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3 years ago

mark wants to order a pizza. which is the better deal 10.00 with a diameter of 12 or 20.00 with a diameter of 18

1 answer:

7 0

I'm gonna calculate by the area of pizza to see how much food you get for 10 and 20 dollars

A...area
r...radius
d...diameter

d1=12 r1=6
d2=18 r2=9

A1=r1^2×3.14
A1=6^2×3.14
A1=36×3.14
A1=113.04

A2=r2^2×3.13
A2=9^2×3.14
A2=81×3.14
A2=254.34

if you bought 2, 12 diameter pizzas for 20 dollars you would get 226.08 area od food,still less if you bought 1, 18 diameter pizza for 20 dollars. because 223.08 < 254.34

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tigry1 [53]

Answer:

46.66 or if rounded 46.7

Step-by-step explanation:

First you want to do 28x5 that would get you 140

Then, you divide 140 by 3. That would get you ur answer, Hope this helped!! :)

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2 years ago

Raymond is making stir-fry rice for dinner. The recipe says to add 3/4 tablespoon of ground ginger, but he only has a 1/8 tables

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Answer:

Step-by-step explanation:

3/4 = 6/8.

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2 years ago

PLEASE HELP ASAP

maria [59]

Answer:

First option: $3x^2-5x=-8$

Second option: $2x^2=6x-5$

Fourth option: $-x^2-10x=34$

Step-by-step explanation:

Rewrite each equation in the form $ax^2+bx+c=0$ and then use the Discriminant formula for each equation. This is:

$D=b^2-4ac$

1) For $3x^2-5x=-8$ :

$3x^2-5x+8=0$

Then:

$D=(-5)^2-4(3)(8)=-71$

Since $D$ this equation has no real solutions, but has two complex solutions.

2) For $2x^2=6x-5$ :

$2x^2-6x+5=0$

Then:

$D=(-6)^2-4(2)(5)=-4$

Since $D$ this equation has no real solutions, but has two complex solutions.

3) For $12x=9x^2+4$ :

$9x^2-12x+4=0$

Then:

$D=(-12)^2-4(9)(4)=0$

Since $D=0$ this equation has one real solution.

4) For $-x^2-10x=34$ :

$-x^2-10x-34=0$

Then:

$D=(-10)^2-4(-1)(-34)=-36$

Since $D$ , this equation has no real solutions, but has two complex solutions.

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3 years ago

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lutik1710 [3]

Answer:

todavía no me la sé pero al rato te mando la respuesta

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3 years ago

There are infinitely How many pairs of numbers of which the sum of their cube roots is zero give two of these pairs

ivanzaharov [21]

Answer:

Infinite pairs of numbers

1 and -1

8 and -8

Step-by-step explanation:

Let x³ and y³ be any two real numbers. If the sum of their cube roots is zero, then the following must be true:

$\sqrt[3]{x^3}+ \sqrt[3]{y^3}=0\\ \sqrt[3]{x^3}=- \sqrt[3]{y^3}\\x=-y$

Therefore, any pair of numbers with same absolute value but different signs fit the description, which means that there are infinite pairs of possible numbers.

Examples: 1 and -1; 8 and -8; 27 and -27.

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3 years ago