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DerKrebs [107]
3 years ago
9

a large pizza is advertised to have 14 inch diameter. If a costumer orders two large pizzas, how many squares inches of pizza wi

ll he receive?
Mathematics
2 answers:
kari74 [83]3 years ago
5 0
Radius ^2 * pi = 1 pizza * 2 = area of 2 pizzas = 307.876 in ^2
kirza4 [7]3 years ago
5 0
In order to find the total square inches of the pizza we just need to find the area of one of the pizzas then multiply that by 2.

A pizza is a circle so we can use the Area of a Circle formula which is:
A = πr^2

A: Area
π: pi (3.14......)
r: radius (half of the diameter)

The diameter of the pizza is 14 inches so the radius is 14/2 = 7 inches.

Now we plug the radius into the equation.

A = π(7)^2
Simplify
7^2 = 49
A = π(49)
This means the area of one of the pizzas is 49π square inches.

To find total square inches for both pizza, multiply 49π by 2.

49π x 2 = 98π 

This means the customer will receive 98π square inches of pizza.
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Use the integers that are closest to the number in the middle.
IrinaK [193]

Answer:

14<√215<15

Step-by-step explanation:

√215=14.6628

integer is a number doesn't contain decimal places, it would be 14 and 15

5 0
3 years ago
In how many ways can we seat 3 pairs of siblings in a row of 7 chairs, so that nobody sits next to their sibling
monitta

Answer:

1,968

Step-by-step explanation:

Let x₁ and x₂, y₁ and y₂, and z₁ and z₂ represent the 3 pairs of siblings, and let;

Set X represent the set where the siblings x₁ and x₂ sit together

Set Y represent the set where the siblings y₁ and y₂ sit together

Set Z represent the set where the siblings z₁ and z₂ sit together

We have;

Where the three siblings don't sit together given as X^c∩Y^c∩Z^c

By set theory, we have;

\left | X^c \cap Y^c \cap Z^c  \right | = \left | X^c \cup Y^c \cup Z^c  \right | =  \left | U  \right | - \left | X \cup Y \cup Z  \right |

\left | U  \right | - \left | X \cup Y \cup Z  \right | = \left | U  \right | - \left (\left | X \right | +  \left | Y\right | +  \left | Z\right | -  \left | X \cap Y\right | -  \left | X \cap Z\right | -  \left | Y\cap Z\right | +  \left | X \cap Y \cap Z\right | \right)

Therefore;

\left | X^c \cap Y^c \cap Z^c  \right | = \left | U  \right | - \left (\left | X \right | +  \left | Y\right | +  \left | Z\right | -  \left | X \cap Y\right | -  \left | X \cap Z\right | -  \left | Y\cap Z\right | +  \left | X \cap Y \cap Z\right | \right)

Where;

\left | U\right | = The number of ways the 3 pairs of siblings can sit on the 7 chairs = 7!

\left | X\right | = The number of ways x₁ and x₂ can sit together on the 7 chairs = 2 × 6!

\left | Y\right | = The number of ways y₁ and y₂ can sit together on the 7 chairs = 2 × 6!

\left | Z\right | = The number of ways z₁ and z₂ can sit together on the 7 chairs = 2 × 6!

\left | X \cap Y\right | = The number of ways x₁ and x₂ and y₁ and y₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | X \cap Z\right | = The number of ways x₁ and x₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | Y \cap Z\right | = The number of ways y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | X \cap Y \cap Z\right | = The number of ways x₁ and x₂,  y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 2 × 4!

Therefore, we get;

\left | X^c \cap Y^c \cap Z^c  \right | = 7! - (2×6! + 2×6! + 2×6! - 2 × 2 × 5! - 2 × 2 × 5! - 2 × 2 × 5! + 2 × 2 × 2 × 4!)

\left | X^c \cap Y^c \cap Z^c  \right | = 5,040 - 3072 = 1,968

The number of ways where the three siblings don't sit together given as \left | X^c \cap Y^c \cap Z^c  \right |  = 1,968

5 0
2 years ago
If you wanted to mix pure methane with water and end up with 70 gallons of 20% methane, how many gallons of each should you use?
PtichkaEL [24]

Answer:

You want 40% of 140 gallons to be methane,

so 60% of 140 gallons is water

methane:  0.4(140) = 56 gallons

water:  0.6(140)  = 84 gallons

Step-by-step explanation:

3 0
2 years ago
Which set of shapes could be used to form a net for a square tvyranni?
Bumek [7]
Okay i’m pretty sure it’s the first one, 1 square and 4 triangles.
8 0
3 years ago
What is the answer to this solution. -7r−4≥ 4r+2
tia_tia [17]

Answer:

The solution is:

-7r-4\ge \:\:4r+2\quad \::\quad \:\begin{bmatrix}\mathrm{Solution:}\:&\:r\le \:\:-\frac{6}{11}\:\\ \:\:\mathrm{Decimal:}&\:r\le \:\:-0.54545\dots \:\\ \:\:\mathrm{Interval\:Notation:}&\:(-\infty \:\:,\:-\frac{6}{11}]\end{bmatrix}

Please check the attached line graph below.

Step-by-step explanation:

Given the expression

-7r-4\ge \:4r+2

Add 4 to both sides

-7r-4+4\ge \:4r+2+4

Simplify

-7r\ge \:4r+6

Subtract 4r from both sides

-7r-4r\ge \:4r+6-4r

Simplify

-11r\ge \:6

Multiply both sides by -1 (reverses the inequality)

\left(-11r\right)\left(-1\right)\le \:6\left(-1\right)

Simplify

11r\le \:-6

Divide both sides by 11

\frac{11r}{11}\le \frac{-6}{11}

Simplify

r\le \:-\frac{6}{11}

Therefore, the solution is:

-7r-4\ge \:\:4r+2\quad \::\quad \:\begin{bmatrix}\mathrm{Solution:}\:&\:r\le \:\:-\frac{6}{11}\:\\ \:\:\mathrm{Decimal:}&\:r\le \:\:-0.54545\dots \:\\ \:\:\mathrm{Interval\:Notation:}&\:(-\infty \:\:,\:-\frac{6}{11}]\end{bmatrix}

Please check the attached line graph below.

7 0
2 years ago
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