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Artyom0805 [142]
3 years ago
10

Which of the four representations are incorrect?

Mathematics
2 answers:
ruslelena [56]3 years ago
8 0
The second one is correct because it does not match the rest
Schach [20]3 years ago
4 0
Y=6x + 27 isn’t right
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What is 0.2 as a fraction
Vikentia [17]
It’s 2/10 there you go since there’s no 0 infort of the 2 it’s always a 10
3 0
3 years ago
Read 2 more answers
If the farmers market sold 35% of its produce and has 280 peaches remaining, how many peaches did it have?
PSYCHO15rus [73]

100-35 = 65%

280/0.65 = 430

 he had 430 peaches


3 0
3 years ago
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
4 years ago
Negative five is three more than twice a number(n). what is n?
artcher [175]

N is -4 because -4x2=-8 and -8+3=-5

4 0
3 years ago
Can someone help me please <br> And please show your work
kondaur [170]
A)1:2
b)21:2
c)6:44 or 3:22
d)2:5
5 0
3 years ago
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