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Whitepunk [10]
3 years ago
8

A man walked steadily from 11.00am to 2.30pm at 5 kilometres per hour. how far did he walk

Mathematics
2 answers:
Pavlova-9 [17]3 years ago
7 0
Time taken = 14 : 30 - 11 :00 = 3.5 hrs
Speed = 5 km /hr
Distance = time * speed 
= 3.5 * 5 
= 17.5 km
DerKrebs [107]3 years ago
5 0
Taken time is from 11.00 am to 2.30 pm (14.30) = 14.30 - 11 = 3.30 hours

We know that D = S* T (D - distance, S - speed , T- time)

So, D= \frac{5km}{h}*3.5h=\boxed {17.5 hours}
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3 years ago
Can someone help me please?
Masja [62]

Given:

A line segment AB.

A point is \dfrac{3}{10} of the way from A to B.

To find:

The coordinates of that point.

Solution:

Section formula: If a point divide a line segment in m:n, then

Point=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)

From the given figure, it is clear that the two end points of the line segment are A(-4,-5) and B(12,4).

Let the unknown point be P. The point is \dfrac{3}{10} of the way from A to B. It means,

\dfrac{AP}{AB}=\dfrac{3}{10}

Let AP and AB are 3x and 10x, then

\dfrac{AP}{PB}=\dfrac{AP}{AB-AP}

\dfrac{AP}{PB}=\dfrac{3x}{10x-3x}

\dfrac{AP}{PB}=\dfrac{3x}{7x}

\dfrac{AP}{PB}=\dfrac{3}{7}

It means, point P divided the line segment in 3:7.

Using section formula, we get

P=\left(\dfrac{3(12)+7(-4)}{3+7},\dfrac{3(4)+7(-5)}{3+7}\right)

P=\left(\dfrac{36-28}{10},\dfrac{12-35}{10}\right)

P=\left(\dfrac{8}{10},\dfrac{-23}{10}\right)

P=\left(0.8,-2.3\right)

Therefore, the coordinates of the required point are (0.8, -2.3).

6 0
3 years ago
Plz help!! Fast it's porportional
densk [106]
It would be h 
hope this helped
8 0
3 years ago
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