Answer:
7(7x + 2)
Step-by-step explanation:
when distributed it will be 49x + 14
Completing the square is a process to find the solutions, or the x-values, to a quadratic equation. This method can only work if it is in the format: x^2 + bx = c
In this equation, the b value is -12 and the c value is -6. The process for completing the square goes like this:
x^2 + bx + (b/2)^2 = c + (b/2)^2
Now let’s solve the equation above using this method.
Step 1: x^2 - 12x + (-12/2)^2 = -6 + (-12/2)^2
Step 2: x^2 - 12x + (-6)^2 = -6 + (-6)^2
Step 3: x^2 - 12x + 36 = -6 + 36
Step 4: x^2 - 12x + 36 = 30
Now, to factor it. After doing the process until now, the left side of the equation can ALWAYS be in the format: (x + a)^2
Step 5: x^2 - 12x + 36 can be factored in this format as (x - 6)^2
Step 6: (x - 6)^2 = 30
Step 7: x - 6 = √30
Step 8: x = 6 ±√30
The solution is:A mole of gas occupies 22.4 L A liter is 1000 cubic centimeters because 1 cubic centimeter = 1 mL
One Helium molecule (essentially one helium atom) has atomic mass 4 g/mol So for every 22400 cubic centimeters of volume, we have 4 grams of helium Density of helium = 4g / 22400 cm^3 = 1g / 5600 cm^3
Volume of a sphere = (4/3)(pi)r^3 Volume of the outside sphere (the entire sphere) is (4/3)(pi)(R+T)^3 Volume of the inside sphere (the hollow region) is (4/3)(pi)R^3
The difference for the volume of silver. (4/3)(pi)(R+T)^3 - (4/3)(pi)R^3 = (4/3)(pi)(3R^2T + 3RT^2 + T^3)
The density of silver is 10.5g/cm^3
So the mass of the silver is computed by:10.5*(4/3)(pi)(3R^2T + 3RT^2 + T^3) = (14*pi)(3R^2T + 3RT^2 + T^3) = (14pi)T(3R^2 + 3RT + T^2)
Now for the mass of helium: volume x density = (4/3)(pi)R^3 (1/5600) = (pi/4200)R^3
Set the two masses equal: (pi/4200)R^3 = (14pi)T(3R^2 + 3RT + T^2) R^3 = 58800*T(3R^2 + 3RT + T^2) R / T = 58800*(3R^2 + 3RT + T^2) / R^2 = 58800*( 3+T/R^2+(T/R)^2)
then solve for xx = T / R 1/x = 58800*( 3+x/R+x^2) 1/58800 = x (3 + x/R + x^2) 1/58800 = 3x + x^3 x^3 + 3x - 1/58800= 0 x = ~ 5.66893x10^(-6)
The quadrants are as follows:
The 1st quadrant has the points which have both x and y positive and the 3rd quadrant has the points which have both x and y negative. If the ordered pair and the same x and y value, if ons is positive, the other also is, and the same for negative.
So, at first we see that there are point where the x and y are the same and that are in the 1st or 3rd quadrant.
However, there is one special case:
When x and y are 0, that is, the ordered pair is (0, 0).
Since this point is the origin, it doesn't lie on any of the quadrants.
Thus, this affirmative is sometimes true. Every point but (0, 0) that have same x and y values are in the 1st or 3rd quadrant except for (0, 0).
