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3 years ago

Solve the equation: 4 1/5 + b = 9 3/5

2 answers:

7 0

$4 \frac{1}{5}+ b = 9 \frac{3}{5} \ \ | (-4\frac{1}{5})\\ \\4 \frac{1}{5 }-4 \frac{1}{5}+ b = 9 \frac{3}{5}-4 \frac{1}{5}\\ \\b= 5\frac{2}{5}=5.4$

kvasek [131]3 years ago

3 0

$4 \frac{1}{5} + b = 9 \frac{3}{5} \\\\b= 9 \frac{3}{5}-4 \frac{1}{5} \\\\b=5 \frac{2}{5}$

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2 years ago

What is the remainder R when the polynomial p(x) is divided by (x + 1)? Is (x + 1) a factor of p(x)? p(x) = -3x4 + 2x3 - x2 + 6

vodka [1.7K]

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(x+1) is a factor with remainder 0.

Step-by-step explanation:

We divide (x+1) into the polynomial $-3x^4+2x^3-x^2+6$ through long division or synthetic. We choose long division and look for what will multiply with (x+1) to make the polynomial $-3x^4+2x^3-x^2+6$ .

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We subtract this from the original $-3x^4-(-3x^4)+2x^3-(-3x^3)-x^2+6$ .

This leaves $5x^3-x^2+6$ . We repeat the step above.

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We subtract this from $5x^3-(5x^3)-x^2-(5x^2)+6=-6x^2+6$ . We repeat the step above.

$(x+1)(-6x)=-6x^2-6x$ .

We subtract this from $-6x^2-(-6x^2)+0x-(-6x)+6=6x+6$ . We repeat the step above.

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We subtract this from $6x-(6x)+6-(6)=0$ . There is no remainder. This means (x+1) is a factor.

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3 years ago