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mezya [45]
3 years ago
14

HELPPPPPP PLEASEEEEEEEEEEEEEEEEEEEEE

Mathematics
1 answer:
Sholpan [36]3 years ago
7 0
M = 2
b = 5
equation = y = 2x + 5
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Z=-3x2+2ey-2y+3 cực trị tự do
topjm [15]

Answer:

z=( -2 , 5/3) can you speak English

5 0
2 years ago
14
Nuetrik [128]

Answer:

Distance LM = 5.20 unit (Approx.)

Step-by-step explanation:

Given coordinates;

L(1, 4, 7) and M(2, 9, 8)

Find:

Distance LM

Computation:

Distance between three-dimensional plane = √(x2 - x1)² + (y2 - y1)² + (z2 - z1)²

Distance LM = √(2 - 1)² + (9 - 4)² + (8 - 7)²

Distance LM = √(1)² + (5)² + (1)²

Distance LM = √1 + 25 + 1

Distance LM = √27

Distance LM = 3√3 unit

Distance LM = 3(1.732)

Distance LM = 5.196

Distance LM = 5.20 unit (Approx.)

5 0
3 years ago
ABC is isosceles.<br>m &lt; a = 3x + 40 and m &lt; c = x + 50<br>what is m &lt; a ??​
il63 [147K]

Answer:

  • 55°

Step-by-step explanation:

<u>It is assumed the angles A and C are congruent:</u>

  • 3x + 40 = x + 50
  • 3x - x = 50 - 40
  • 2x = 10
  • x = 5

m∠A = 3*5 + 40 = 55°

7 0
3 years ago
Read 2 more answers
How many different perfect cubes are among the positive actors of 2021^2021
9966 [12]

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.

8 0
3 years ago
How many solutions does this equation have?<br> (15x + 21) / 3<br> = 5x + 7
Lesechka [4]

Answer:

3*(5*x-7)-(15*x-21)=0

Step-by-step explanation:

3 • (5x - 7) -  (15x - 21)  = 0

Step  2  :

Equation at the end of step  2  :

 0  = 0

Step  3  :

Equations which are always true :

7 0
3 years ago
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