Answer:
A = 3
B = 1
C = 2
D = 10
E = 2
F = Not listed
G = 1
Step-by-step explanation:
![\sqrt[3]{270x^5y^7} =\sqrt[3]{27*10x^5y^7}=3xy^2\sqrt[3]{10x^2y}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B270x%5E5y%5E7%7D%20%3D%5Csqrt%5B3%5D%7B27%2A10x%5E5y%5E7%7D%3D3xy%5E2%5Csqrt%5B3%5D%7B10x%5E2y%7D)
Answer: ill try looking this one up to help u just add me on snap its johni_spurgeon
Step-by-step explanation:
The first step is to write this equation into general form. The
general form of an equation is:
ax^2 + bx + c = 0
To make this equation to general form, you have to simplify
the equation first.
2/3(x-4) (x+5) = 1
2/3 (x^2 + 5x – 4x – 20) = 1
2/3(x^2 + x -20) = 1
2/3x^2 + 2/3x – 40/3 = 1
2/3x^2 + 2/3x – 40/3 – 1 = 0
2/3x^2 +2/3x – 43/3 = 0
Therefore, a = 2/3 ; b = 2/3 ; c = -43/3
Answer:
a. (3x^4 + 6 ) + (2x^2)
b. (X^3 ) + (-x -7)
c. (4.6x^4) + (-1.5x^2)
Step-by-step explanation:
For the first one, we have to write to polynomials, which equal 3x^4 + 2x^2 +6
One possible solution is (3x^4 + 6 ) + (2x^2)
For b, we can do,
(X^3 ) + (-x -7)
And finally for c, we can write,
(4.6x^4) + (-1.5x^2)
Answer:
The top left graph.
Step-by-step explanation:
That is because it goes through y=-2