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3 years ago

Can someone help me with this please

Mathematics

2 answers:

kirill [66]3 years ago

7 0

Answer:

Answer and steps is in the picture.

pochemuha3 years ago

3 0

Answer: No

Step-by-step explanation:

What we have to do is to substitute the values of x and y that we are given into the inequality:

x = 8

y = 7

y > 4x - 6

7 > 4(8) - 6

7 > 32 - 6

7 > 26

This is not true so it isn't a solution

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2 years ago

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Verdich [7]

Answer:

$x=-1,\:x=-7,\:x=i,\:x=-i$

Step-by-step explanation:

Considering the equation

$x^4+8x^3+8x^2+8x+7=0$

Solving

$x^4+8x^3+8x^2+8x+7$

$\mathrm{Factor\:}x^4+8x^3+8x^2+8x+7:\quad \left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=7,\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:7,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:7}{1}$

$-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1$

$=\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

Solving

$\frac{x^4+8x^3+8x^2+8x+7}{x+1}$

$=x^3+7x^2+x+7$

Putting $\frac{x^4+8x^3+8x^2+8x+7}{x+1}$ = $x^3+7x^2+x+7$ in equation [A]

So,

$\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

$=\left(x+1\right)x^3+7x^2+x+7$

$x^3+7x^2+x+7=\left(x+7\right)\left(x^2+1\right)$

So,

Equation [A] becomes

$=\left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

So, the polynomial equation becomes

$\left(x+1\right)\left(x+7\right)\left(x^2+1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$ $\mathrm{Solve\:}\:x+1=0:\quad x=-1$