Ask question

Ask question

All categories

3 years ago

6

sarah is training for a bike race. she rides her bike 5 3/4 miles in 1/3 hour. what is sarahs's rate in miles per hour?

2 answers:

Helga [31]3 years ago

8 0

Hey there!

Since it is 1/3 of an hour for 5 3/4 miles, multiply 5 3/4 x 3 . . .

5 3/4 = 23/4

3 = 3/1

23 x 3 = 69
4 x 1 = 4

Simplifying 69/4 . . .

69/4 = 17 1/4 <-----------

Sarah's rate is 17 1/4 miles per hour.

Hope this helps you.
Have a great day!

dolphi86 [110]3 years ago

7 0

Sarah covers 17 1/4 miles per hour.

Hope this helps!

You might be interested in

Find the diameter of a circle whose circumference is 74cm

jenyasd209 [6]

<em>Answer:</em>

<em>≈ 24 cm</em>

<em>Step-by-step explanation:</em>

<em>Hi there !</em>

<em>c = 2πr } => c = πd => d = c/π</em>

<em>2r = d</em>

<em>replace c ; π</em>

<em>π =3.14</em>

<em>d = 74cm/3.14</em>

<em>= 23.5668</em>

<em>≈ 24 cm</em>

<em>Good luck !</em>

4 0

3 years ago

4(0.5f−0.25)=6+f what is f<br><br><br> I am so lost when we were doing this in class

muminat

4 (0.5f - 0.25) = 6 + f
2.0f - 1.0 = 6.0 + f
1.00f = 6.0 + f
f = -5.0
hope this is right,
~ Harley Quinn~

8 0

3 years ago

Read 2 more answers

I need help i'm really still confused on the Tan thing in algebra

AysviL [449]

Answer:6/8

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the soultion(s) to the system of equations. Select all that apply

neonofarm [45]

Answer:

(0,-3)

(3,0)

Step-by-step explanation:

The solutions to the system of equations are where the two graphs cross

The first is at x=0 and y=-3

The second is at x=3 and y=0

3 0

3 years ago

A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula

Ilia_Sergeevich [38]

Answer: A. $A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}$

B. A'(5) = 1.76 cm/s

Step-by-step explanation: <u>Rate</u> <u>of</u> <u>change</u> measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

$A=\pi.r^{2}$

So to find the rate of the area:

$\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}$

$\frac{dA}{dr} =2.\pi.r$

Using $r(t)=3-\frac{363}{(t+11)^{2}}$

$\frac{dr}{dt}=\frac{726}{(t+11)^{3}}$

Then

$\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]$

$\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}$

Multipying and simplifying:

$\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$

The rate at which the area is increasing is given by expression $A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$ .

B. At t = 5, rate is:

$A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}$

$A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}$

$A'(5)=\frac{2408693760\pi}{4294967296}$

$A'(5)=1.76268$

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

5 0

2 years ago