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2 years ago

Find the area of the circle. Round your answer to the nearest tenth.

Mathematics

1 answer:

kotegsom [21]2 years ago

8 0

Step-by-step explanation:

area of a circle

A = pi x r^2

r=9

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Jimmy wants to serve iced tea at his birthday party. He purchases 14 containers filled with 1.5 liters of tea. Approximately how

jenyasd209 [6]

Answer:

5.5 gallons

Step-by-step explanation:

set up a proportion; Liters in numerator and Gallons in denominator

let x = approx number of gallons

x 1

3.8x = 21

x = 21 ÷ 3.8

x ≈ 5.5

4 0

2 years ago

Solve the inequality for w.

ahrayia [7]

Answer

Example 1 Solve for x and check: x + 5 = 3

Solution

Using the same procedures learned in chapter 2, we subtract 5 from each side of the equation obtaining

Example 2 Solve for x and check: - 3x = 12

Solution

Dividing each side by -3, we obtain

Step-by-step explanation: here is your anserw the let me know if it right

8 0

2 years ago

Eli lives 0.7 mile from his school. He lives 0.25 mile from his friends house. Does Eli live closer to his school or his friends

Hitman42 [59]

Eli lives closer to his friends house than his school

5 0

2 years ago

How do I find out a mortgage payment on a home

kozerog [31]

Answer:

Step-by-step explanation:by finding how much it costs

4 0

2 years ago

What is the form of the two squares identity?

zvonat [6]

Answer:

Step-by-step explanation:

Consider all options:

A. False, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ab-cd)^2+(ac+bd)^2=a^2b^2-2abcd+c^2d^2+a^2c^2+2abcd+b^2d^2=\\ =a^2b^2+c^2d^2+a^2c^2+b^2d^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2\neq a^2b^2+c^2d^2+a^2c^2+b^2d^2$

B. False, because

$(a^2-b^2)(c^2+d^2)=a^2c^2+a^2d^2-b^2c^2-b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2-b^2c^2-b^2d^2\neq a^2c^2+b^2d^2+a^2d^2+b^2c^2$

C. False, because

$(a^2+b^2)(c^2-d^2)=a^2c^2-a^2d^2+b^2c^2-b^2d^2\\ \\(ac+bd)^2-(ad+bc)^2=a^2c^2+2abcd+b^2d^2-a^2d^2-2abcd-b^2c^2=\\=a^2c^2+b^2d^2-a^2d^2-b^2c^2\\ \\a^2c^2-a^2d^2+b^2c^2-b^2d^2\neq a^2c^2+b^2d^2-a^2d^2-b^2c^2$

D. True, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2= a^2c^2+b^2d^2+a^2d^2+b^2c^2$

6 0

3 years ago