Subtract four from both sides. That'll put both numbers without a variable on one side, and the number with the variable is by itself.
I think the perimeter is at least 6m
Step-by-step explanation:
part A:
part B:
![Area_{_{surface}}=3.14*(15+9)=3.14*24=75.36[units^2].](https://tex.z-dn.net/?f=Area_%7B_%7Bsurface%7D%7D%3D3.14%2A%2815%2B9%29%3D3.14%2A24%3D75.36%5Bunits%5E2%5D.)
Answer:
i need this toooooooooooooooooooo
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
