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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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
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Hit me with em (;
I'm quite sure the answer is C on that one

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Other answers:

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

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