Tutors Answer Your Questions about Quadratic Equations (FREE)
Question 111036: The Height in feet of a baseball above ground is given by h(t)=-16t^2+32t+3 where t represents the time in seconds after the baseball is thrown.
If you throw the ball into the air, how many seconds have passed until the ball hits the ground??
I really don't understand how to put this problem in the equations it's supposed to go into and I don't understand how to solve them...help!
Click here to see answer by scott8148(6628)  |
Question 111036: The Height in feet of a baseball above ground is given by h(t)=-16t^2+32t+3 where t represents the time in seconds after the baseball is thrown.
If you throw the ball into the air, how many seconds have passed until the ball hits the ground??
I really don't understand how to put this problem in the equations it's supposed to go into and I don't understand how to solve them...help!
Click here to see answer by ankor@dixie-net.com(15652)  |
Question 111052: Please help! Solve the following exponential equation:
26 over(/)1-2e^6x = 50
I tried multiplying the denominator on the left & right, and dividing by the 50 to get this far: 26/50 = 1-2e^6x
I tried subtracting the one and multiplying the -2, but I wound up with Ln48/50 = 6x. My answers to that are wrong. What am I missing? How do you do this problem if I'm wrong?
Click here to see answer by scott8148(6628)  |
Question 111186: Hello,
In the proof of the quadractic equation, you begin by dividing both parts by "a".
ax2 (2 is the square)+ bx + c = 0
(divide both sides by a)
x2(2 is the square) + b/a.x + c/a = 0
(then suddenly)
x2(2 is the square) 2.(b/2.a).x = c/a = 0
Then all of a sudden, a "2" appears. Where did that come from?
Thank you for your help.
Barbara
Click here to see answer by Earlsdon(6287) |
Question 111223: I am graphing a parabola and I have all of my points done but I'm having a problem getting the x & y vertex's....the equation is 3x²-x-1=0 ...I have 1.5 for x but I'm told this is wrong and I have no idea how to solve it for the y vertex. Please help!
Thanks so much!
B. Cook
Click here to see answer by solver91311(16885)  |
Question 111223: I am graphing a parabola and I have all of my points done but I'm having a problem getting the x & y vertex's....the equation is 3x²-x-1=0 ...I have 1.5 for x but I'm told this is wrong and I have no idea how to solve it for the y vertex. Please help!
Thanks so much!
B. Cook
Click here to see answer by checkley71(8403) |
Question 111414: Could you please help me with an application using a quadratic equation?
The question is: The width of a rectangular parking lot is 51 ft less than its length. Determine the dimensions of the parking lot if it measures 250 ft diagonally.
What I've got so far:
x^2+(x-51)^2=250^2
x^2+x^2-102x+2601=62500
2x^2-102x-59899=0
factor out 2
x^2-51x-29949.5=0
I stopped there because I got a decimal number (29949.5) and that seems wrong.
Please help!
Thank you in advance!
Brenda
Click here to see answer by Earlsdon(6287) |
Question 111413: Could you please help me with this application using a quadratic equation?
W=sqrt1/(LC) solve for L (an electricty formula)
I honestly don't even know where to begin...please help!
Thank you in advance!
Brenda
Click here to see answer by Earlsdon(6287) |
Question 111421: Could you please help me with an application question involving a quadratic equation?
The question is: On a sales trip, Gail drives the 600 mi to Richmond at a certain speed. The return trip is made at a speed that is 10 mph slower. Total time for the round trip is 22 hr. How fast did Gail travel on each part of the trip?
This is what I've got so far:
t[1]=600/r
t[2]=600/(r-10)
(600/r)+(600/(r-10))=22
r(r-10)(600/r)+r(r-10)(600/(r-10))=r(r-10)22
600(r-10)+600r=22r(r-10)
600r-6000+600r=22r^2-220r
22r^2-1420r+6000=0
I know at this point to plug it into quadratic equation form... but when factoring this I get decimal numbers, so it seems wrong.
-(-64.545)+-(sqrt(-64.545)^2-4(1)(272.727))/2(1)
64.545+-(sqrt4166.057-1090.908)/2
64.545+-(sqrt3075.149)/2
(64.545+55.454)/2=60
or
(64.545-55.454)/2=4.5
Please help!
Thank you in advance for your help!
Brenda
Click here to see answer by jim_thompson5910(28595) |
Question 111388: How do I write a quadratic equation for a parabola given the vertex and the y-intercept?
I am also attempting to determine the values of "a" and "b" in the equation.
Ordered pairs are as follows:
(0,53)- y-intercept
(2,35)- vertex
(4,42)
(6,54)
(8,82)
My textbook ISBN is 0-13-175204-9
Thanks in advance!
Click here to see answer by scott8148(6628)  |
Question 111412: Could you please help me with this application involving a quadratic equation?
s=v[0]t+(gt^2)/2 solve for t (a motion formula)
First I got rid of the fraction by mulitplying the equation by 2
2s=2v[0]t=gt^2
2s/2v[0]t-g=t^2
sqrt 2s/2v[0]t - g = sqrt t^2
sqrt 2s/2v[0]t - g = t
I have a feeling I'm completely off... but I tried. Please help!
Thank you in advance!
Brenda
Click here to see answer by Fombitz(13828)  |
Question 111604: Will you please help me solve the following quadratic equation?
m=(m[0])/sqrt(1-(v^2/c^2)) solving for c
This is what I've attempted...
multiply both sides by the LCM sqrt1-(v^2/c^2)to get rid of the fraction
(m)qrt1-(v^2/c^2)=m[0]
then multiply both sides again to get rid of the fraction in the sqrt?
(mc^2)sqrtc^2-v^2=m[0]
then divide both sides by (m)sqrt(c^2-v^2)
c^2=(m[0])/(m)sqrt(c^2-v^2)
square both sides with a final answer of
c=sqrt(m[0])/(m)sqrt(c^2-v^2)
Thank you in advance for your help!!
Brenda
Click here to see answer by stanbon(57347) |
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