$ax^2+ay^2-bx^2-by^2+b-a\\\\=(ax^2-bx^2)+(ay^2-by^2)+(b-a)\qquad\text{use the distributive property}\\\\=x^2(a-b)+y^2(a-b)-1(a-b)\\\\=(a-b)(x^2+y^2-1)$

4 0

3 years ago

Respond to all of the questions within the topic (see that there are six questions below). 1. Explain how you can determine if t

Arturiano [62]

Answer:

Step-by-step explanation:

6 0

3 years ago

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What is PA <br> Enter your answer in the box

SSSSS [86.1K]

Answer:

Step-by-step explanation:

P is the in-center

⇒PA=PE=PD because they are in-radius of the in-circle

We know that, tangent segments drawn from a point outside the circle are always equal in length

⇒DK=EK=7.2

In right triangle PKE,

using Pythagoras' Theorem : $PK^{2}=PE^{2}+KE^{2}$

⇒ $PE^{2}=PK^{2}-KE^{2}$

⇒ $PE=\sqrt{PK^{2}-KE^{2} }$

⇒ $PE^{2}=\sqrt{7.8^{2}-7.2^{2}} }$

⇒ $PE=3$

Therefore, $PA=3$

3 0

3 years ago