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3 years ago

Tim can mow 4 lawns in 6 hours. How many lawns can be mow in 15 hours?

Mathematics

2 answers:

Illusion [34]3 years ago

8 0

The answer is 10 lawns in 15 hours

xxTIMURxx [149]3 years ago

3 0

Its 4 lawns per 6 hours so to make it easier divide both sides by 2, then 2 lawns per 3 hours. So if he lawn 15 hours then 15/3(2) is 5(2) then he can mow 10 hours per 15 hours

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kayla took $36.75 to the state fair. each ticket in the fair cost x dollars. kayla bought 3 tickets. write an expression that re

Gennadij [26K]

F(x)=y=the amount of money, in dollars , that Kad had after she bought the tickets.
x=price of one ticket

f(x)=-3x+36.75 or y=-3x+36.75

7 0

3 years ago

Rational numbers with distribute property

Svetradugi [14.3K]

I'm not sure what this means. If you have choices you should list them.

(1/2)*(1/4 + 1/6) is an example of what should be given. There are two ways to solve this.

1. Use the distributive property.

1/2*1/4 + 1/2* 1/6

1/8 + 1/12 Which can be added using the LCD of 24

3/24 + 2/24 = 5/24

Method 2

Add what is inside the brackets first.

1/2 ( 1/4 + 1/6)

1/2(3/12 + 2/12 = 5/12

Now multiply by 1/2

1/2(5/12) = 5*1/(12 * 2) = 5 / 24 Same answer.

4 0

3 years ago

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago

How do you do number 16?

MA_775_DIABLO [31]

The answer to this would be 1/5.
hope this helps

7 0

3 years ago

At 8:30 A.M.., Briana started filling a 2,800-gallon pond. At 10:30 A.M., she had filled 1,400 gallons. At what time will the po

Sati [7]

Answer:

12:30 pm

Step-by-step explanation:

8:30 am pond is empty

10:30 am pound is 1,400 gallons full

time? pound is 2,800 gallons full

from 8:30 to 10:30 are 2 hours so the rate of filling up is

1,400 gallons in 2 hours

to have a whole pond filled we need another 2 hours because

in another 2 hours will have another 1,400 gallons so a pound full. 8:30 am + 2hours = 12:30 pm

8:30 am pond is empty

10:30 am pound is 1,400 gallons full

12:30 pm pound is 2,800 gallons full.

3 0

3 years ago