Answer:
could we have a picture or the equation?? plz
Answer:
Step-by-step explanation:
Remark
Complete the square for x and for y.
complete for x
x^2 + 2x + (2/2)^2
(x + 1)^2
Complete square for y
y^2 + 10y + (10/2)^2
(y + 5)^2
Now put these two together
(x + 1)^2 + (y + 5)^2 = 15 + (10/2)^2 + (2/2)^2
(x + 1)^2 + (y + 5)^2 = 15 + 25 + 1
(x + 1)^2 + (y + 5)^2 = 41
Answer
Center: (-1,-5)
radius: (sqrt(41) = 6.403
The third choice:
f(0) = g(-2)
Given that <span>AB
∥ DE. Because the lines are parallel and segment CB crosses both lines,
we can consider segment CB a transversal of the parallel lines. Angles
CED and CBA are corresponding angles of transversal CB and are therefore
congruent, so ∠CED ≅ ∠CBA. We can state ∠C ≅ ∠C using the reflexive
property.
From the proof, we have been able to establish two congruencies, that is that angles CED and CBA are congruent and angle C is congruent to angle C.
Therefore, we conclude that △ACB ~ △DCE by the AA similarlity theorem.</span>
Answer:
x = 15
Step-by-step explanation:
Using the Altitude on Hypotenuse theorem
( leg of large Δ )² = (part of hypotenuse below it ) × ( whole hypotenuse )
Thus
x² = 9 × (9 + 16) = 9 × 25 = 225 ( take the square root of both sides )
x = 
= 15
