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finlep [7]
3 years ago
10

Write an equation of the line that is:

Mathematics
1 answer:
Marysya12 [62]3 years ago
4 0
a)\\y=4x+1\\a=4\\new\ y=bx+c\\if\ parallel a=b\\y=4x+c\\substituting\point\ (2,3)\\3=4*2+c\\c=-5\\y=4x-5\\b)\\2y-6x=9\\2y=6x+9\\y=3x+4,5\\a=3\\new\ y=bx+c\\if\ parallel a=b\\y=3x+c\\substituting\ point\ (-2,1)\\1=4*(-2)+c\\c=9\\y=3x+9\\ c)\\y=4x+3\\a=4\\new\ y=bx+c\\if\ parallel a=b\\y=4x+c\\c=-3\\y=4x-3\\e)\\2y-4x=8\\2y=4x+8\\y=2x+4\\a= 2 \\new\y=bx+c\\if\ perpendicular b=-1/a\\y=- \frac{1}{2} x+c\\point(6,-6)\\-6=-1/2*6+c\\c=-3\\y=-1/2x-3\\f)\\2y-5x=15\\2y=5x+15\\y=\frac{5}{2}x+\frac{15}{2}
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Approximate the real zeros of f(x) = 2x^4 - x^3 + x-2 to the nearest tenth
Vlad1618 [11]

Answer:

-1.0, 1.0

Step-by-step explanation:

The given polynomial function is

f(x) = 2 {x}^{4} - {x}^{3} + x - 2

According to the Rational Roots Theorem, the possible roots of this function are;

\pm1,\pm \frac{1}{2}

We now use the Remainder Theorem to obtain;

f(1) = 2 {(1)}^{4} - {(1)}^{3} + 1 - 2

f(1) = 2 - 1+ 1 - 2 = 0

f( - 1) = 2 {( - 1)}^{4} - {( - 1)}^{3} - 1 - 2  

f( - 1) = 2 + 1 - 1 - 2 = 0

But;

f( \frac{1}{2} ) = - 1.5

f( - \frac{1}{2} ) = - 2.25

Since f(1)=0 and f(-1)=0, the real zeros to the nearest tenth are:

-1.0 and 1.0

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3 years ago
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Answer:

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Step-by-step explanation:

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$34.00 + .08x = total

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Answer:31557600

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