• Solve 5x-19y?

    14 answers · 1 day ago
  • I need to learn to read a tape measure - advice?

    I'm aware that this is elementary school match, but then again, a fourth grade likely couldn't take those fractions and apply them to a tape measure to use rapid-fire without a lot of practice. So in a way, it is different. Plus, elementary school was decades ago. My previous jobs haven't require the... show more
    I'm aware that this is elementary school match, but then again, a fourth grade likely couldn't take those fractions and apply them to a tape measure to use rapid-fire without a lot of practice. So in a way, it is different. Plus, elementary school was decades ago. My previous jobs haven't require the use of a tape measure before and I have had some pretty technical jobs. But I'd like to gain that skill, "basic" as it may be. What's my best place to start? Obviously go back and study fractions all over again. I know that everything is in eighths. So 1/8, 2/8 (which is 1/4), 3/8, 4/8 (which is 1/2), 5/8, 6/8 (which is 3/4), 7/8 and 1". I've got that. I just can't seem to place them rapid-fire with the appropriate lines on the tape measure. I have trouble recognizing if a tape measure is down to 1/16" or 1/32", for instance. Then what confuses me even more is when people say, "It's all in eights" and it isn't. 1/16. 9/16. 17/16. 25/16, etc. Those are not eights. Then learning which one to skip in order to count up or count back from space to space with those in between makes it confusing. Got any advice?
    11 answers · 2 days ago
  • How to find derivative of (8x)^(2/3)?

    The answer sheet steps are: f (x) = (2/3)(8x)^(-1/3)(8) f (x) = 16/3(8)^(1/3)(x)^(1/3) <-- stuck here I don t understand, so the exponent of 1/3 can apply seperately to the 8 as well as the x instaead of being 8x^(1/3)? The answer is 8/13x^(1/3)
    The answer sheet steps are: f (x) = (2/3)(8x)^(-1/3)(8) f (x) = 16/3(8)^(1/3)(x)^(1/3) <-- stuck here I don t understand, so the exponent of 1/3 can apply seperately to the 8 as well as the x instaead of being 8x^(1/3)? The answer is 8/13x^(1/3)
    6 answers · 16 hours ago
  • How many yards is 44 inches?

    8 answers · 2 days ago
  • If 2*3 =812 and 4*5 =1620.then 6*7 is?

    5 answers · 13 hours ago
  • Math questions please help!?

    1. You get paid $20 for 4hrs of work. What is your hourly rate? 2. You drive a distance of 242 miles and use 11 gallons of gas. What is the average miles per gallon of your car? If possible, could you solve them and explain how to do them? Thank you!!!
    1. You get paid $20 for 4hrs of work. What is your hourly rate? 2. You drive a distance of 242 miles and use 11 gallons of gas. What is the average miles per gallon of your car? If possible, could you solve them and explain how to do them? Thank you!!!
    12 answers · 4 days ago
  • Find the minimum distance of the circle from the point 0?

    Find the minimum distance of the circle from the point 0?

    Best answer: . ( 1 ) Method 1: If this is not a Calculus topic Using Geometry: Check image attached Pythagora's Theorem (d + 6)² = 6² + 6² d = 6( √2 - 1 ) ≈ 2.49 ━━━━━━━━━ ( 2 ) Method 2: Using Calculus ➤ Center and equation of the circle Center = (6, 6) (x - 6)² + (y - 6)² = 6² x² - 12x + y² - 12y + 36 =... show more
    Best answer: .
    ( 1 )
    Method 1: If this is not a Calculus topic
    Using Geometry: Check image attached

    Pythagora's Theorem
    (d + 6)² = 6² + 6²
    d = 6( √2 - 1 ) ≈ 2.49
    ━━━━━━━━━

    ( 2 )
    Method 2: Using Calculus
    ➤ Center and equation of the circle
    Center = (6, 6)
    (x - 6)² + (y - 6)² = 6²
    x² - 12x + y² - 12y + 36 = 0
    —————————————

    Distance from (0, 0) to the point (x, y)
    d² = x² + y²
    d = √( x² + y² )
    ———————

    Minimize √( x² + y² ) where x² - 12x + y² - 12y + 36 = 0
    √( x² + y² ) = λ( x² - 12x + y² - 12y + 36 )

    Differentiate both sides using partial differentiation
    x / √( x² + y² ) = 2λ( x - 6 ), then rearrange =====> x / ( x - 6 ) = 2λ √( x² + y² )
    y / √( x² + y² ) = 2λ( y - 6 ), then rearrange =====> y / ( y - 6 ) = 2λ √( x² + y² )

    Therefore
    x / ( x - 6 ) = y / ( y - 6 ) where x - 6 ≠ 0 or y - 6 ≠ 0
    x = y
    ———

    x² - 12x + y² - 12y + 36 = 0
    x² - 12x + 18 = 0
    x = 6 ± 3√2
    ——————

    d = √( x² + y² )
    d = √( x² + x² )
    d = √( 2x² )
    d = x√2

    Minimum distance:
    d = x√2
    d = ( 6 - 3√2 ) √2
    d = 6√2 - 6
    d = 6( √2 - 1 ) ≈ 2.49
    ━━━━━━━━━
    6 answers · 2 days ago
  • The ball left my hand 1.61 meters, and after 0.6 seconds reached the maximum height of 3.34 meter. equation for the height, after t seconds.?

    Best answer: Horizontal and vertical motions are independent so only the vertical movement of the ball is relevant. In a constant acceleration field the equation of motion of a particle is given by: y(t) = y₀ + v₀t + at²/2 where y₀ is the initial coordinate of the particle v₀ = dy/dt is the initial velocity of the particle in... show more
    Best answer: Horizontal and vertical motions are independent so only the vertical movement of the ball is relevant. In a constant acceleration field the equation of motion of a particle is given by:
    y(t) = y₀ + v₀t + at²/2
    where
    y₀ is the initial coordinate of the particle
    v₀ = dy/dt is the initial velocity of the particle in the direction of the acceleration
    a = acceleration of the particle
    t = time in units consistent with the coordinate system and acceleration units.
    y(t) = coordinate of the particle at time t.

    Given in this case:
    y₀ = 1.61 m
    y(0.6) = 3.34 m
    y(0.6) is the maximum coordinate → dy/dt(0.6) = y'(0.6) = 0

    y'(0.6) = v₀ + 0.6a = 0
    y(0.6) = 1.61 + 0.6v₀ + a(0.6²)/2 = 3.34

    v₀ = -0.6a
    v₀ = (3.34 - 1.61 - 0.18a)/0.6 = (1.73/0.6 - 0.3a

    -0.6a = 1.73/0.6 - 0.3a
    0.3a = -1.73/0.6
    a = -1.73/0.18 = -173/18 m/s² ≈ -9.6111
    v₀ = -0.6(-173/18) = 173/30 m/s ≈ 5.7667

    Ans:
    Exact: y(t) = 1.61 + (173/30)t - (173/36)t²
    Appx: y(t) = 1.61 + 5.767t - 4.806t²

    Note:
    The initial height = 1.61, and maximum height = 3.34 at 0.6 seconds, are incompatible with a gravitational field of 1 g ≈ 9.8 m/s². The gravity at the location of this activity is ≈9.611 m/s²
    6 answers · 1 day ago