Single Gay boys onlyI’m 137th gradeand i need help with this problem and I’m yours! and maybe brainlistwhat’s 2 7/9

Single Gay boys only I'm 13
7th grade
and i need help with this problem and I'm yours! and maybe brainlist
what's
2 7/9 times 3/2 times 7/9 ?

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  1. Step-by-step explanation:

    Question 6)

    sin Y= m

    sin Y = m/1

    So, hypotenuse is 1

    Since sine is opposite over hypotenuse

    So XZ= m and YZ = 1

    Similarly, cos Y = k

    cos Y = k/1

    So adjacent side of angle Y is k

    So XY = k

    cos z - sin z = [tex]\frac{XZ }{YZ } - \frac{XY}{YZ}[/tex]

    cos z - sin z = [tex]\frac{m }{1 } - \frac{k}{1}[/tex]

    cos z - sin z = m - k

    Question 7)

    the relationship between sine, cosine, and tangent.

    tan(x) = sin(x)/cos(x) = (11/61)/(60/61)

    tan(x) = 11/60

    Question 8)

    Start with where the shorter leg is. It must be opposite the smallest angle.

    In a 30 - 60 - 90 degree triangle you have the hypotenuse to be twice as long as the shortest side. You have to read that a couple of times to make sure you understand it. 

    That being said, if the shortest side is x, the hypotenuse will be 2x.

    Since in this case the shortest side is 11, the hypotenuse will be 2*11 = 22

    The answer is 22

  2. Step-by-step explanation:

    I believe you meant y = f(x) = a(x - h)^2 + k.  This is the vertex form of the equation of a vertical parabola (vertex is located at (h, k) ).

    We start with the simplest vertical parabola, y = x^2.  This parabola has its vertex at the origin, (0, 0).

    13. Translated 2 units left and 7 units down:  Substitute -7 for k and -2 for h:

         y - (-7) = a (x - (-2) )^2, or (in simpler form), y + 7 = a(x + 2)^2

    14. Translated 5 unit right, 2 units up, and stretched by a factor of 3.

    Substitute 5 for h and 2 for k, and let z = 3:

    y - 2 = 3(x - 5)^2

    Please leave out "I'm horrible at math" comments.  Certainly you have the ability to learn this material.

  3. btw I am straighand I have bf. I thought u needed help on this math question so answered it for u. Wish u luck and I hope u find your Valentine's love

    Explanation:

    Exact form: 175/54

    Decimal form: 3.2407

    Mixed number: 3 13/54

  4. 1a) Bill and the dog must have a speed of 13.0 m/s

    1b) The speed of the dog must be 22.5 m/s

    2a) The ball passes over the outfielder's head at 3.33 s

    2b) The ball passes 1.2 m above the glove

    2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

    2d) One solution is when the player is jumping up, the other solution is when the player is falling down

    Explanation:

    1a)

    The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

    - A uniform motion (constant velocity) along the horizontal direction

    - An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

    In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

    The ball's initial speed is

    u = 15 m/s

    And the angle of projection is

    [tex]\theta=30^{\circ}[/tex]

    So, the ball's horizontal velocity is

    [tex]v_x = u cos \theta = (15)(cos 30)=13.0 m/s[/tex]

    And therefore, Bill and the dog must have this speed.

    1b)

    For this part, we have to consider the vertical motion of the ball first.

    The vertical position of the ball at time t is given by

    [tex]y=u_yt+\frac{1}{2}at^2[/tex]

    where

    [tex]u_y = u sin \theta = (15)(sin 30) = 7.5 m/s[/tex] is the initial vertical velocity

    [tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity

    The ball is at a position of y = 2 m above the ground when:

    [tex]2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0[/tex]

    Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

    The horizontal distance covered by the ball during this time is

    [tex]d=v_x t =(13.0)(1.19)=15.5 m[/tex]

    The dog must be there 0.5 s before, so at a time

    t' = t - 0.5 = 0.69 s

    So, the speed of the dog must be

    [tex]v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s[/tex]

    2a)

    Here we just need to consider the horizontal motion of the ball.

    The horizontal distance covered is

    [tex]d=98 m[/tex]

    while the horizontal velocity of the ball is

    [tex]v_x = u cos \theta = (34)(cos 30)=29.4 m/s[/tex]

    where u = 34 m/s is the initial speed.

    So, the time taken for the ball to cover this distance is

    [tex]t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s[/tex]

    2b)

    Here we need to calculate the vertical position of the ball at t = 3.33 s.

    The vertical position is given by

    [tex]y= h + u_y t + \frac{1}{2}at^2[/tex]

    where

    h = 1.2 m is the initial height

    [tex]u_y = u sin \theta = (34)(sin 30)=17.0 m/s[/tex] is the initial vertical velocity

    [tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity

    Substituting t = 3.33 s,

    [tex]y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m[/tex]

    And sinc the glove is at a height of y' = 2.3 m, the difference in height is

    y - y' = 3.5 - 2.3 = 1.2 m

    2c)

    In order to intercept the ball, he jumps upward at a vertical speed of

    [tex]u_y' = 7 m/s[/tex]

    So its position of the glove at time t' is

    [tex]y'= h' + u_y' t' + \frac{1}{2}at'^2[/tex]

    where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

    y' = y = 3.5 m (the height of the ball)

    Substituting and solving for t', we find

    [tex]3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0[/tex]

    Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

    [tex]t'' = t -t'[/tex]

    So we have two solutions:

    [tex]t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s[/tex]

    So, the player can jump after 2.10 s or after 3.13 s.

    2d)

    The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he  falls down.

    Therefore, the two solutions corresponds to the two different part of the motion.

    The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

    On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

    Learn more about projectile motion:

    #LearnwithBrainly

  5. I don’t know the answer and I’m only here for the point hahaha free
    Oof incorrect answer
    Oofofof
    I Hi wait to
    I need to make a point to max out of the game and then I can get emerald green green and crystal purple orange brown orange orange green
    I need to

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