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Mnenie [13.5K]
3 years ago
15

Select the correct answer. You have this information about ΔABC, ΔFDE, and ΔGIH: AB = DF AB = GI BC = HI DE = HI m∠B = m∠D = m∠I

Which triangles must be congruent? A. ΔABC, ΔFDE, and ΔGIH B. ΔABC and ΔFDE only C. ΔGIH and ΔABC only D. none of the triangles
Mathematics
1 answer:
choli [55]3 years ago
3 0

Answer:

c

Step-by-step explanation:

plato

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Solve for C.<br> -5/8c=20
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-5/8c = 20....multiply both sides by -8/5 (this cancels out the -5/8 on the left
c = 20(-8/5)
c = -160/5
c = - 32

check..
-5/8(-32) = 20
160/8 = 20
20 = 20 (correct)

so c = -32
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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

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 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

=364

Hence required number of factors is 364.

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