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3 years ago

Solve x + 10 = 3x + 4 *

Mathematics

2 answers:

shtirl [24]3 years ago

5 0

It’s 3 if you want me to go in depth tell me

Mashutka [201]3 years ago

4 0

Answer:

x= 3 hope it helps thanks

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4 years ago

The polynomial of degree 5, P ( x ) has leading coefficient a=1, has roots of multiplicity 2 at x = 3 and x = 0 , and a root of

ser-zykov [4K]

Answer:

$p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}$ , for $r_{1} = 0$

Step-by-step explanation:

The general form of quintic-order polynomial is:

$p_{5}(t) = a\cdot x^{5} + b\cdot x^{4} + c\cdot x^{3} + d\cdot x^{2} + e \cdot x + f$

According to the statement of the problem, the polynomial has the following roots:

$p_{5} (t) = (x - r_{1})\cdot (x-3)^{2}\cdot x^{2} \cdot (x+1)$

Then, some algebraic handling is done to expand the polynomial:

$p_{5} (t) = (x - r_{1}) \cdot (x^{3}-6\cdot x^{2}+9\cdot x) \cdot (x+1)\\p_{5} (t) = (x - r_{1}) \cdot (x^{4}-5\cdot x^{3} + 3 \cdot x^{2} + 9 \cdot x)$

$p_{5} (t) = x^{5} - (5+r_{1})\cdot x^{4} + (3 + 5\cdot r_{1})\cdot x^{3} +(9-3\cdot r_{1})\cdot x^{2} - 9 \cdot r_{1}\cdot x$

If $r_{1} = 0$ , then:

$p_{5} (t) = x^{5} - 5\cdot x^{4} + 3\cdot x^{3} +9\cdot x^{2}$

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3 years ago

Will give BRAINEST

victus00 [196]

For the first equation, -2x+2y=-6 so you take the second equation minus this new equation that would get you 5x=15 :)

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