Answer:
Let's solve your equation step-by-step.
|−x−15|=|5x−23|
Solve Absolute Value.
|−x−15|=|5x−23|
We know either−x−15=5x−23or−x−15=−(5x−23)
−x−15=5x−23(Possibility 1)
−x−15−5x=5x−23−5x(Subtract 5x from both sides)
−6x−15=−23
−6x−15+15=−23+15(Add 15 to both sides)
−6x=−8
−6x
−6
=
−8
−6
(Divide both sides by -6)
x=
4
3
−x−15=−(5x−23)(Possibility 2)
−x−15=−5x+23(Simplify both sides of the equation)
−x−15+5x=−5x+23+5x(Add 5x to both sides)
4x−15=23
4x−15+15=23+15(Add 15 to both sides)
4x=38
4x
4
=
38
4
(Divide both sides by 4)
x=
19
2
Check answers. (Plug them in to make sure they work.)
x=
4
3
(Works in original equation)
x=
19
2
(Works in original equation)
Answer:
x=
4
3
or x=
19
2
Step-by-step explanation:
Answer:
(x+6)² + (y-6)² = 181
Step-by-step explanation:
equation of the circle in general is (x - a)² + (y - b)² = r²
center is (a, b) in our task a=-6 b=6
so equation of the circle is (x+6)² + (y-6)² = r²
we have point (3,-4) ⇒ x=3 y=-4
put x y in equation of the circle ⇒ (3+6)² + (-4-6)² = r² ⇒r² =181
⇒The equation of a circle is (x+6)² + (y-6)² = 181
The usefulness is that this data allows you to understand how well students perform with their instruction, based on a standardized test. The limitation is that the scores don't account for different situations that may affect the results you receive.
