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aniked [119]
3 years ago
7

Please help me with the problem

Mathematics
2 answers:
masya89 [10]3 years ago
5 0
The answer is 1/4. 1/2 times 2 is 0.25 or 1/4. hope this helps :)
mixas84 [53]3 years ago
3 0
Let's see what we're working on here 

\frac{1}{2}  - 2(m) =?
            ? = 0
Simplify this → \frac{1}{2}

\frac{1}2y} - 2(m) = ?

\frac{2(m)(2)}{2}

\frac{1 - 4(m)}{2} = ?

\frac{1 - 4(m)}{2} = 2

-4(m) + 1 = 0
 Subtract ( - )the number 1 from each of your sides on this part

4(m) = 1 
Multiply( ×) the number -1 to your sides for this part of the equation 

Therefore, the value of m is \frac{1}{4}
m =  \frac{1}{4}


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Find a polynomial P(x) with real coefficients having a degree 4, leading coefficient 4, and zeros 3-i and 4i.
Alisiya [41]

now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.

let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.

\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0

\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill

3 0
3 years ago
Please help me
Flauer [41]
Probability of heads or tails is .5
so you do (.5)^3 and you get D = 0.125
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