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bobobox
 one year ago
I have seven math questions I need help answering please, It's my very very last assignement of the year and I need to pass.
bobobox
 one year ago
I have seven math questions I need help answering please, It's my very very last assignement of the year and I need to pass.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm quite sure the answer is C on that one

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3dw:1435638657119:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.31. \((3 + 2x)(5 + 2x) = 19.25\) \((2x + 3)(2x + 5) = 19.25\) \(4x^2 + 16x + 15 = 19.25\) \(4x^2 + 16x  4.25 = 0\)

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2Yay thanks so much for the explanation! It helps!

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2Could you help me with the soccer one?

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2@mathstudent55 @willb1220

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3The acceleration due to gravity is g in the down direction, so let;'s call it g. \(a = g\) The velocity is \(v(t) = gt + v_0\) where \(v_0\) is the initial velocity. The position (height) is \(h = \dfrac{1}{2}gt^2 + v_0t + h_0\) where \(h_0\) is the initial height.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3We are told that the initial position is 0, so \(h_0 = 0\) We are also told the initial velocity is 42 ft/sec, so \(v_0 = 42\)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3Also, g = 32 ft/sec^2

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3\(h(t) = 16t^2 + 42t \) This is an inverted parabola. We can take the first derivative to find where the height is maximum: \(h'(t) = 32t + 42\) Now we set the derivative equal to zero and solve for t: \(32t + 42 = 0\) \(32t = 42\) \(t = 1.3125~sec\) At 1.3125 sec the ball reaches maximum height.

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2So a since you round it

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2Thank you for being awesome and walking me through it, do you think you could help me with the next?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3Now we plug in that time in the position (height) equation to find at what height the ball is at 1.3125 sec: \(h(t) = 16t^2 + 42t\) \(h(1.3125) = 16(1.3125)^2 + 42(1.3125) = 27.56\)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3Notice we first found the time was 1.3 sec. The question is what is the height, not the time, so we need to take the time in the position equation and find the position at that time. The answer is 27.6 ft, not 1.3 sec.

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2COuld you help with another? It's simelar!

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3It's so similar it's exactly the same. The answer is the same.

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2it's 27.6? It's not the same question though

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3Sorry. I read it too fast. The answer is the 1.3 sec we already got before.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3I finally read the whole question correctly.

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2okay thanks! Can I pester you to help me with the last four questions? I really appreciate How hard you work to help others.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3Look at this graph: dw:1435640162113:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3dw:1435640215914:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3The ball is kicked at (0, 0), where time is zero, and the height of the ball is zero.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3As time goes by, the ball goes higher. It reaches a maximum height at 1.3 sec. dw:1435640295295:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3Notice that the shape of the curve is an inverted parabola. That means it is symmetric left to right. The amount of time it took for the ball to reach maximum height is the same amount of time it will take for the ball to hit the ground on the way down.

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2okay so in total the ball was in the air for 2.6 seconds

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3dw:1435640433043:dw

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2thanks can I pester you to help with the last four?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3We can get this answer this way bec we already solved the part about maximum height. It's simply double the time it took to reach maximum height.

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2Cool ill post the next

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2A golf ball is hit from the ground with an initial velocity of 208 feet per second. Assume the starting height of the ball is 0 feet. How long will it take the golf ball to hit the ground? (1 point) 13 secs 21 secs 42 secs 6 secs

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3If we had not solved that part, then we could solve this problem on its own. The equation of the position is \(h(t) = 16t^2 + 42t\) The height is 0 at the beginning (when the ball is kicked) and at the end (when the ball falls on the ground) \(0 = 16t^2 + 42t\) \(t(16t + 42) = 0\) \(t = 0\) or \(16t + 42 = 0\) \(t = 0\) or \(16t = 42\) \(t = 0\) or \(t = 2.6\) The ball is at zero height at 0 sec and at 2.6 sec. That means it was in the air for 2.6 sec.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3Use the general equation of height we have: \(h(t) = 16t^2 + v_0t + h_0\)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3\(h_0 = 0\) \(v_0 = 208\) Plug these values in and solve the equation for t.

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2I tried I got a fraction so I am not even sure I did it right

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3\(h(t) = 16t^2 + v_0t + h_0\) \(h(t) = 16t^2 + 208t = 0\) \(16t^2 + 208t = 0\) \(2t^2 + 26t = 0\) \(t(2t + 26) = 0\) \(t = 0\) or \(2t + 26 = 0\) \(~~~~~~~~~~~~~~~~~~2t = 26\) \(~~~~~~~~~~~~~~~~~~~~~~~~~t = 13\)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3The ball is at height zero at 0 seconds, when it is hit, and at 13 seconds when if falls on the ground.

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2cool thanks! only three more left!

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2A particle is moving along a projectile path where the initial height is 160 feet with an initial speed of 144 feet per second. What is the maximum height of the particle? (1 point) 128 feet 224 feet 272 feet 484 feet

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2@mathstudent55 please help me I need you without you I might fail please please please

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2@mathstudent55 I will fan write a testimony and I will give you a medal please

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2@willb1220 help please help

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3We use the same height equation again, and we plug in the info of this problem. Here we have an initial position.

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2is the answer A? because I tried figuring it out

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3\(h(t) = 16t^2 + v_0t + h_0\) \(h(t) = 16t^2 + 144t + 160\) \(h'(t) = 32t + 144\) \(32t + 144 = 0\) \(32t = 144\) \(t = 4.5~sec\) Maximum height is achieved at 4.5 sec. Now we plug in 4.5 sec in the height equation to find the actual height.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3\(h(t) = 16(4.5)^2 + 144(4.5) + 160\)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3What DDDDDDDdo you get?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3You DDDDDDid get the correct answer.

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2yay! Okay just two more questions okay?

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2I think it's the one that I choose

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2Yay the last one I am not to sure about

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3dw:1435642461182:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3Can you solve it for c^2? Since c^2 is being multiplied by m, divide both sides by m. dw:1435642705027:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3dw:1435642763475:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3Now you want c but you have c squared. The opposite of squaring is taking the square root, so you take the square root of both sides: dw:1435642840101:dw

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2I just want to say thank you, I don't think I could have said it enough, really

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3You're welcome. I hope you pass & graduate.

bobobox
 one year ago
Best ResponseYou've already chosen the best response.2I got a 100%!!!!!!!!!!!!!!!!!!!!!!!!! @mathstudent55
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