Question

Please help and show work!! I really need help please!!

Answer 1

Answer Left Panel: 3/5 or B

Answer Right Panel: D

Explanation:

Left Panel

• 5x + 3y = 8 Subtract 5x from both sides
• 5x - 5x + 3y = - 5x + 8 Combine
• 3y = - 5x+ 8 Divide by 3
• 3y/3 = -5x/3 + 8/3
• y = - 5x/3 + 8/3 The slope of any line is the number with the x
• Slope of this line = -5/3
• The slope of two lines that are perpendicular when multiplied = - 1
• slope of the given line(m1) * slope of the perpendicular(m2) = - 1
• m1 * m2 = - 1
• -5/3 * m2 = - 1 Multiply by 3
• -5x/3 *3 = - 1 * 3
• -5x = - 3 Divide by -5
• -5x/-5 = -3 / 5
• x = 3/5

Right Panel

The slope of a line parallel to another line is the same as the given line.

• The given line is 3x - 5y = 10 Subtract 3x from both sides
• 3x - 3x - 5y = -3x + 10 Combine
• -5y = -3x + 10 Divide by - 5
• -5y/-5 = -3x/-5 + 10/-5 Do the division
• y = 3x/5 - 2 So look for a line with a slope of 3/5
• y= 3x /5 Answer D

Answer 2

`\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-(1)/(m_1)\\-------------------------\\\\\text{We have}\ 5x+3y=8\qquad\text{subtract 5x from both sides}\\\\3y=-5x+8\qquad\text{divide both sides by 3}\\\\y=-(5)/(3)x+(8)/(3)\to m_1=-(5)/(3)\\\\\text{Therefore}\ m_2=-(1)/(-(5)/(3))=(3)/(5)\\\\Answer:\ \boxed{b.\ (3)/(5)}`

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`\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \parallel\ k\iff m_1=m_2\\\\\text{We have}\ 3x-5y=10\qquad\text{subtract 3x from both sides}\\\\-5y=-3x+10\qquad\text{divide both sides by (-5)}\\\\y=(3)/(5)x-2\to m_1=(3)/(5)\\\\\text{Therefore we have}\ m_2=(3)/(5).\\\\Answer:\ \boxed{D.\ y=(3)/(5)x}`