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MAXImum [283]
2 years ago
9

Take a square sheet of paper 20 cm by 20 cm, cut identical squares from each corner, and fold up the flaps to make a box (withou

t a lid). What is the volume of the box? What different volumes can you make by varying the size of the squares you cut of?
Task: what is the maximum possible volume of this type of box that can be made from a 20 cm by 20 cm square of paper?
Mathematics
1 answer:
iris [78.8K]2 years ago
3 0

The maximum possible volume of this type of box that can be made from a 20 cm by 20 cm square of paper is 512cm³.

<h3>How to calculate the volume?</h3>

From the information, when the side of 1cm are cut, this will become an open box. The volume will be:

= 18 × 18 × 1

= 324cm³

Also, when 2cm is cut, the volume will be:

= 16 × 16 × 2

= 512cm³

Learn more about volume on:

brainly.com/question/1972490

#SPJ1

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Emma borrowed $300 to repair her car. the finance (interest) charge on the loan was $20, and the term on the loan was 14 days. w
Murrr4er [49]
Given:
loan amount = 300
finance charge = 20
term = 14 days.

To solve for APR.
<span>1. Divide the finance charge by the loan amount.
20/300 = 0.0667

2. Multiply the result by 365. 
0.0667 x 365 = 24.35

3. Divide the result by the term of the loan.
24.35/14 = 1.74 (APR in decimal format)
<span>
4. Multiply the result by 100.
1.74 x 100 = 174% APR</span></span>
7 0
3 years ago
How do I find this, what’s the formula to find the 20th term and can anyone explain the formula? Will award brainliest
harina [27]

Answer:

  • f(n) = n^2 +1
  • f(20) = 401

Step-by-step explanation:

The first differences of the sequence are ...

  • 5-2 = 3
  • 10-5 = 5
  • 17-10 = 7
  • 26-17 = 9
  • 37-26 = 11

Second differences are ...

  • 5 -3 = 2
  • 7 -5 = 2
  • 9 -7 = 2
  • 11 -9 = 2

The second differences are constant, so the sequence can be described by a second-degree polynomial.

We can write and solve three equations for the coefficients of the polynomial. Let's define the polynomial for the sequence as ...

  f(n) = an^2 + bn + c

Then the first three terms of the sequence are ...

  • f(1) = 2 = a·1^2 + b·1 + c
  • f(2) = 5 = a·2^2 +b·2 + c
  • f(3) = 10 = a·3^2 +b·3 +c

Subtracting the first equation from the other two gives ...

  3a +b = 3

  8a +2b = 8

Subtracting the first of these from half the second gives ...

  (4a +b) -(3a +b) = (4) -(3)

  a = 1 . . . . . simplify

Substituting into the first of the 2-term equations, we get ...

  3·1 +b = 3

  b = 0

And substituting the values for a and b into the equation for f(1), we have ...

  1·1 + 0 + c = 2

  c = 1

So, the formula for the sequence is ...

  f(n) = n^2 + 1

__

The 20th term is f(20):

  f(20) = 20^2 +1 = 401

_____

<em>Comment on the solution</em>

It looks like this matches the solution of the "worked example" on your problem page.

4 0
3 years ago
????????????????????
Shtirlitz [24]
It’s a diagram ur welcome
3 0
3 years ago
High of the following transformations does not result in a congruent figure
LuckyWell [14K]
Any transformation involving change of scale will not result in a congruent figure. Rotations and reflections and translations maintain congruence.
7 0
3 years ago
jon used a gift card to pay for an $18 shirt.The remaining balance on the card is $28.write and solve an equation that represent
denis-greek [22]

Take note that they spent $18 afterwards having $28 on the card so they're spending money off the card so...

x-18=28

would be the equation for this and to solve do the following.

First add 18 to both sides making the equation.

x=46

7 0
3 years ago
Read 2 more answers
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