All categories

2 years ago

Joe travelled 60 miles in 1 hour 30 minutes.Work out Joe's average speed.

Mathematics

1 answer:

Vilka [71]2 years ago

6 0

Answer:

1.5

Step-by-step explanation:

1hour and 30 mins = 90 mins

60miles:90mins that is the ratio

90/60=1.5miles per minute

i think this is the right answer

pls let me know and i hope it helps

You might be interested in

An exercise machine with an original selling price of $1,275 is marked down by 20%. What is the sale price of the exercise machi

krek1111 [17]

Answer:

1,020

Step-by-step explanation:

If it is marked down by 20%, then 80% is left. What is 80% of 1275 you ask? .8*1275=1020!

5 0

3 years ago

1)Sheyna drive to the lake and back. It took two hours less time to get there than it did to get back. The average speed on the

ipn [44]

Answer:

8 hours

Step-by-step explanation:

Given:

Sheyna drives to the lake with average speed of 60 mph and

$v_1 = 60\ mph$

Sheyna drives back from the lake with average speed of 36 mph

$v_2 = 36\ mph$

It took 2 hours less time to get there than it did to get back.

Let $t_1$ be the time taken to drive to lake.

Let $t_2$ be the time taken to drive back from lake.

$t_2-t_1 = 2$ hrs ..... (1)

To find:

Total time taken = ?

$t_1+t_2 = ?$

Solution:

Let D be the distance to lake.

Formula for time is given as:

$Time =\dfrac{Distance}{Speed }$

$t_1 = \dfrac{D}{60}\ hrs$

$t_2 = \dfrac{D}{36}\ hrs$

Putting in equation (1):

$\dfrac{D}{36}-\dfrac{D}{60} = 2\\\Rightarrow \dfrac{5D-3D}{180} = 2\\\Rightarrow \dfrac{2D}{180} = 2\\\Rightarrow D = 180\ miles$

So,

$t_1 = \dfrac{180}{60}\ hrs = 3 \ hrs$

$t_2 = \dfrac{180}{36}\ hrs = 5\ hrs$

So, the answer is:

$t_1+t_2 = \bold{8\ hrs}$

8 0

3 years ago

If two parallel planes are cut by a third plane, then the lines of intersection are _________.

maxonik [38]

<span>If two parallel planes are cut by a third plane, then the lines of intersection are parallel and cannot intersect one another.</span>

6 0

3 years ago

Read 2 more answers

Let a, b, c and x elements in the group G. In each of the following solve for x in terms of a, b, and c.

alina1380 [7]

Answer:

The answer is $x=a^{-1}cb^{-1}$ .

Step-by-step explanation:

First, it is important to recall that the group law is not commutative in general, so we cannot assume it here. In order to solve the exercise we need to remember the axioms of group, specially the existence of the inverse element, i.e., for each element $g\in G$ there exist another element, denoted by $g^{-1}$ such that $gg^{-1}=e$ , where $e$ stands for the identity element of G.