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Verdich [7]
2 years ago
12

Multiply (x^2+3x+5)(2x^2-2x+1

Mathematics
1 answer:
yawa3891 [41]2 years ago
3 0

Answer:

2x^4+4x^3+5x^2-7x+5

Step-by-step explanation:

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Math help Please!!!!
GalinKa [24]

Part A. What is the slope of a line that is perpendicular to a line whose equation is −2y=3x+7?

Rewrite the equation  −2y=3x+7 in the form y=-\dfrac{3}{2}x-\dfrac{7}{2}. Here the slope of the given line is  m_1=-\dfrac{3}{2}. If m_2 is the slope of perpendicular line, then

m_1\cdot m_2=-1,\\ \\m_2=-\dfrac{1}{m_1}=\dfrac{2}{3}.

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Since -2\cdot \left(-\dfrac{3}{2}\right)=3\neq -1\quad \text{and}\quad -2\cdot \dfrac{2}{3}=-\dfrac{4}{3}\neq -1, both lines are not perpendicular to line a.

Answer 2: Neither parallel nor perpendicular to line a

Part C. The line parallel to the line 2x+5y=10 has the equation 2x+5y=b. This line passes through the point (5,-4), then

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10-20=b,

b=-10.

Answer 3: 2x+5y=-10.

Part D. The slope of the line y=\dfrac{x}{4}+5 is \dfrac{1}{4}. Then the slope of perpendicular line is -4 and the equation of the perpendicular line is y=-4x+b. This line passes through the point (2,7), then

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b=7+8,

b=15.

Answer 4: y=-4x+15.

Part E. Consider vectors \vec{p}_1=(-c-0,0-(-d))=(-c,d)\quad \text{and}\quad \vec{p}_2=(0-b,a-0)=(-b,a). These vectors are collinear, then

\dfrac{-c}{-b}=\dfrac{d}{a},\quad \text{or}\quad -\dfrac{a}{b}=-\dfrac{d}{c}.

Answer 5: -\dfrac{a}{b}=-\dfrac{d}{c}.

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