the distance from a boy's house to his school is 10km. he left home at 6.30am, cycling at the rate of 4km/h. if he spent 15 minutes on the way waiting for his friend, when did he get to school?

9:15 am

Step-by-step explanation:

First you know the speed and you know the distance.

So now we have to find the time it took for him to get to school(ignore the waiting time first)

time=10/4=2.5h

now we add the waiting time,

2.5h+0.25h(15minutes is the same as 0.25h)=2.75h

convert 0.75h to minuts,

0.75*60=45minutes

6.30+2.45=9.15am

9:35

Step-by-step explanation:

6 : 30 min + 15 min = 6 : 45

10 km ÷ 4 km/h = 2 h and 50 min

2h and 50 min + 6 : 45 = 9 : 34

step-by-step explanation:

a rhombus is a parallelogram enjoying his symmetry.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!