( f + g ) (x) = –2x + 6
( f – g ) (x) = 8x – 2
( f × g ) (x) = –15x2 + 2x + 8
(
f
g
)
(
x
)
=
3
x
+
2
4
−
5
x
For a triangle the area is given as:
![A=\frac{bh}{2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7Bbh%7D%7B2%7D)
Re-arrange / solve for 'h':
![h=\frac{2A}{b}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B2A%7D%7Bb%7D)
Substitute known values:
![h=\frac{2(60m^{2})}{15m}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B2%2860m%5E%7B2%7D%29%7D%7B15m%7D)
Compute:
![h=\frac{120m^{2}}{15m}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B120m%5E%7B2%7D%7D%7B15m%7D)
Answer:
-225
Step-by-step explanation:
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