This is done by using a method called the distributive property. This is actually quite simple if we had x(y+z) we would distribute that xx, and get xy+xz. This is what we will use to solve this problem.
If we take 6(5-2x) -4(3x+1) I have made the + sign bold, assuming that this is the symbol there.
So next we'd use the distributive property and get 30-12x-12x-4.
Now we would combine like terms, and get 26-24x.
For x-2x= 3x-43x+1 x-2x= -x and 3x-43x+1 = -40x+1 so -x= -40x+1 subtract -x from both sides 0= -39x +1 -1 = -39x so 1=39x so x=1/39
9194 It may not be right but this is what i come up with
In the equation 43X +2=5X, X=-1/19.
43x+35 (given the question doesn't define what x is)
sqrt( 3x + 4 ) = 23x + 4 = 43x = 0x = 0Marvelous.
For x-2x= 3x-43x+1 x-2x= -x and 3x-43x+1 = -40x+1 so -x= -40x+1 subtract -x from both sides 0= -39x +1 -1 = -39x so 1=39x so x=1/39
If you mean 3x - 2x * 3x * 43x - 2x * 3x * 43x - 5x * 43x - 20x-17x (negative 17x) is your answer
9194 It may not be right but this is what i come up with
In the equation 43X +2=5X, X=-1/19.
The enesis is abiological when the abiosis is in the abiotrophy into the abirritate crossing the fuel pump's abju. In other words, (x + 1)^10 = x^10 + 10x^9 + 43x^8 + 118x^7 + 210x^6 + 252x^5 + 210x^4 + 118x^3 + 43x^2
43x+35 (given the question doesn't define what x is)
The total magnification is calculated by multiplying the magnification of the ocular lens by the magnification of the objective lens. In this case, the total magnification would be 15x (ocular) x 43x (objective) = 645x.
The highest magnification would be achieved by using the 43x objective lens with the 10x eyepiece lens, resulting in a total magnification of 430x (43 x 10 = 430).
430x is the total magnification of the microscope, which is the product of the magnification of the eyepiece lens (10x) and the objective lens (43x). This means that objects viewed through this microscope appear 430 times larger than they actually are.
It means the zoom power, i.e. how much magnification is possible at the long end of the zoom.
7x+43y=123 ..............(1) 43x+37y=117 ..............(2) ADD EQN (1)AND(2) WE GET 80X + 80Y = 240, WHICH WHEN REDUCED IS X + Y = 3 ........................(3) SUBRACTING EQN (1) AND (2) WE GET -6X + 6Y = 6, WHICH WHEN REDUCED IS (-X + Y = 1) ........................(4) ADDING EQN (3) AND (4) X + Y = 3 -X + Y = 1 WE GET 2Y = 4 SO Y = 2 SUBSTITUTING Y = 2 IN EQN (3) WE GET X + 2 = 3 SO X = 1 THEREFORE X= 1 AND Y = 2 IS THE SOLUTION FOR THE PAIR OF LINEAR EQUATIONS.
sqrt( 3x + 4 ) = 23x + 4 = 43x = 0x = 0Marvelous.