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Paha777 [63]
3 years ago
13

Algebra 2 need help on answer ( A,B,C )

Mathematics
1 answer:
statuscvo [17]3 years ago
6 0

Answer:  A) when Celsius (c) = 0, then Fahrenheit (F) = 32

               B) F(100) = 212°

               C) F(25) = 77°

<u>Step-by-step explanation:</u>

A) F(c) = 32 means that when c = 0 then F = 32

B) when c = 100°, then F = 212° can be written as: F(100) = 212°

C) \dfrac{F(100)-F(0)}{100-0}=\dfrac{212-32}{100}=\dfrac{180}{100}=1.8

   F(c) = 1.8c + 32

 F(25) = 1.8(25) + 32

           =   45  +   32

           =         77°

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A x B

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3 years ago
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What is the value of x?<br><br><br><br> Enter your answer in the box.<br><br> x =
FinnZ [79.3K]

Answer:

x = 12

Step-by-step explanation:

IMPORTANT: Both answers should add up to 180 if they dont you did it wrong!

(10x - 20) + (6x + 8 ) = 180 plug in 12 for both of them

(10 * 12 - 20) + 6 * 12 + 8) = 120

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Y = -5x -5

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3 years ago
How many different perfect cubes are among the positive actors of 2021^2021
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Answer:

hope this helps :D

Step-by-step explanation:

Perfect cube factors:

If a number is a perfect cube, then the power of the prime factors should be divisible by 3.

Example 1:Find the number of factors of293655118 that are perfect cube?

Solution: If a number is a perfect cube, then the power of the prime factors should be divisible by 3. Hence perfect cube factors must have

2(0 or 3 or 6or 9)—– 4 factors

3(0 or 3 or 6)  —–  3  factors

5(0 or 3)——- 2 factors

11(0 or 3 or 6 )— 3 factors

Hence, the total number of factors which are perfect cube 4x3x2x3=72

Perfect square and perfect cube

If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.

Example 2: How many factors of 293655118 are both perfect square and perfect cube?

Solution: If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.Hence both perfect square and perfect cube must have

2(0 or 6)—– 2 factors

3(0 or 6) —– 2 factors

5(0)——- 1 factor

11(0 or 6)— 2 factors

Hence total number of such factors are 2x2x1x2=8

Example 3: How many factors of293655118are either perfect squares or perfect cubes but not both?

Solution:

Let A denotes set of numbers, which are perfect squares.

If a number is a perfect square, then the power of the prime factors should be divisible by 2. Hence perfect square factors must have

2(0 or 2 or 4 or 6 or 8)—– 5 factors

3(0 or 2 or 4 or 6)  —– 4 factors

5(0 or 2or 4 )——- 3 factors

11(0 or 2or 4 or6 or 8 )— 5 factors

Hence, the total number of factors which are perfect square i.e. n(A)=5x4x3x5=300

Let B denotes set of numbers, which are perfect cubes

If a number is a perfect cube, then the power of the prime factors should be divisible by 3. Hence perfect cube factors must have

2(0 or 3 or 6or 9)—– 4 factors

3(0 or 3 or 6)  —–  3  factors

5(0 or 3)——- 2 factors

11(0 or 3 or 6 )— 3 factors

Hence, the total number of factors which are perfect cube i.e. n(B)=4x3x2x3=72

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2(0 or 6)—– 2 factors

3(0 or 6) —– 2 factors

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11(0 or 6)— 2 factors

Hence total number of such factors are i.e.n(A∩B)=2x2x1x2=8

We are asked to calculate which are either perfect square or perfect cubes i.e.

n(A U B )= n(A) + n(B) – n(A∩B)

=300+72 – 8

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Hence required number of factors is 364.

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