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barxatty [35]
3 years ago
9

Calculate 3n-1 for n equals 2 to 10 explain how you know that the numbers are composite

Mathematics
1 answer:
Troyanec [42]3 years ago
5 0

composite number means a number that can be formed by multiplying two smaller numbers

3n-1 for n=2, 3n-1=3*2-1=6-1=5

3n-1 for n=10, 3n-1=3*10-1=30-1=29

5 is composite because 2*2.5=5

29 is composite because 5*5.8=29

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Simplify the expressions. Show your work.
Hunter-Best [27]
7) Use distributive property
3x(4x⁴ - 5x)=3x*4x⁴- 3x*5x = 12x⁵ - 15x²
8) When you open parenthesis with minus sigh in front of it, you need change sings of the terms on opposite. Then find like terms.
(5x⁴-3x³+6x)-(3x³+11x²-8x)= 5x⁴-3x³+6x -3x³-11x²+8x =5x⁴-6x³ - 11x²+14x
9)(x-2)(3x-4)=x*3x-4x-6x+8=3x²-10x+8
10) (x+6)²=(x+6)(x+6)=x²+6x+6x+36=x²+12x+36
For (x+6)², you can also use the formula of the perfect square
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For (x+6)² =x²+2*x*6 +6²=x²+12x+36


5 0
3 years ago
The sum of all the interior angles in a triangle is blank degrees. Please HELP
ozzi

Answer:

180º

Step-by-step explanation:

All the interior angles add up to 180º degrees.

3 0
2 years ago
Read 2 more answers
Is the point (3, 11) a solution for the linear equation 1/3x + y = 12? Explain.
PtichkaEL [24]

9514 1404 393

Answer:

  yes

Step-by-step explanation:

The point is on the line.

The coordinate values satisfy the equation:

  1/3x + y = 12

  1/3(3) + 11 = 12 . . . substitute for x and y

  1 + 11 = 12 . . . . . . . true

Any point that satisfies the equation is a solution. (3, 11) is a solution.

5 0
2 years ago
3. If<br> 15<br> x+20'<br> what is the value of?<br> X
guapka [62]
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6 0
3 years ago
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If m is a common divisor of x and y, then x/y = (x ÷ k)/______
myrzilka [38]

Answer:

\frac{x}{y} = \left(\frac{x}{y} \right)\cdot 1 = \left(\frac{x}{y}\right)\cdot \left(\frac{m}{m} \right) = \frac{\frac{x}{m} }{\frac{y}{m} }

Step-by-step explanation:

The common divisor is the greatest natural number so that two or more natural numbers have no remainder. If m is a common divisor of x and y, then p = \frac{x}{m} and q = \frac{y}{m}, where p, q \in \mathbb{Z}. Then, we have the following expression:

\frac{x}{y} = \left(\frac{x}{y} \right)\cdot 1 = \left(\frac{x}{y}\right)\cdot \left(\frac{m}{m} \right) = \frac{\frac{x}{m} }{\frac{y}{m} }

3 0
2 years ago
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