Question

can anybody help me with the connexus algebra 2 lesson 11 unit 4?

Answer 1

if you post a question, it would be easier to help ....

Answer 2

a football is kicked into the air from an initial height of 3 feet. the height, in feet, of the football above the ground is given by s(t)=-16t^2+35t+3... where t is time in seconds and t>=0. which is closest to the time when the football will be 20ft above the ground??

Answer 3

2.72 seconds

0.73 seconds or 1.46 seconds

0.53 seconds

0.53 seconds or 2.72 seconds

Answer 4

do you have any ideas on how to approach the solution?

Answer 5

no i do not, im struggling BIG TIME with algebra

Answer 6

one way to approach the solution will depend on knowing the quadratic formula

Answer 7

i dont understand how the question is, and how to use the quadratic formula with it.

Answer 8

using the quadratic formula is a simple plug and compute operation ....

the question gives us a quadratic equation the defines the height at any moment in time (t).

to determine when that equation is equal to 20, set it equal to 20...

s(t)=-16t^2+35t+3 ; when does s(t)= 20?

20 = -16t^2+35t+3

subtract 20 from both sides: 0 = -16t^2+35t -17

now we have a format that we can use the quadratic equation with

Answer 9

given a setup: \( ax^2+bx+c=0\)

the quadratic formula defines the variable x as:\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 10

thats pretty good, missed a b^2 is all i see

Answer 11

yea i noticed that lol

Answer 12

but that confuses me still .
i understand how to do it when my teacher goes over it,
but then when i do it myself i dont get it

Answer 13

0 = -16t^2+35t -17


(-35+sqrt(35^2-4(-16)(-17)))\(2(-16))

http://www.wolframalpha.com/input/?i=%28-35%2Bsqrt%2835%5E2-4%28-16%29%28-17%29%29%29%5C%282%28-16%29%29


(-35-sqrt(35^2-4(-16)(-17)))\(2(-16))

http://www.wolframalpha.com/input/?i=%28-35-sqrt%2835%5E2-4%28-16%29%28-17%29%29%29%5C%282%28-16%29%29

Answer 14

the proof behind the quadratic formula is developed from a method called; completing the square

Answer 15

\[0=ax^2+bx+c\]
\[0=a(x^2+\frac bax+\frac ca)\]
\[0=x^2+\frac bax+\frac ca\]
\[\frac{b^2}{4a^2}-\frac ca=x^2+\frac bax+\frac{b^2}{4a^2}\]
\[\frac{b^2}{4a^2}-\frac ca=(x+\frac ba)^2\]
\[\pm\sqrt{\frac{b^2}{4a^2}-\frac ca}=x+\frac ba\]
\[-\frac{b}{a}\pm\sqrt{\frac{b^2}{4a^2}-\frac ca}=x\]
to put it into its "usual" form you can then paly with the adding the fractions and simplifying

Answer 16

i dropped a 2 along the way trying to code it nice :/

Answer 17

\[\frac{b^2}{4a^2}-\frac ca=(x+\frac ba)^2\] ^^
2a


\[\frac{b^2}{4a^2}-\frac ca=(x+\frac b{2a})^2\]
and then it works out as expected :)

Answer 18

ok. ill try it. ty for ur help .

Answer 19

good luck