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3 years ago

14

Water flowing at the rate of 15km/hr through a pipe of diameter 14cm into a cuboidal pond which is 50m long and 44m wide. In whi

ch time will the level of water in the pond rise 21cm?

1 answer:

katen-ka-za [31]3 years ago

5 0

Hello,

We must calculate the volume of water in one hour.

V=15000*PI*0.07²=230,9070600... m^3/h.

Volume of the pond=50*44*.21=462 m^3

Time=462 m^3/ 230.907... m^3/h=2,0008... h≈2 h

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Calculate the slope of each segment (just need a couple example problems and step by step, no need to answer all. Will mark brai

Tpy6a [65]

Answer:

12) Line segment AB

$m_{AB} = \frac{7-4}{10-4}= \frac{1}{2}$

13) Line segment CD

$m_{CD} = \frac{6-11}{15-13}=-\frac{5}{2}$

14) Line segment EF

$m_{EF} = \frac{7-11}{4-3}= -4$

15) Line segment GH

$m_{GH} = \frac{5-2}{8-6}= \frac{3}{2}$

16) Line segment IJ

$m_{IJ} = \frac{0-(-6)}{0-4}=-\frac{3}{2}$

17) Line segment KL

$m_{KL} = \frac{-2-(-3)}{5-(-2)} = \frac{1}{7}$

18) Line segment MN

$m_{MN} = \frac{5-5}{3-(-4)} = 0$

19) Line segment OP

$m_{OP} = \frac{4-(-2)}{13-13} = \frac{1}{0}$

20) Line segment QR

$m_{QR} = \frac{7-11}{-2-(-7)} = -\frac{4}{5}$

21) Line segment ST

$m_{ST} = \frac{-2-(-3)}{10-9} = 1$

Step-by-step explanation:

Coordinates are now described: $A(x,y) = (4,4)$ , $B(x,y) = (10, 7)$ , $I(x,y) = (-6, 4)$ , $J(x,y) = (0,0)$ , $Q(x,y) = (-7,11)$ , $R(x,y) = (-2,7)$ , $C(x,y) = (13,11)$ , $D(x,y) = (15, 6)$ , $K(x,y) = (-2, -3)$ , $L(x,y) = (5,-2)$ , $S(x,y) = (9,-3)$ , $T(x,y) = (10, -2)$ , $E(x,y) = (3, 11)$ , $F(x,y) = (4,7)$ , $M(x,y) = (-4,5)$ , $N(x,y) = (3, 5)$ , $G(x,y) = (6,2)$ , $H(x,y) = (8,5)$ , $O(x,y) = (13,-2)$ , $P(x,y) = (13,4)$

The slope of each line segment is calculated by applying definition of secant line:

12) Line segment AB

$m_{AB} = \frac{7-4}{10-4}= \frac{1}{2}$

13) Line segment CD

$m_{CD} = \frac{6-11}{15-13}=-\frac{5}{2}$

14) Line segment EF

$m_{EF} = \frac{7-11}{4-3}= -4$

15) Line segment GH

$m_{GH} = \frac{5-2}{8-6}= \frac{3}{2}$

16) Line segment IJ

$m_{IJ} = \frac{0-(-6)}{0-4}=-\frac{3}{2}$

17) Line segment KL

$m_{KL} = \frac{-2-(-3)}{5-(-2)} = \frac{1}{7}$

18) Line segment MN

$m_{MN} = \frac{5-5}{3-(-4)} = 0$

19) Line segment OP

$m_{OP} = \frac{4-(-2)}{13-13} = \frac{1}{0}$

20) Line segment QR

$m_{QR} = \frac{7-11}{-2-(-7)} = -\frac{4}{5}$

21) Line segment ST

$m_{ST} = \frac{-2-(-3)}{10-9} = 1$

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