If u (x) = negative 2 x squared and v (x) = StartFraction 1 Over x EndFraction, what is the range of (u circle ma poottay

1 answer:

5 0

The range is negative numbers.
The interval for the range is .
***You might want to look at your functions again because I don't see a choice that matches.
Step-by-step explanation:
Given functions:

We are asked to find the range of .
I'm also going to look at the domain just to see if this possibly might change my range .
is the inner function. So we will consider the domain of that function first.
You only have to worry about division by zero for the function .
Since we are dividing by , we don't want to be zero.
So far the domain is all real numbers except .
Now let's move out.
exists for all numbers, . So we didn't want to include from before.
Now let's put it together:

So the domain is still all real numbers except at since we cannot divide by 0 and is 0 when .
with .
is positive for all numbers except .
So is negative for all numbers since negative divided by positive is negative.
So the range is only negative numbers.
Let's also look at the inverse:

Multiply both sides by :

Divide both sides by :

Take the square root of both sides:
.
So can't be 0 and it also can't be positive because the inside of the square root will be negative (since negative divided by positive results in negative).

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You want to make 500 ml of a 1 N solution of sulfuric acid (H2SO4, MW: 98.1). How many grams of sulfuric acid do you need?

umka21 [38]

Answer:

24.525 g of sulfuric acid.

Explanation:

Hello,

Normality (units of eq/L) is defined as:

$N=\frac{eq_{solute}}{V_{solution}}$

Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:

$eq_{solute}=N*V_{solution}=1\frac{eq}{L}*0.5L=0.5 eq$

Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:

$m_{H_2SO_4}=0.5eqH_2SO_4(\frac{1molH_2SO_4}{2 eqH_2SO_4}) (\frac{98.1 g H_2SO_4}{1 mol H_2SO_4} )\\m_{H_2SO_4}=24.525 g H_2SO_4$