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2 years ago

If a jogger runs 2 miles and burned 185 calories, how many calories when he burned jogging 3 miles?

Mathematics

2 answers:

Tom [10]2 years ago

3 0

Answer:

The jogger will burn 277.5 calories

jok3333 [9.3K]2 years ago

3 0

Answer:

277.5 calories

Step-by-step explanation:

185/2 = 92.5 calories per mile

3 x 92.5 = 277.5 calories when he runs 3 miles.

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At what position does the ball have its maximum potential energy

Anuta_ua [19.1K]

Potential energy is the energy possessed by an object because of its position.
So it is D the highest point of the trajectory.

7 0

2 years ago

Read 2 more answers

Jordan earns $15.35 per hour. on Tuesday , Jordan clocked in at 7:45 Am then clocked out for a break at 12:15 am and working unt

Vaselesa [24]

12:45Answer:

Part A: 9 hours 45 minutes

Part B: $126.1575

Step-by-step explanation:

Part A,

From 7:45 to 12:15 is 4 hours and 30 minutes

From 12:45 which he clocked back in to 6:00 is 5 hours and 15 minutes.

4 hours and 30 minutes plus 5 hours and 15 minutes = 9 hours and 45 minutes.

Part B, 9 hours and 45 minutes can be represented as 9.45

And since he gets 15.35 "per" hour shows we can multiply.

15.35 x 9.45 = 126.1575

Therefor he earned 126.15.

8 0

3 years ago

The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of th

inna [77]

Answer:

Length = 18.099 ft

Width = 11.049 ft

Step-by-step explanation:

let the length of the field be x ft

and the width be y ft

as per the condition given in problem

x=2y-4 -----------(A)

Also the area is given as 200 sqft

Hence

xy=200

Hence from A we get

y(2y-4)=200

taking 2 as GCF out

2y(y-2)=200

Dividing both sides by 2 we get

$y(y-2)=100$

$y^2-2y=100$

subtracting 100 from both sides

$y^2-2y-100=0$

Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form

$ax^2+bx+c=0$

Here in our case

a=1

b=-1

c=-100

Putting those values in the formula and solving them for y

$y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

Solving first

$y=\frac{2+\sqrt{4+400}{2}$

$y=\frac{2+\sqrt{404}{2}$

$y=\frac{2+20.099}{2}$

$y=\frac{22.099}{2}$

$y=11.049$

Solving second one

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{2-\sqrt{4+400}{2}$

$y=\frac{2-\sqrt{404}{2}$

$y=\frac{2-20.099}{2}$

$y=\frac{-18.99}{2}$

$y=-9.045$

Which is wrong as the width can not be in negative

Our width of the field is

y=11.099

Hence the length will be

x=2y-4

x=2(11.049)-4

x=22.099-4

x=18.099

Hence our length x and width y :

Length = 18.099 ft

Width = 11.049 ft

4 0

2 years ago

Ben jogged 4 and 1/2 miles in 1 and 1/4 hours at a constant rate. How far did Ben jog in one hour? *

Mrac [35]

let's firstly convert the mixed fractions to improper fractions and then proceed.

$\stackrel{mixed}{4\frac{1}{2}}\implies \cfrac{4\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{9}{2}}~\hfill \stackrel{mixed}{1\frac{1}{4}}\implies \cfrac{1\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{5}{4}} \\\\[-0.35em] ~\dotfill$

$\begin{array}{ccll} miles&hours\\ \cline{1-2} \frac{9}{2}&\frac{5}{4}\\[1em] x&1 \end{array}\implies \cfrac{~~ \frac{9}{2}~~}{x}=\cfrac{~~ \frac{5}{4}~~}{1}\implies \cfrac{~~ \frac{9}{2}~~}{\frac{x}{1}}=\cfrac{5}{4}\implies \cfrac{9}{2}\cdot \cfrac{1}{x}=\cfrac{5}{4} \\\\\\ \cfrac{9}{2x}=\cfrac{5}{4}\implies 36=10x\implies \cfrac{36}{10}=x\implies \cfrac{18}{5}=x\implies 3\frac{3}{5}=x$

7 0

2 years ago

If two men can paint two rooms in two hours, how long will it take for one man to paint one room

Mrac [35]

Each room would be 1 hour because it say that two rooms get painted in two hours

6 0

3 years ago