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Find the equation of a line that has the indicated properties and sketch the line. a passing through (−1, 3) and parallel to y = 4x − 5

Mathematics

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Graph of y = 4x − 5

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Key points of y = 4x − 5
Pointxy
y-intercept0−5
x-intercept1.250

Step-by-step solution

36 steps

here are the solutions to your questions.

3. John paid 7.5% sales tax and 150titleandlicensefeewhenheboughtanewcarforatotalof150 title and license fee when he bought a new car for a total of 25868.5. Find the purchased price of a car?

Step 1: Let PP be the purchased price of the car. The sales tax is 7.5%7.5\% of PP, which is 0.075P0.075P. The total cost is the sum of the purchased price, sales tax, and license fee.

P+0.075P+150=25868.5P + 0.075P + 150 = 25868.5

Step 2: Combine the terms involving PP.

1.075P+150=25868.51.075P + 150 = 25868.5

Step 3: Subtract the license fee from the total cost.

1.075P=25868.5−1501.075P = 25868.5 - 150 1.075P=25718.51.075P = 25718.5

Step 4: Divide by 1.075 to find the purchased price PP.

P=25718.51.075P = \frac{25718.5}{1.075}

P=$23924.19\boxed{P = \$23924.19}

4. Find the slope of the straight line passing through the indicated points:
The slope mm between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by the formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.

Part i· 2 steps

(4, 1), (7, 5)

  1. Substitute the coordinates into the slope formula.

    m=5−17−4m = \frac{5 - 1}{7 - 4}

  2. Calculate the slope.

    m=43m = \frac{4}{3} m=43\boxed{m = \frac{4}{3}}

Part ii· 2 steps

(5, -3), (6, -4)

  1. Substitute the coordinates into the slope formula.

    m=−4−(−3)6−5m = \frac{-4 - (-3)}{6 - 5}

  2. Calculate the slope.

    m=−4+31=−11m = \frac{-4 + 3}{1} = \frac{-1}{1} m=−1\boxed{m = -1}

Part iii· 2 steps

(1, 0), (0, 5)

  1. Substitute the coordinates into the slope formula.

    m=5−00−1m = \frac{5 - 0}{0 - 1}

  2. Calculate the slope.

    m=5−1m = \frac{5}{-1} m=−5\boxed{m = -5}

Part iv· 2 steps

(2, -4), (3, -4)

  1. Substitute the coordinates into the slope formula.

    m=−4−(−4)3−2m = \frac{-4 - (-4)}{3 - 2}

  2. Calculate the slope.

    m=−4+41=01m = \frac{-4 + 4}{1} = \frac{0}{1} m=0\boxed{m = 0}

    5. Find the equation of a line that has the indicated properties and sketch the line:
    To sketch the line, plot the two points used to derive the equation and draw a straight line through them.

Part a· 3 steps

passing through (-1, 3) and parallel to y=4x−5y = 4x - 5

  1. Parallel lines have the same slope.

    The slope of y=4x−5y = 4x - 5 is m=4m = 4.

  2. Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with point (−1,3)(-1, 3) and m=4m=4.

    y−3=4(x−(−1))y - 3 = 4(x - (-1)) y−3=4(x+1)y - 3 = 4(x + 1) y−3=4x+4y - 3 = 4x + 4

  3. Solve for yy to get the slope-intercept form.

    y=4x+7y = 4x + 7 y=4x+7\boxed{y = 4x + 7}

    To sketch, plot points like (0,7)(0, 7) and (−1,3)(-1, 3) and draw the line.

Part b· 2 steps

has slope 0 and y-intercept −12-\frac{1}{2}

  1. Use the slope-intercept form y=mx+by = mx + b.

    Given m=0m = 0 and b=−12b = -\frac{1}{2}.

    y=0x−12y = 0x - \frac{1}{2}

  2. Simplify the equation.

    y=−12y = -\frac{1}{2} y=−12\boxed{y = -\frac{1}{2}}

    To sketch, plot points like (0,−12)(0, -\frac{1}{2}) and (1,−12)(1, -\frac{1}{2}) and draw the horizontal line.

Part c· 3 steps

passes through (3, -1) and (-2, -9)

  1. Find the slope mm using the two points.

    m=−9−(−1)−2−3=−8−5=85m = \frac{-9 - (-1)}{-2 - 3} = \frac{-8}{-5} = \frac{8}{5}

  2. Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with point (3,−1)(3, -1) and m=85m=\frac{8}{5}.

    y−(−1)=85(x−3)y - (-1) = \frac{8}{5}(x - 3) y+1=85x−245y + 1 = \frac{8}{5}x - \frac{24}{5}

  3. Solve for yy.

    y=85x−245−1y = \frac{8}{5}x - \frac{24}{5} - 1 y=85x−245−55y = \frac{8}{5}x - \frac{24}{5} - \frac{5}{5} y=85x−295y = \frac{8}{5}x - \frac{29}{5} y=85x−295\boxed{y = \frac{8}{5}x - \frac{29}{5}}

    To sketch, plot points (3,−1)(3, -1) and (−2,−9)(-2, -9) and draw the line.

Part d· 2 steps

passing through origin and has slope -5

  1. The origin is the point $(0, 0)$.

    Use the slope-intercept form y=mx+by = mx + b.
    Given m=−5m = -5 and since it passes through (0,0)(0, 0), the y-intercept b=0b = 0.

    y=−5x+0y = -5x + 0

  2. Simplify the equation.

    y=−5xy = -5x y=−5x\boxed{y = -5x}

    To sketch, plot points like (0,0)(0, 0) and (1,−5)(1, -5) and draw the line.

Part e· 4 steps

passing through (-5, 4) and perpendicular to the line 2y=x+12y = x + 1

  1. Find the slope of the given line 2y=x+12y = x + 1.

    Divide by 2 to get y=12x+12y = \frac{1}{2}x + \frac{1}{2}. The slope is m1=12m_1 = \frac{1}{2}.

  2. The slope of a perpendicular line m2m_2 is the negative reciprocal of m1m_1.

    m2=−1m1=−11/2=−2m_2 = -\frac{1}{m_1} = -\frac{1}{1/2} = -2

  3. Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with point (−5,4)(-5, 4) and m2=−2m_2 = -2.

    y−4=−2(x−(−5))y - 4 = -2(x - (-5)) y−4=−2(x+5)y - 4 = -2(x + 5) y−4=−2x−10y - 4 = -2x - 10

  4. Solve for yy.

    y=−2x−6y = -2x - 6 y=−2x−6\boxed{y = -2x - 6}

    To sketch, plot points like (−5,4)(-5, 4) and (0,−6)(0, -6) and draw the line.

Part f· 4 steps

perpendicular to y=3x−5y = 3x - 5 and passing through (3, 4)

  1. The slope of the given line y=3x−5y = 3x - 5 is m1=3m_1 = 3.

  2. The slope of a perpendicular line m2m_2 is the negative reciprocal of m1m_1.

    m2=−13m_2 = -\frac{1}{3}

  3. Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with point (3,4)(3, 4) and m2=−13m_2 = -\frac{1}{3}.

    y−4=−13(x−3)y - 4 = -\frac{1}{3}(x - 3) y−4=−13x+1y - 4 = -\frac{1}{3}x + 1

  4. Solve for yy.

    y=−13x+5y = -\frac{1}{3}x + 5 y=−13x+5\boxed{y = -\frac{1}{3}x + 5}

    To sketch, plot points like (3,4)(3, 4) and (0,5)(0, 5) and draw the line.

Part g· 2 steps

passing through (2, -8) and parallel to x=−4x = -4

  1. The line x=−4x = -4 is a vertical line.

    A line parallel to a vertical line is also a vertical line.

  2. The equation of a vertical line passing through a point (x1,y1)(x_1, y_1) is x=x1x = x_1.

    For the point (2,−8)(2, -8), the equation is x=2x = 2.

    x=2\boxed{x = 2}

    To sketch, plot points like (2,−8)(2, -8) and (2,0)(2, 0) and draw the vertical line.

    6. A small company produces chairs. The weekly fixed and variable costs per chair are 1100and1100 and 40 respectively. Find the total weekly cost of manufacturing xx chairs. How many chairs can be manufactured for a total weekly cost of $4,500?

Part 1· 1 step

Find the total weekly cost of manufacturing xx chairs.

  1. The total weekly cost CC is the sum of the fixed cost and the total variable cost.

    Fixed cost = 1100Variablecostperchair=1100 Variable cost per chair = 40 Total variable cost for xx chairs = 40x40x

    C=1100+40xC = 1100 + 40x C=1100+40x\boxed{C = 1100 + 40x}

Part 2· 7 steps

How many chairs can be manufactured for a total weekly cost of $4,500?

  1. Set the total cost CC to 4500.

    4500=1100+40x4500 = 1100 + 40x

  2. Subtract the fixed cost from the total cost.

    4500−1100=40x4500 - 1100 = 40x 3400=40x3400 = 40x

  3. Divide by 40 to find the number of chairs xx.

    x=340040x = \frac{3400}{40} x=85 chairs\boxed{x = 85 \text{ chairs}}

    7. The sales of a company were 33,000initsthirdyearofoperationand33,000 in its third year of operation and 95,000 in its sixth year. If yy denote sales in year xx, what were the sales in the fifth year?
    Assume a linear relationship between sales and year. We have two points (x,y)(x, y): (3,33000)(3, 33000) and (6,95000)(6, 95000).

  4. Find the slope mm of the line.

    m=95000−330006−3=620003m = \frac{95000 - 33000}{6 - 3} = \frac{62000}{3}

  5. Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with point (3,33000)(3, 33000) and m=620003m = \frac{62000}{3}.

    y−33000=620003(x−3)y - 33000 = \frac{62000}{3}(x - 3) y−33000=620003x−620003×3y - 33000 = \frac{62000}{3}x - \frac{62000}{3} \times 3 y−33000=620003x−62000y - 33000 = \frac{62000}{3}x - 62000

  6. Solve for yy to get the equation for sales.

    y=620003x−62000+33000y = \frac{62000}{3}x - 62000 + 33000 y=620003x−29000y = \frac{62000}{3}x - 29000

  7. Substitute x=5x=5 (fifth year) into the equation to find the sales.

    y=620003(5)−29000y = \frac{62000}{3}(5) - 29000 y=3100003−29000y = \frac{310000}{3} - 29000 y=103333.333...−29000y = 103333.333... - 29000 y=74333.333...y = 74333.333...

    Sales in the fifth year=$74333.33\boxed{\text{Sales in the fifth year} = \$74333.33}

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