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Select the correct answer from the drop-down menu.
Consider the equations y=√x and y=x² -1
The system of equations is equal at approximately
A.(1.5,1.2)
B.(-1.5,-1.2)
C.(1.5,-1.2)
D.(-1.5,1.2)

Respuesta :

Answer:

Choice A.

Step.-by-step explanation:

y=√x

y=x² -1

Substituting the values in choice A:

1.2 = √1.5 = 1.2247     Approximately equal.

1.2 = (1.5)^2 - 1 = 1.25   Approximately equal.

Choice B.

-1.2 = √-1.5  which is imaginary so NOT this one.

Choice C

-1.2 = √1.5 = -1.2247

-1.2 = (1.5)^2 - 1 = 1.25  NO.

Choice D

We have the non real value √-1.5 again so NOT this one.

Answer: Option A

A.(1.5,1.2)

Step-by-step explanation:

We have the following system of equations

[tex]y=\sqrt{x}[/tex]

[tex]y=x^2 -1[/tex]

the solution of the system will be all points that satisfy both equations at the same time

For (1.5,1.2)

[tex]y=\sqrt{1.5}=1.2[/tex]

[tex]y=(1.5)^2 -1=1.2[/tex]

Both equations are satisfied

Note that we can discard options B and D because the domain of the equation [tex]y =\sqrt{x}[/tex] does not include the negative numbers.

We can discard option C because the range of the function [tex]y =\sqrt{x}[/tex] does not include the negative numbers.

Finally the answer is the option A