1/4 cups = 1lb mince divide both by three
1/12 cups = 1/3lb hamburgers
1 1/3 cups = 12/12 cups + 4/12 cups
= 16/12 cups
=16 * 1/3lb hamburgers
- as 1/12 cups = 1/3lb burger, 16/12 cups = 16/3 lb burgers
897+450+328= 1675
(Not sure if this is a joke?)
Answer:
1 i think
Step-by-step explanation:
hope this helps :) have a nice day :) :)

- Given - <u>A </u><u>trapezium</u><u> </u><u>ABCD </u><u>with </u><u>non </u><u>parallel </u><u>sides </u><u>of </u><u>measure </u><u>1</u><u>5</u><u> </u><u>cm </u><u>each </u><u>!</u><u> </u><u>along </u><u>,</u><u> </u><u>the </u><u>parallel </u><u>sides </u><u>are </u><u>of </u><u>measure </u><u>1</u><u>3</u><u> </u><u>cm </u><u>and </u><u>2</u><u>5</u><u> </u><u>cm</u>
- To find - <u>Area </u><u>of </u><u>trapezium</u>
Refer the figure attached ~
In the given figure ,
AB = 25 cm
BC = AD = 15 cm
CD = 13 cm
<u>Construction</u><u> </u><u>-</u>

Now , we can clearly see that AECD is a parallelogram !
AE = CD = 13 cm
Now ,

Now , In ∆ BCE ,

Now , by Heron's formula

Also ,

<u>Since </u><u>we've </u><u>obtained </u><u>the </u><u>height </u><u>now </u><u>,</u><u> </u><u>we </u><u>can </u><u>easily </u><u>find </u><u>out </u><u>the </u><u>area </u><u>of </u><u>trapezium </u><u>!</u>

hope helpful :D
Distance between points

and

is

Distance from H to B:
![[tex]d=\sqrt{(10-(-3))^2+(1-(-9))^2}=\sqrt{169+100}=\sqrt{269}](https://tex.z-dn.net/?f=%5Btex%5Dd%3D%5Csqrt%7B%2810-%28-3%29%29%5E2%2B%281-%28-9%29%29%5E2%7D%3D%5Csqrt%7B169%2B100%7D%3D%5Csqrt%7B269%7D)
d=\sqrt{(1-(-3))^2+(10-(-9))^2}=\sqrt{16+361}=\sqrt{376}[/tex] units.
Distance from Z to B:

units.
Horse Z is closer to the barn.
(The conversion to meters is not required; the question does not ask for actual distances, so "units" is OK.)