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)
Factor out the common term; 3
(3(x + 1))^2 = 36
Use the Multiplication Distributive Property; (xy)^a = x^ay^a
3^2(x + 1)^2 = 36
Simplify 3^2 to 9
9(x + 1)^2 = 36
Divide both sides by 9
(x + 1)^2 = 36/9
Simplify 36/9 to 4
(x + 1)^2 = 4
Take the square root of both sides
x + 1 = √4
Since 2 * 2 = 4, the square root of 2 is 2
x + 1 = 2
Break down the problem into these 2 equations
x + 1 = 2
x + 1 = -2
Solve the first equation; x + 1 = 2
x = 1
Solve the second equation; x + 1 = -2
x = -3
Collect all solutions;
<u>x = 1, -3</u>
To solve this equation by elimination, what you would do is multiply one of the equations by -1, or distribute -1 to each term in the equation, any of the 2 equations. Then align the equations and add them together.
-(X + 3y = 3)
-X - 3y = -3
-X - 3y = -3
X + 6y = 3
__________
3y = 0
y = 0/3 = 0.
Now we can solve for x, by simply plugging the value of y into any of the 2 equations.
X + 6y = 3
X + 6(0) = 3
X + 0 = 3
X = 3.
The solution to your system of equations would be (3,0).
Check this by plugging in the point to the other equation and see if it is true.
X + 3y = 3
(3) + 3(0) = 3
3 + 0 = 3
3 = 3.
Thus it is the solution.
So lets just assume that y = 1 since 2x is most likely an even number.
Then we can say that 2x = 8
8 divided by 2 is 4
so a point on this line could be (4,1)
Hope this helped
Answer:
<u>Apply the property above:</u>
- 1. ln 618 = p ⇔ e^p = 618
- 2. ln q = 2 ⇔ e^2 = q
- 3. ln 100 = t ⇔ e^t = 100
- 4. ln (e^3) = 3 ⇔ e^3 = e^3
