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Keith_Richards [23]
1 year ago
5

Deon has a pizza with a diameter

Mathematics
1 answer:
OverLord2011 [107]1 year ago
7 0

The square box is enough to fit the pizza with a diameter of 10 inches inside. Since the area of the square box is more than the area of the pizza, the pizza fits easily in the square box.

<h3>What is the area of the circle and the square?</h3>

The area of the circle is

Ac = πr² = πd²/4 sq. units

Where r is the radius and d is the diameter of the circle.

The area of the square is given by

As = s² sq. units

Where s is the length of the side of a square.

<h3>Calculation:</h3>

It is given that a pizza(in a circular shape) with a diameter d = 10 in is to be placed in a square box of the same length as the diameter of the pizza.

So,

The area of pizza is

Ap = Ac = πd²/4 sq. units

     = π(10)²/4

     = 25π

     = 78.54 sq. in

Then, the area of the square box with the length same as the diameter of the pizza is,

As = d²

    = 10²

    = 100 sq. in

Since the area of the square is more than the area of the pizza (100 sq. inch > 78.54 sq. inch), the pizza easily fits into the square box.

Learn more about the area of a circle here:

brainly.com/question/15673093

#SPJ1

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Step-by-step explanation:

This one is simple substitution... at least, substitution is the easiest method. The first equation is 3<em>x</em> – 30 = <em>y</em>  and the second is 7<em>y</em> – 6 = 3<em>x</em>

As I look, I see 3 ways to use substitution to solve this:

  1. substitute 7<em>y</em> – 6 for 3<em>x</em>
  2. substitute 3<em>x</em> – 30 for <em>y</em>
  3. solve 3<em>x</em> – 30 = <em>y</em> for 3<em>x</em> and make it equal to 7<em>y</em> – 6

We're going to only use 1 method for the sake of time. Try the other two on your own. Assuming you don't make any mistakes, they will work.

<u>Method 1</u>:

3<em>x</em> – 30 = <em>y</em>

7<em>y</em> – 6 = 3<em>x</em>  — initial system of equations

7<em>y</em> – 6 – 30 = <em>y</em>  — substitute 7<em>y</em> – 6 for 3<em>x</em>

<u>7</u><u><em>y</em></u> – 6 – 30 = <u><em>y</em></u>  — marking like terms, bold for constants, <u>underlined</u> for variables

7<em>y</em> – 36 = <em>y</em>  — combining the constants and simplifying

Here, you could diverge into multiple paths: add 36 to both sides, subtract <em>y</em> from both sides, divide by 6 OR subtract 7<em>y</em> from both sides and divide by –6 . For the sake of time, I'm subtracting 7<em>y</em>, though I don't like dealing with negatives.

7<em>y</em> – 7<em>y</em> – 36 = <em>y</em> – 7<em>y</em>  — subtract 7<em>y</em> from both sides

–36 = –6<em>y</em>  — simplify

–36 ÷ –6 = –6<em>y</em> ÷ –6  — divide by –6 on both sides

<em>y</em> = 6  — simplify

Again, we can diverge here: substitute <em>y</em> into 3<em>x</em> – 30 = <em>y</em> or substitute <em>y</em> into 7<em>y</em> – 6 = 3<em>x</em>

I'm going to choose 3<em>x</em> – 30 = <em>y</em> but it will work either way, should you take the time (if you have it) to chase down every path this problem can take.

3<em>x</em> – 30 = <em>y</em>  — initial equation

3<em>x</em> – 30 = 6  — substitute 6 for <em>y</em>

3<em>x</em> – 30 + 30 = 6 + 30  — add 30 to both sides to isolate 3<em>x</em>

3<em>x</em> = 36  — simplify the expression

3<em>x</em> ÷ 3 = 36 ÷ 3  — divide both sides by 3 to isolate <em>x</em>

<em>x</em> = 12  — simplify

So, we have <em>x</em> = 12 and <em>y</em> = 6 . We know they work for 3<em>x</em> – 30 = <em>y</em>  but not if they work for 7<em>y</em> – 6 = 3<em>x</em> . Let's substitute those in to see if (12, 6) really is the solution point.

7<em>y</em> – 6 = 3<em>x</em>  — original equation

7(6) – 6 ≟ 3(12)  — substitute 6 for <em>y</em> and 12 for <em>x</em>

42 – 6 ≟ 36  — simplify by multiplying

36 = 36 ✔  — simplify by combining like terms on left side

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