Math Semester 1 Unit 2 Flashcards
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Questions and Answers

What is the equation solved for C when converting Kelvin to Celsius?

C = K - 273.15

What is the solution for w in the equation 5w + 9z = 2z + 3w?

w = -7/2z

Which are equivalent equations for the lateral area of a right cone, LA = pi rs? (Select two correct answers)

  • s = r/LA pi
  • s = LA/pi r (correct)
  • r = LA/(pi s)
  • r = LA/pi r (correct)
  • What is the equivalent equation solved for r from the formula C = 2 pi r?

    <p>r = c/2 pi</p> Signup and view all the answers

    Which are equivalent equations for the area of a rhombus, A = 1/2 d1 d2? (Select two correct answers)

    <p>d1 = 2A/d2</p> Signup and view all the answers

    What is the acceleration of an object with F = 7.92 newtons and m = 3.6 kilograms?

    <p>2.2 m/s^2</p> Signup and view all the answers

    What is the result when solving the equation S = 2πrh + 2πr^2 for h?

    <p>h = s/(2 pi r) - r</p> Signup and view all the answers

    What is the equivalent equation solved for t in the equation f = v + at?

    <p>t = (f - v)/a</p> Signup and view all the answers

    What is the value of r if the circumference of a circle is C = 2 pi r?

    <p>r = 8</p> Signup and view all the answers

    What is the equivalent equation solved for the slope, m, from the slope-intercept form y = mx + b?

    <p>m = (y - b)/x</p> Signup and view all the answers

    Study Notes

    Temperature Conversion

    • Temperature in Kelvin (K) can be calculated from Celsius (C) using K = C + 273.15.
    • To convert from K to C, the equation is C = K - 273.15.

    Algebraic Manipulation

    • To solve for w in the equation 5w + 9z = 2z + 3w, the result is w = -7/2z.

    Lateral Area of a Cone

    • The lateral area (LA) of a right cone is given by LA = πrs, where r is the radius and s is the slant height.
    • Equivalent expressions for s are: s = LA / (πr) and re-arranging yields proper forms.

    Circle Circumference

    • Circumference (C) of a circle is calculated via C = 2πr.
    • When solving for the radius (r), the equivalent equation is r = C / (2π).

    Area of a Rhombus

    • The area (A) of a rhombus is given by A = 1/2 * d1 * d2, where d1 and d2 are diagonal lengths.
    • Solving for d1 gives d1 = 2A / d2, and for d2 gives d2 = 2A / d1.

    Newton's Second Law

    • Force (F) is defined by the equation F = ma, where m is mass and a is acceleration.
    • For an object with F = 7.92 N and m = 3.6 kg, the calculated acceleration is 2.2 m/s².

    Surface Area of a Cylinder

    • The equation for the surface area of a cylinder is S = 2πrh + 2πr².
    • Solving for height (h) results in h = S / (2πr) - r.

    Final Velocity Formula

    • The final velocity formula is f = v + at, where f is final velocity, v is initial velocity, a is acceleration, and t is time.
    • Rearranging to solve for time (t) yields t = (f - v) / a.

    Circle Radius Confirmation

    • The radius can also be determined using the circumference formula C = 2πr, leading to r = C / (2π).
    • When C is specifically given as 16π, then r = 8.

    Slope-Intercept Form

    • The slope-intercept form of a linear equation is y = mx + b.
    • To isolate the slope (m), the equivalent equation is m = (y - b) / x.

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    Description

    Test your knowledge with these flashcards covering key concepts from Math Semester 1, Unit 2. The flashcards include definitions and equations related to temperature conversions, solving linear equations, and formulas for geometric shapes. Perfect for quick review and retention of important mathematical principles.

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