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defon
3 years ago
8

Lin has a small baking pans shaped like a rectangular prism.

Mathematics
2 answers:
Rzqust [24]3 years ago
7 0

Answer:

2 1/2

Step-by-step explanation:

( 11 1/4 ) / ( 4 1/2 ) = 2 1/2

( 2 1/2 ) = 5/2

Hope this helps!

lara31 [8.8K]3 years ago
6 0

Answer:

15 in^3.

Step-by-step explanation:

First convert the mixed numbers to improper fractions:

2 2/3 = 8/3, 3 3/4 = 15/4 and 1 1/2 =  3/2.

Volume = 8/3 * 15/4 * 3/2

= 360/24

= 15 in^3.

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A jewellery shop sells 240 necklaces in a month. 18 of the necklaces were sold via the shop's website, the rest were sold in hig
ICE Princess25 [194]

Answer: 3 : 37

Step-by-step explanation:

The shop sold 240 necklaces in a month.

Out of these, 18 were sold online.

The number that was sold offline was therefore:

= 240 - 18

= 222 shop sales

The ratio of  online sales to shop sales is therefore:

18 : 222

Take it to simplest form by dividing both sides by their highest common factor of 6:

3 : 37

5 0
3 years ago
Find the dimensions of the rectangle with largest area that can be inscribed in an equilateral triangle with sides of 1 unit, if
prohojiy [21]
<span>Maximum area = sqrt(3)/8 Let's first express the width of the triangle as a function of it's height. If you draw an equilateral triangle, then a rectangle using one of the triangles edges as the base, you'll see that there's 4 regions created. They are the rectangle, a smaller equilateral triangle above the rectangle, and 2 right triangles with one leg being the height of the rectangle and the other 2 angles being 30 and 60 degrees. Let's call the short leg of that triangle b. And that makes the width of the rectangle equal to 1 minus twice b. So we have w = 1 - 2b b = h/sqrt(3) So w = 1 - 2*h/sqrt(3) The area of the rectangle is A = hw A = h(1 - 2*h/sqrt(3)) A = h*1 - h*2*h/sqrt(3) A = h - 2h^2/sqrt(3) We now have a quadratic equation where A = -2/sqrt(3), b = 1, and c=0. We can solve the problem by using a bit of calculus and calculating the first derivative, then solving for 0. But since this is a simple quadratic, we could also take advantage that a parabola is symmetrical and that the maximum value will be the midpoint between it's roots. So let's use the quadratic formula and solve it that way. The 2 roots are 0, and 1.5/sqrt(3). The midpoint is (0 + 1.5/sqrt(3))/2 = 1.5/sqrt(3) / 2 = 0.75/sqrt(3) So the desired height is 0.75/sqrt(3). Now let's calculate the width: w = 1 - 2*h/sqrt(3) w = 1 - 2* 0.75/sqrt(3) /sqrt(3) w = 1 - 2* 0.75/3 w = 1 - 1.5/3 w = 1 - 0.5 w = 0.5 The area is A = hw A = 0.75/sqrt(3) * 0.5 A = 0.375/sqrt(3) Now as I said earlier, we could use the first derivative. Let's do that as well and see what happens. A = h - 2h^2/sqrt(3) A' = 1h^0 - 4h/sqrt(3) A' = 1 - 4h/sqrt(3) Now solve for 0. A' = 1 - 4h/sqrt(3) 0 = 1 - 4h/sqrt(3) 4h/sqrt(3) = 1 4h = sqrt(3) h = sqrt(3)/4 w = 1 - 2*(sqrt(3)/4)/sqrt(3) w = 1 - 2/4 w = 1 -1/2 w = 1/2 A = wh A = 1/2 * sqrt(3)/4 A = sqrt(3)/8 And the other method got us 0.375/sqrt(3). Are they the same? Let's see. 0.375/sqrt(3) Multiply top and bottom by sqrt(3) 0.375*sqrt(3)/3 Multiply top and bottom by 8 3*sqrt(3)/24 Divide top and bottom by 3 sqrt(3)/8 Yep, they're the same. And since sqrt(3)/8 looks so much nicer than 0.375/sqrt(3), let's use that as the answer.</span>
7 0
3 years ago
Read 2 more answers
What is this? <br><br><br><br>What would be an equation for the graph?​
Travka [436]
Y=3000x+44000
You find slope by choosing two points and subtracting them.
Also try using Desmond
6 0
3 years ago
Cally made a rectangular sandbox. It is 8 feet long and 10 feet
Mnenie [13.5K]

Answer:

160 pounds of sand.

Step-by-step explanation:

7 0
3 years ago
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Find the volume. Help please it’s due tomorrow
Dmitry_Shevchenko [17]

Answer:

2cm

Step-by-step explanation:

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