54 = 2(4x-3) + 2(x)
54 = 8x-6 + 2x
60 = 10x
6 = x
so two sides are 6 meters, and the other 2 sides are 21 meters
Answer:
-50
Step-by-step explanation:
If the drink is descending by 50 feet, hypothetically I would say that the drinks at 0 ft. so it would be -50
Answer:
the volume of sand is ![50 ft^{3}](https://tex.z-dn.net/?f=50%20ft%5E%7B3%7D)
Step-by-step explanation:
This problem bothers on the mensuration of solid shapes, rectangular prism.
Given data
![Length l= 4 ft Width w= 5ft\\Height h = 3ft](https://tex.z-dn.net/?f=Length%20%20l%3D%204%20ft%20Width%20w%3D%205ft%5C%5CHeight%20h%20%3D%203ft)
if the sandbox is filled with sand of
deep the volume of the sand in the box will be calculated based on the the depth of the sand
converting the depth of sand from mixed fraction to proper fraction we have ![2\frac{1}{2} = \frac{5}{2}](https://tex.z-dn.net/?f=2%5Cfrac%7B1%7D%7B2%7D%20%3D%20%5Cfrac%7B5%7D%7B2%7D)
the expression for the volume of a rectangular prism is
![volume of sand = length * width* height](https://tex.z-dn.net/?f=volume%20of%20sand%20%3D%20length%20%2A%20width%2A%20height)
substituting our data into the expression we have
![volume of sand= 5*4 * \frac{5}{2} \\volume of sand = \frac{100}{2} \\volume of sand = 50 ft^{3}](https://tex.z-dn.net/?f=volume%20of%20sand%3D%205%2A4%20%2A%20%5Cfrac%7B5%7D%7B2%7D%20%5C%5Cvolume%20of%20sand%20%3D%20%5Cfrac%7B100%7D%7B2%7D%20%5C%5Cvolume%20of%20sand%20%3D%2050%20ft%5E%7B3%7D)
It helps you know what to multiply and the product when you subtracting and adding fractions.
<h3>
<u>Answer:</u></h3>
![\boxed{\boxed{\pink{\sf \leadsto Hence \ the \ area \ of \ shaded \ region \ is 264 cm^2}}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7B%5Cpink%7B%5Csf%20%5Cleadsto%20Hence%20%5C%20the%20%5C%20area%20%5C%20of%20%5C%20shaded%20%5C%20region%20%5C%20is%20264%20cm%5E2%7D%7D%7D)
<h3>
<u>Step-by-step explanation:</u></h3>
Here , two circles are given which are concentric. The radius of larger circle is 10cm and that of smaller circle is 4cm . And we need to find thelarea of shaded region.
From the figure it's clear that the area of shaded region will be the difference of areas of two circles.
Let the,
- Radius of smaller circle be r .
- Radius of smaller circle be r .
- Area of shaded region be
<h3>
<u>Hence </u><u>the</u><u> </u><u>area</u><u> </u><u>of</u><u> the</u><u> </u><u>shaded </u><u>region</u><u> is</u><u> </u><u>2</u><u>6</u><u>4</u><u> </u><u>cm²</u><u>.</u></h3>