Ask question

Ask question

All categories

4 years ago

6

Janice aud Thomas are roller blading around Sampson Pond It takes them 4 minutes to complete a lap. Write and solve an equation

to how many laps they can complete in one hour if they continue roller at this pace. Brainly

1 answer:

grigory [225]4 years ago

8 0

The total number of laps taken by Janice and Thomas in 1 hour is 15

<h3><u>Solution:</u></h3>

We have been given the following data, Janice and Thomas are roller blading around Sampson Pond it takes them 4 minutes to complete a lap.

We need to write and solve an equation to how many laps they can complete in one hour if they continue roller at this pace.

We know that, 1 lap = 4mins

$\text {The total number of laps }=\frac{\text { total time spent roller blading }}{\text { time taken to complete each lap }}$

Plugging in values, we get

$\text { Total number of laps }=\frac{1 \text { hour }}{4 \text { mins }}$

We know that 1 hour = 60 minutes

$=\frac{60 \mathrm{mins}}{4 \mathrm{mins}}=15$

Hence, the total number of laps taken by Janice and Thomas in 1 hour is 15

You might be interested in

Find the coordinate of K' after a reflection of the triangle about the x-axis. Write your answer in the form (a,b)

Elodia [21]

Answer:

K = (3,3)

After a reflection across the x axis (Vertical reflection), the coordinate becomes (3, -3)

Let me know if this helps!

5 0

3 years ago

Read 2 more answers

1 +2 +3 + ... + 620=

Lina20 [59]

Answer:626

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Which shows the lengths from greatest to least?

astraxan [27]

A because the length is greater

5 0

3 years ago

If sinx = p and cosx = 4, work out the following forms :<br><br><br>

Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16}$

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

$\boxed{\sin(-x)=-\sin(x)}$

$\boxed{\cos(-x)=\cos(x)}$

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

$\sin(x)=p$

$\cos(x)=4$

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]$

Using the sum and difference identities:

$\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}$

$\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}$

$\sin(\pi -x) = \sin(x)$

$\sin(2\pi +x)=\sin(x)$

$\cos(2\pi-x)=\cos(x)$

Therefore,

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]$

$\implies [p+4] \cdot [p \cdot 4]=4p^2+16p$

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16}$

8 0

3 years ago

Factorise completely:<br> (a) 8m - 50 n2

Sever21 [200]

Answer:

2(4m-25n2)

Step-by-step explanation:

To factorise,we will have to know the value that can divide both variables given

But based on this question,we have different variables that is m and n

So we don't have to deal with that but with the numbers which means that we will only deal with the numbers ignoring the letters

8m-50n2

So the only number that can divide the two is 2

2(4m-25n2)

This cannot be solved further so the final answer is 2(4m-25n2)

I hope this will help you

7 0

4 years ago