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k0ka [10]
2 years ago
11

If the graph of y = a x 2 + b x + c does not intersect the x -axis, then what is true about the roots? Both are real roots. Both

are real roots. Both are imaginary roots. Both are imaginary roots. One is a real root, and one is irrational. One is a real root, and one is irrational. One is a real root, and one is rational.
Chemistry
1 answer:
Reptile [31]2 years ago
8 0

Answer:

Both roots are imaginary roots.

Explanation:

Consider these things:

If we try to solve x²+1 = 0, notice that we aren't able to solve the equation in Real Number system because there are no negative outputs for quadratic function.

Remember that quadratic function has range greater or equal to the max-min value.

x-axis plane represents the solutions of that equation. If a graph intersects x-axis plane then it has a solution.

While a graph that doesn't have any intersects on x-plane, it means that the equation for that graph doesn't have real solutions but imaginary solutions.

As you may notice some of parabola graph has one intersect, two intersects or none. One intersect is one solution to the equation — Two intersects are two solutions of the equation and lastly, no intersects mean that no real solutions and remain only imaginary solution.

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If the relative activities of two metals are known which metal is more easily oxidized
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Answer: The metal that has a greater reactivity is more easily oxidized.

Explanation:

Oxidation is when the elements lose electrons and increase their oxidation state.

The metals tend to react by losing electrons and form the corresponding cation.

For expample, sodium (an alkalyne metal) loses one elecron and form the cation Na¹⁺ , then this cation combine with an anion and form compounds like NaCl, NaOH. The same do the other alkalyne metals.

Magnesium (an alkalyne earth metal) loses two electrons and form the cation Mg²⁺, then it combines with some anions to form compounds, like MgSO₄, Mg(OH)₂.

So, the easier the metal gets oxidized the greater its reactivity.




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3 years ago
You are given a solution that is 518 mM lactose. You need to make up 4.5 L of 16.7 mM solution. What volume do you need to trans
OLga [1]

Answer:

The volume you need to transfer from the stock solution is 0.145 l

Explanation:

Since the number of moles of lactose in the volume of stock solution that you transfer will be the same as the number of moles of lactose in the final solution, you can use this expression:

number of moles in volume to transfer = number of moles in the final solution

Since number of moles = concentration * volume (if the concentration is expressed in molarity), then:

Ci * Vi = Cf * Vf

where:

Ci = concentration of the stock solution.

Vi = volume of the stock solution to be transferred.

Cf = concentration of the final solution

Vf = volume of the final solution

Then, replacing with the data:

518 mM * Vi = 16.7 mM * 4.5 l

Vi = 16.7 mM * 4.5 l / 518 mM

<u>Vi = 0.145 l or 145 ml</u>

Notice that any concentration unit can be used, as long as the units of the concentration of the stock and final solution are the same.

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