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den301095 [7]
3 years ago
11

Can every even number be written as a sum of two primes?

Mathematics
1 answer:
astraxan [27]3 years ago
5 0
The Goldbach Conjecture is a yet unproven conjecture stating that every even integer greater than two is thesum of two prime numbers. The conjecture has been tested up to 400,000,000,000,000. Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics.
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Tehtävä 44<br><br>b) 12,2,10,4, ,​
Tpy6a [65]

I am not sure it is 2i think

7 0
3 years ago
Please help quickly! 20 points
scZoUnD [109]

Answer:

y+3 = 4(x+3)

Step-by-step explanation:

its this answer because to find point slope form you have to do y1 - y2 and in the coordinate provided, -3 is y so y- (-3) or y+3 hope this helped ^u^

5 0
3 years ago
Use the quadratic formula to solve for the roots in the following equation. 4x 2 + 5x + 2 = 2x 2 + 7x – 1
Natali5045456 [20]

Answer:

So, The roots are x= \frac{1+\sqrt{5}i}{2} \,\, and \,\,x= \frac{1-\sqrt{5}i}{2}

Step-by-step explanation:

4x^2 + 5x + 2 = 2x^2 + 7x - 1

We need to solve the equation to find the roots using quadratic formula.

The quadratic formula is:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Rearranging the above equation:

4x^2 -2x^2+ 5x-7x + 2+1 =0

2x^2 -2x + 3 =0

Where a =2 , b=-2 and c =3 Putting values in quadratic equation and solving:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-2)\pm\sqrt{(-2)^2-4(2)(3)}}{2(2)}\\x=\frac{2\pm\sqrt{4-24}}{4}\\x=\frac{2\pm\sqrt{-20}}{4}\\\sqrt{-20} \,\,can\,\,be\,\, written\,\, as\,\, 2\sqrt{-5}\\ x=\frac{2\pm2\sqrt{-5}}{4}\\x=\frac{2+2\sqrt{-5}}{4} \,\, and \,\, x=\frac{2-2\sqrt{-5}}{4}\\x=\frac{2(1+\sqrt{-5})}{4} \,\, and \,\, x=\frac{2(1-\sqrt{-5})}{4}\\ x=\frac{1+\sqrt{-5}}{2} \,\, and \,\, x=\frac{1-\sqrt{-5}}{2}\\As \,\,we\,\, know\,\, \sqrt{-1} = i \\

x= \frac{1+\sqrt{5}i}{2} \,\, and \,\,x= \frac{1-\sqrt{5}i}{2}

So, The roots are x= \frac{1+\sqrt{5}i}{2} \,\, and \,\,x= \frac{1-\sqrt{5}i}{2}

6 0
3 years ago
Factor the following<br>(a²+b²)²-18(a²+b²)-88​
navik [9.2K]

Answer:

\rm\displaystyle(  {a}^{2}  +  {b}^{2} - 22)( {a}^{2}  +  {b}^{2}   +  4)

Step-by-step explanation:

we would like to factor the following:

\rm\displaystyle ( {a}^{2}  +  {b}^{2}  {)}^{2}  - 18( {a}^{2}  +  {b}^{2} ) - 88

let a²+b²=x

thus substitute:

\rm\displaystyle x {}^{2}  - 18x- 88

rewrite the middle term as 4x-22x:

\rm\displaystyle x^{2}  + 4x - 22x -  88

factor out x:

\rm\displaystyle x( x^{}  + 4)- 22x -  88

factor out -22:

\rm\displaystyle x( x^{}  + 4)- 22(x  +  4)

group:

\rm\displaystyle( x- 22)(x  +  4)

substitute back:

\rm\displaystyle(  {a}^{2}  +  {b}^{2} - 22)( {a}^{2}  +  {b}^{2}   +  4)

and we are done!

6 0
3 years ago
Read 2 more answers
Solve each equation. <img src="https://tex.z-dn.net/?f=32%5E%7Bx-1%7D" id="TexFormula1" title="32^{x-1}" alt="32^{x-1}" align="a
Sonja [21]

Answer:

  x = 5/4

Step-by-step explanation:

32^(x-1) = 16^(x/4)

(2^5)^(x-1) = (2^4)^(x/4) . . . . . express as powers of 2

2^(5x -5) = 2^x . . . . . . simplify

5x -5 = x . . . . . . equate exponents

4x = 5 . . . . . . . . add 5-x

x = 5/4

__

Applicable rules of exponents are ...

  (a^b)^c = a^(bc)

_____

Since this solution does not match any offered answer choice, we suggest you ask your teacher to show you how to work this problem.

7 0
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