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gumstopper99 Group TitleBest ResponseYou've already chosen the best response.0
Oops . 7t(t4) ??
 one year ago

help123please. Group TitleBest ResponseYou've already chosen the best response.1
Hi,batman.
 one year ago

DDCamp Group TitleBest ResponseYou've already chosen the best response.0
No, that would multiply to 7t²  28t. Have you learned the quadratic equation yet?
 one year ago

ladyrosebud Group TitleBest ResponseYou've already chosen the best response.0
(2t+3)(3t5)
 one year ago

gumstopper99 Group TitleBest ResponseYou've already chosen the best response.0
@ladyrosebud how do you figure it out?
 one year ago

galacticwavesXX Group TitleBest ResponseYou've already chosen the best response.0
yeah @ladyrosebud is correct
 one year ago

gumstopper99 Group TitleBest ResponseYou've already chosen the best response.0
I made an error :( its 7t^228t
 one year ago

ladyrosebud Group TitleBest ResponseYou've already chosen the best response.0
7t(t4)
 one year ago

rajee_sam Group TitleBest ResponseYou've already chosen the best response.0
Ok can I help?
 one year ago

gumstopper99 Group TitleBest ResponseYou've already chosen the best response.0
Yes, thats what I thought
 one year ago

kyusakazaki Group TitleBest ResponseYou've already chosen the best response.0
it's (t5/3)(t+3/2)
 one year ago

rajee_sam Group TitleBest ResponseYou've already chosen the best response.0
We can do this using a method called "Product  Sum" Method The standard form of a Quadratic Function is \[Ax ^{2} + Bx + C\] We have to find two numbers such that their Product is A.C and their Sum is B In the given function \[6t ^{2}t15\] A = 6; B = 1 and C = 15 I have to find two numbers that I multiply should give me 90 and add them should give me 1. So Product is 6 x 15 = 90 Sum is 1 dw:1368671725295:dw Now set up a table like this and list all product pairs for 90.
 one year ago

rajee_sam Group TitleBest ResponseYou've already chosen the best response.0
dw:1368671986360:dw
 one year ago

rajee_sam Group TitleBest ResponseYou've already chosen the best response.0
Now my product is negative so one number should be positive and the other number should be negative. now when I look at all the product pair \[\pm 9 \space and \space \pm 10\] will give me 1 if '1' if I add them one being negative and one being positive. dw:1368672235371:dw So the numbers are either +9 and 10 or 9 and +10. Since the sum is 1 the numbers are +9 and 10.
 one year ago

rajee_sam Group TitleBest ResponseYou've already chosen the best response.0
Since my A is not equal to 1 I have to rewrite the function with these two numbers. \[6t ^{2}  t 15 = 6t ^{2} + 9t + 10t  15\] Now I group them. \[(6t ^{2} + 9t ) + ( 10t  15 )\] Now the GCF of the first group is 3t if I take that out, it will be \[3t(2t + 3) + (10t 15)\] Now I have to take a GCF out of the second group such that I get a (2t + 3) with them too. so if I take a '5' out I will get a 2t + 3 . So I do\[3t(2t + 3) + 5 (2t + 3) = 3t (2t + 3) 5 (2t + 3)\] Now the common factor in both terms is (2t + 3), Doing distributive property in reverse, you will get \[(2t + 3) ( 3t  5 )\]
 one year ago

galacticwavesXX Group TitleBest ResponseYou've already chosen the best response.0
the answer (2t+3)(3t5) was answered wayyy long ago. factoring this isn't that complexed
 one year ago

rajee_sam Group TitleBest ResponseYou've already chosen the best response.0
Gummy asked how did lady figured it out? that is why I tried explaining my bad.
 one year ago

galacticwavesXX Group TitleBest ResponseYou've already chosen the best response.0
lol my bad i just kept seeing people solving and solving for the expression so i was like it shouldn't take this long
 one year ago
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