Mathematics: The temperature of water in a lake at 9 a.m. is -2.6 by noon the temperature of the water rises by 5.1 celsius by 9 p.m. the temperature of the waterfalls by 12.8 from what it had been at noon right in addition problem to represent the change in temperature of the water and find the temperature of the water at 9 p.m.

Distance to halfway point=45 kmWilson s Time=5hrsJoseph s time=6.26hrsStep-by-step explanation:-Let x be the number of hours it takes for the two to meet halfway when Wilson starts his ride.#The time already utilized prior to Wilson s start is:Joseph s remaining distance to the hallway point is less by 10kmSince, the distance to meet is equal, we equate their times as:Hence, the distance ridden by each to the halfway point is 45 km#We now use this distance to solve for individual times:Hence, Joseph...

The temperature of water in a lake at 9 a.m. is

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sipstick9411

25.04.2021

Mathematics

Distance to halfway point=45 km

Wilson's Time=5hrs

Joseph's time=6.26hrs

Step-by-step explanation:

-Let x be the number of hours it takes for the two to meet halfway when Wilson starts his ride.

#The time already utilized prior to Wilson's start is:

$Speed=\frac{Distance}{Time}\\\\Time=\frac{10}{7}\\\\=\frac{10}{7}\ hrs$

Joseph's remaining distance to the hallway point is less by 10km

Since, the distance to meet is equal, we equate their times as:

$Time=\frac{Distance}{Speed}\\\\\frac{d-10}{7}=\frac{d}{9}\\\\9d-90=7d\\\\2d=90\\\\d=45\ km\\\\(d-10)=35\ km$

Hence, the distance ridden by each to the halfway point is 45 km

#We now use this distance to solve for individual times:

$Time=\frac{Distance}{Speed}\\\\\#Joseph\\\\T_j=\frac{10}{7}+\frac{35}{7}\\\\=6\frac{3}{7}\ hrs \ or \ 6.26\ hrs\\\\\#Wilson\\\\T_w=\frac{45}{9}\\\\=5\ hrs$

Hence, Joseph takes 6.26hrs while Wilson takes 5hrs to get to the halfway point.

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