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
Given:
Consider the equation is:
To prove:
by using the properties of logarithms.
Solution:
We have,
Taking left hand side (LHS), we get
![\left[\because \log_ab=\dfrac{\log_x a}{\log_x b}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbecause%20%5Clog_ab%3D%5Cdfrac%7B%5Clog_x%20a%7D%7B%5Clog_x%20b%7D%5Cright%5D)
![[\because \log x^n=n\log x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Clog%20x%5En%3Dn%5Clog%20x%5D)
Hence proved.
The mode would be 17, it’s the number that appears the most. I’m not rewriting it though
Answer:
pls stop deleting my answers brainly thx. also its b! ^-^
Step-by-step explanation:
