Answer:
1/7, 3/7, 4/7, 6/7
Step-by-step explanation:
There's like denominators(bottom number) so you would list the numerators(top number) from least to greatest.
The number of grams present in 0.54 moles of calcium is; 22 g
We are given;
Number of moles; n = 0.54 moles
Now, Molar mass of calcium is;
M = 40.1 g
Formula for number of moles is;
n = m/M
Where;
n is number of moles
m is mass in grams
M is molar mass
Thus;
0.54 = m/40.1
m = 40.1 × 0.54
m = 21.654 g
Approximately m = 22 g
Read more at; brainly.com/question/5397233
Answer and explanation:
The expression P-2P+3P-4P cannot have the same values for different values of P. Let us illustrate this by substituting different values of P in the expression:
If P= 2
Substitute P=2 in the expression
2-2×2+3×2-4×2
=2-4+6-8
=-4
If P= 3
Substitute P=3 in the expression
3-2×3+3×3-4×3
=3-6+9-12
=-6
If P = 1
Substitute P=1 in the expression
1-2×1+3×1-4×1
= 1-2+3-4
= -2
If P = -2
Substitute P=1 in the expression
-2-2×-2+3×-2-4×-2
= -2+4-6+8
=4
Therefore we can see from the above that the expression has different values for different values of P
T=48
You just need to find out how many times 12 goes into a number, 4 times.
Which is 48 because 12x4 is 48 and you can check your answer by dividing 48/12=4
the idea behind the completion of the square is simply using a perfect square trinomial, hmmm usually we do that by using our very good friend Mr Zero, 0.
if we look at the 2nd step, we have a group as x² - x, hmmm so we need a third element, which will be squared.
keeping in mind that the middle term of the perfect square trinomial is simply the product of the roots of "a" and "b", so in this case the middle term is "-x", and the 1st term is x², so we can say that
so that means that our missing third term for a perfect square trinomial is simply 1/2, now we'll go to our good friend Mr Zero, if we add (1/2)², we have to also subtract (1/2)², because all we're really doing is borrowing from Zero, so we'll be including then +(1/2)² and -(1/2)², keeping in mind that 1/4 - 1/4 = 0, so let's do that.
![-3~~ = ~~-2\left[ x^2-x+\left( \cfrac{1}{2} \right)^2 ~~ - ~~\left( \cfrac{1}{2} \right)^2\right]\implies -3=-2\left(x^2-x+\cfrac{1}{4}-\cfrac{1}{4} \right) \\\\\\ -3=-2\left(x^2-x+\cfrac{1}{4} \right)+(-2)-\cfrac{1}{4}\implies -3=-2\left(x^2-x+\cfrac{1}{4} \right)+\cfrac{1}{2} \\\\\\ -3-\cfrac{1}{2}=-2\left(x^2-x+\cfrac{1}{4} \right)\implies -\cfrac{7}{2}=-2\left(x-\cfrac{1}{2} \right)^2\implies \cfrac{7}{4}=\left(x-\cfrac{1}{2} \right)^2](https://tex.z-dn.net/?f=-3~~%20%3D%20~~-2%5Cleft%5B%20x%5E2-x%2B%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%20~~%20-%20~~%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%5Cright%5D%5Cimplies%20-3%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D-%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20-3%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%2B%28-2%29-%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20-3%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%2B%5Ccfrac%7B1%7D%7B2%7D%20%5C%5C%5C%5C%5C%5C%20-3-%5Ccfrac%7B1%7D%7B2%7D%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%5Cimplies%20-%5Ccfrac%7B7%7D%7B2%7D%3D-2%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%5Cimplies%20%5Ccfrac%7B7%7D%7B4%7D%3D%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2)
