I hope my answer help you in your question
What I meant was, which ones do you need and could you take a picture in better lighting? I can't see it all the way.
For this case we must indicate which of the equations shown can be solved using the quadratic formula.
By definition, the quadratic formula is applied to equations of the second degree, of the form:
Option A:
Rewriting we have:
This equation can be solved using the quadratic formula
Option B:
Rewriting we have:
It can not be solved with the quadratic formula.
Option C:
Rewriting we have:
This equation can be solved using the quadratic formula
Option D:
Rewriting we have:
It can not be solved with the quadratic formula.
Answer:
A and C
Answer:
1. a 3d rectangle
2. 40 in cubed
3. repeat
4. another repeat
Step-by-step explanation:
In fact, this problem belongs to the chemistry section. Recall that many other sciences require mathematical calculations. The problem will belong to Mathematics only if no knowledge of other sciences are required to solve the problem.
Solubility for the given substances is measured in grams per 100 g of water at a particular temperature (20 deg.C).
This means that the mass (assumed to be the solute) will not change the solubility, just the minimum quantity of solvent (water) will.
Thus the solubility of sodium chloride will remain L=36 g/100g H2O for any quantity of solute. Similarly, the solubility of lead nitrate will remain as K=54 g/100 g H2O.
The reason that they remain constant is because the quantity of solvent (water) is fixed at 100 g. Varying amount of solute will affect the quantity of solvent required, but not the solubility.
I'll leave it to you to calculate the difference between K & L.
