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3 years ago

In the figure shown PS = RQ. which theorem can be used to prove Triangle SQR = Triangle QSP?

Mathematics

1 answer:

Nuetrik [128]3 years ago

3 0

Answer:

HL (hypotenuse leg)

Step-by-step explanation:

From the diagram, we have two sides that are congruent

The side QS is common to both triangles

Also, we have it that the side PS and QR are congruent

What this mean is that we have two sides that are already the same

Now, looking at the right angle position in both, we can see that QR and PS represents the hypotenuse of individual right angles

Using the HL congruence statement, since we have it that the hypotenuse one side is already congruent, we can conclude that the last side too is congruent and as such both triangles are congruent to each one another

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Step-by-step explanation:

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Let's find $\frac{-b}{2a}$ first.

$\frac{-2}{2(1)}=-1$

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$g(x)=x^2+2x-5$

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$g(-1)=1-2-5$

$g(-1)=-1-5$

$g(-1)=-6$ .

The vertex is (-1,-6).

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