We have x^2 + 2 · x · (11/2) + (11/2)^2 = - 24 + (11/2)^2;
Then, ( x + 11/2 )^2 = -24 + 121/4;
( x + 11/2 )^2 + 96/4 - 121/4 = 0;
( x + 11/2 )^2 - 25 / 4 = 0;
( x + 11/2 )^2 - (5/2)^2 = 0;
( x + 11/2 - 5/2)·( x + 11/2 + 5/2 ) = 0;
( x + 6/2 )·( x + 16/2 ) = 0;
( x + 3 )· ( x + 8 ) = 0;
x = - 3 or x = -8;
The first choice is the correct answer.
Answer:
(-5/2, 5)
Step-by-step explanation:
Apply x-y intercepts theorem
(x,y)=(0,y) and (x,0).
then substitutes.
Answer:
404 cm³ Anyway... Look down here for my explanation.
Step-by-step explanation:
Let's Draw a line from the center of the circle to one of the ends of the chord (water surface) and another to the point at greatest depth on your paper. A right-angled triangle is formed too. The Length of side to the water-surface is 5 cm, the hospot is 7 cm.
We Calculate the angle θ in the corner of the right-angled triangle by: cos θ = 5/7 ⇒ θ = cos ˉ¹ (5/7)
44.4°, so the angle subtended at the center of the circle by the water surface is roughly 88.8°
The area shaded will then be the area of the sector minus the area of the triangle above the water in your diagram.
Shaded area 88.8/360*area of circle - ½*7*788.8°
= 88.8/360*π*7² - 24.5*sin 88.8°
13.5 cm²
(using area of ∆ = ½.a.b.sin C for the triangle)
Volume of water = cross-sectional area * length
13.5 * 30 cm³
404 cm³
