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sergejj [24]
3 years ago
10

Plz help me with this

Mathematics
2 answers:
Ostrovityanka [42]3 years ago
8 0

An example is n+2 (n being number).

0,2,4,6,8,10..

k0ka [10]3 years ago
5 0
Just follow the patterns and you'll get the answer
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How many of each number are there 2, 13, 3, 5, 6, 2, 5, 12, 16, 2, 21, 0, 8, 4, 18, 10, 18, 11, 17, 8, 9, 20, 15, 15, 6, 3, 20,
mars1129 [50]

Answer:

Step-by-step explanation:

4 0
3 years ago
PLEASE HELP ASAP QUESTIONS ARE IN PICTURE
ArbitrLikvidat [17]

Answer:

1. Last option

2.  third option

Step-by-step explanation:

1.  Each term is formed by multiplying the previous one by 3

so its  729, 2187

2.  Working from right to left we multiply by 5  so its 500.

5 0
3 years ago
Solve x2 = 12x – 15 by completing the square. Which is the solution set of the equation? Solve x2 = 12x – 15 by completing the s
Svetlanka [38]

Answer: 6+\sqrt{21}\ \text{and}\ 6-\sqrt{21}

Step-by-step explanation:

Given

Quadratic equation is

x^2-12x+15=0

Solving by completing the square method

\Rightarrow x^2-2\cdot 6\cdot x+15=0\\\text{add and subtract 36}\\\Rightarrow x^2-2\cdot 6\cdot x+15+36-36=0\\\Rightarrow x^2-2\cdot 6\cdot x+36+(-36+15)=0\\\text{using}\ (a-b)^2=a^2+b^2-2ab\\\Rightarrow (x-6)^2-21=0\\\Rightarrow (x-6)^2=21\\\Rightarrow x-6=\pm \sqrt{21}\\\Rightarrow x=6\pm \sqrt{21}

The solution set of the equation is 6+\sqrt{21}\ \text{and}\ 6-\sqrt{21}

6 0
3 years ago
Read 2 more answers
How do you solve -3(4x+3)+4(6x+1)=43 ?
Anastaziya [24]

Answer:

x = 4

Step-by-step explanation:

Solution:

-3(4x + 3) + 4(6x + 1) = 43

-12x - 9 + 24x + 4 = 43

12x - 5 = 43

12x = 43 + 5

12x = 48

x = 48 / 12

Therefore, x = 4

PLEASE MARK ME AS BRAINLIEST!!!

4 0
3 years ago
Hi, can someone please help, and possibly explain the answer please.
mr_godi [17]

Answer:

x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}

Step-by-step explanation:

Quadratic formula: x=\frac{-b\±\sqrt{b^2-4ac} }{2a} when the equation is 0=ax^2+bx+c

The given equation is 1=-2x+3x^2+1. Let's first arrange this so its format looks like y=ax^2+bx+c:

1=-2x+3x^2+1

1=3x^2-2x+1

Subtract 1 from both sides of the equation

1-1=3x^2-2x+1-1\\0=3x^2-2x+0

Now, we can easily identify 3 as a, -2 as b and 0 as c. Plug these into the quadratic formula:

x=\frac{-b\±\sqrt{b^2-4ac} }{2a}\\x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}

I hope this helps!  

8 0
3 years ago
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