(5)+5^{2}=16a^{2}+40a+25](https://tex.z-dn.net/?f=%284a%2B5%29%5E%7B2%7D%3D%284a%29%5E%7B2%7D%2B2%284a%29%285%29%2B5%5E%7B2%7D%3D16a%5E%7B2%7D%2B40a%2B25)
The coefficient of the second term is the number in front of the variable <em>a.</em> Then, the answer is: 40
5) Apply the Distributive property and the Product property of exponents, then, oyou must add the like terms:
![(x-4)(x^{2}-3x-9)=x^{3}-3x^{2}-9x-4x^{2}+12x+36=x^{3}-7x^{2}+3x+36](https://tex.z-dn.net/?f=%28x-4%29%28x%5E%7B2%7D-3x-9%29%3Dx%5E%7B3%7D-3x%5E%7B2%7D-9x-4x%5E%7B2%7D%2B12x%2B36%3Dx%5E%7B3%7D-7x%5E%7B2%7D%2B3x%2B36)
You would have to find the height.
To do this, you cut the triangle from the upper vertex to the base to create two new triangles.
Then you use the Law of Cosines. You could use the google calculator for this by searching “law of cosines calculator” and inputting the values, but in case of a test, the formula is a^2 = b^2 + c^2 - 2bc cosA
If you could provide an example of a triangle, I will gladly help you with the steps! Hope this helped!
Let dimes = d and quarters =q
D + q = 50
D= 50-q
0.10d + 0.25Q = 7.70
Replace d with 50-q:
0.10(50-q) + 0.25q = 7.70
Simplify:
5 -0.10q + 0.25q = 7.70
5 + 0.15q = 7.70
Subtract 5 from both sides:
0.15q = 2.70
Divide both sides by 0.15:
Q = 2.70 / 0.15
Q = 18
D = 50-18 = 32
There are 32 dimes and 18 quarters