Most quadratic functions(which is what you have there, to a degree of 2) are solved using factoring and the zero product law. If you can not factor then you have to use the quadratic formula or graph it. However this one can be factored.
It's pretty simple to just factor it by inspection but I use the chart method, if you know decomposition that works as well.
Factoring gives us,
Then you set each factor to 0 and solve for x,
And the second one,
The solutions to this equation are
x = -1/2, 3
One polynomial identity that crops up often in various areas is the difference of squares identity:
A2-b2=(a-b) (a+b)
We meet this in the context of rationalising denominators.
Answer:
Hello,
If f(x)=5x-13, then f^-1(x)= 
have good day
Step-by-step explanation:
Answer:
Claim 2
Step-by-step explanation:
The Inscribed Angle Theorem* tells you ...
... ∠RPQ = 1/2·∠ROQ
The multiplication property of equality tells you that multiplying both sides of this equation by 2 does not change the equality relationship.
... 2·∠RPQ = ∠ROQ
The symmetric property of equality says you can rearrange this to ...
... ∠ROQ = 2·∠RPQ . . . . the measure of ∠ROQ is twice the measure of ∠RPQ
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* You can prove the Inscribed Angle Theorem by drawing diameter POX and considering the relationship of angles XOQ and OPQ. The same consideration should be applied to angles XOR and OPR. In each case, you find the former is twice the latter, so the sum of angles XOR and XOQ will be twice the sum of angles OPR and OPQ. That is, angle ROQ is twice angle RPQ.
You can get to the required relationship by considering the sum of angles in a triangle and the sum of linear angles. As a shortcut, you can use the fact that an external angle is the sum of opposite internal angles of a triangle. Of course, triangles OPQ and OPR are both isosceles.
