bobobox
  • bobobox
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
bobobox
  • bobobox
anonymous
  • anonymous
Hit me with em (;
anonymous
  • anonymous
I'm quite sure the answer is C on that one

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bobobox
  • bobobox
okay
bobobox
  • bobobox
mathstudent55
  • mathstudent55
|dw:1435638657119:dw|
mathstudent55
  • mathstudent55
1. \((3 + 2x)(5 + 2x) = 19.25\) \((2x + 3)(2x + 5) = 19.25\) \(4x^2 + 16x + 15 = 19.25\) \(4x^2 + 16x - 4.25 = 0\)
bobobox
  • bobobox
Yay thanks so much for the explanation! It helps!
bobobox
  • bobobox
Could you help me with the soccer one?
bobobox
  • bobobox
@mathstudent55 @willb1220
mathstudent55
  • mathstudent55
The 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.
bobobox
  • bobobox
okay
mathstudent55
  • mathstudent55
We are told that the initial position is 0, so \(h_0 = 0\) We are also told the initial velocity is 42 ft/sec, so \(v_0 = 42\)
bobobox
  • bobobox
yeah
mathstudent55
  • mathstudent55
Also, g = 32 ft/sec^2
bobobox
  • bobobox
yup
mathstudent55
  • mathstudent55
\(h(t) = -16t^2 + 42t \) 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
  • bobobox
So a since you round it
bobobox
  • bobobox
Thank you for being awesome and walking me through it, do you think you could help me with the next?
mathstudent55
  • mathstudent55
Now we plug in that time in the position (height) equation to find at what height the ball is at 1.3125 sec: \(h(t) = -16t^2 + 42t\) \(h(1.3125) = -16(1.3125)^2 + 42(1.3125) = 27.56\)
bobobox
  • bobobox
oh wow
mathstudent55
  • mathstudent55
Notice 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
  • bobobox
Thank you
bobobox
  • bobobox
COuld you help with another? It's simelar!
mathstudent55
  • mathstudent55
It's so similar it's exactly the same. The answer is the same.
bobobox
  • bobobox
it's 27.6? It's not the same question though
mathstudent55
  • mathstudent55
Sorry. I read it too fast. The answer is the 1.3 sec we already got before.
mathstudent55
  • mathstudent55
No. Wait.
mathstudent55
  • mathstudent55
I finally read the whole question correctly.
bobobox
  • bobobox
okay thanks! Can I pester you to help me with the last four questions? I really appreciate How hard you work to help others.
mathstudent55
  • mathstudent55
Look at this graph: |dw:1435640162113:dw|
mathstudent55
  • mathstudent55
|dw:1435640215914:dw|
mathstudent55
  • mathstudent55
The ball is kicked at (0, 0), where time is zero, and the height of the ball is zero.
bobobox
  • bobobox
okay
mathstudent55
  • mathstudent55
As time goes by, the ball goes higher. It reaches a maximum height at 1.3 sec. |dw:1435640295295:dw|
bobobox
  • bobobox
so is it 1.3?
mathstudent55
  • mathstudent55
Notice 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
  • bobobox
okay so in total the ball was in the air for 2.6 seconds
mathstudent55
  • mathstudent55
|dw:1435640433043:dw|
mathstudent55
  • mathstudent55
Correct.
bobobox
  • bobobox
thanks can I pester you to help with the last four?
mathstudent55
  • mathstudent55
We 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
  • bobobox
Cool ill post the next
bobobox
  • bobobox
A 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
  • mathstudent55
If 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.
bobobox
  • bobobox
yup :)
mathstudent55
  • mathstudent55
Use the general equation of height we have: \(h(t) = -16t^2 + v_0t + h_0\)
mathstudent55
  • mathstudent55
\(h_0 = 0\) \(v_0 = 208\) Plug these values in and solve the equation for t.
bobobox
  • bobobox
okay one sec
bobobox
  • bobobox
I tried I got a fraction so I am not even sure I did it right
mathstudent55
  • mathstudent55
\(h(t) = -16t^2 + v_0t + h_0\) \(h(t) = -16t^2 + 208t = 0\) \(-16t^2 + 208t = 0\) \(-2t^2 + 26t = 0\) \(t(-2t + 26) = 0\) \(t = 0\) or \(-2t + 26 = 0\) \(~~~~~~~~~~~~~~~~~~-2t = -26\) \(~~~~~~~~~~~~~~~~~~~~~~~~~t = 13\)
bobobox
  • bobobox
oohhhhh
bobobox
  • bobobox
so is 13 our answer?
mathstudent55
  • mathstudent55
The ball is at height zero at 0 seconds, when it is hit, and at 13 seconds when if falls on the ground.
bobobox
  • bobobox
cool thanks! only three more left!
bobobox
  • bobobox
A 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
  • bobobox
@mathstudent55 please help me I need you without you I might fail please please please
bobobox
  • bobobox
@mathstudent55 I will fan write a testimony and I will give you a medal please
bobobox
  • bobobox
@willb1220 help please help
mathstudent55
  • mathstudent55
We use the same height equation again, and we plug in the info of this problem. Here we have an initial position.
bobobox
  • bobobox
is the answer A? because I tried figuring it out
bobobox
  • bobobox
crap no I meant d
mathstudent55
  • mathstudent55
\(h(t) = -16t^2 + v_0t + h_0\) \(h(t) = -16t^2 + 144t + 160\) \(h'(t) = -32t + 144\) \(-32t + 144 = 0\) \(-32t = -144\) \(t = 4.5~sec\) 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
  • mathstudent55
\(h(t) = -16(4.5)^2 + 144(4.5) + 160\)
mathstudent55
  • mathstudent55
What DDDDDDDdo you get?
bobobox
  • bobobox
let me see
bobobox
  • bobobox
484
mathstudent55
  • mathstudent55
You DDDDDDid get the correct answer.
bobobox
  • bobobox
yay! Okay just two more questions okay?
mathstudent55
  • mathstudent55
ok
bobobox
  • bobobox
I think it's the one that I choose
1 Attachment
bobobox
  • bobobox
@mathstudent55
mathstudent55
  • mathstudent55
correct
bobobox
  • bobobox
Yay the last one I am not to sure about
mathstudent55
  • mathstudent55
|dw:1435642461182:dw|
bobobox
  • bobobox
Help
1 Attachment
bobobox
  • bobobox
@mathstudent55
mathstudent55
  • mathstudent55
Can you solve it for c^2? Since c^2 is being multiplied by m, divide both sides by m. |dw:1435642705027:dw|
mathstudent55
  • mathstudent55
|dw:1435642763475:dw|
mathstudent55
  • mathstudent55
ok so far?
bobobox
  • bobobox
omg yes
mathstudent55
  • mathstudent55
Now 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
  • bobobox
I just want to say thank you, I don't think I could have said it enough, really
mathstudent55
  • mathstudent55
You're welcome. I hope you pass & graduate.
bobobox
  • bobobox
I got a 100%!!!!!!!!!!!!!!!!!!!!!!!!! @mathstudent55
mathstudent55
  • mathstudent55
Great!!!

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