a math problem and plz show ur work!?

Question

its a quadratic:
10t^2-9t=1
what would t be?
and once again, pleas show your work.

Answer 1

10t^2-9t=1

10t^2 - 9t - 1 = 0

factor

1 t - 1

10 t + 1


(t-1)*(10t+1) = 0

t = 1 , -1/10

Answer: t = 1 , -1/10

*still confused? write back

Answer 2

10t^2-9t=1..........subtracting 1 from both sides:
10t^2 - 9t - 1 = 0
The product of the coefficient of t^2 and the constant in the equation is -10. The factors of -10 that when added together will give the middle coefficient -9 is: -10 and +1. We use these coefficients to break - up the equation. We have:
10t^2 - 10t + t - 1 = 0.
Grouping these terms and factoring out common terms we have:
10t(t - 1) +1(t - 1) = 0.
Since t - 1 is common in the two brackets we choose one of them and combine the terms outside of the parenthesis:
(t - 1)(10t + 1) = 0.
t - 1 = 0 or 10t + 1 = 0.
t = 1. or t = -1/10 = -0.1
t = 1 or t = -0.1.

Answer 3

Hello,

Put all the terms on the left side giving us.

10t^2 - 9t -1 = 0

Now to factor we must have the numbers that multiply to be -10, (10)*(-1) and add to be -9. These are -10 and 1 so rewrite and we have 10t^2 -10t + 1t -1 now factor (10(t - 1)+1(t-1)) = 0 so (10t +1) (t-1) = 0 then t = -1/10 and t = 1.


Checking these in the original equation
10*(-1/10)^2 - 9*(-1/10)
1/10 + 9/10 = 1 This one checks
10*(1)^2 - 9*1
10 -9 = 1 and This one checks.

Hope This Helps!!

Answer 4

10t^2-9t-1 = 0

Delta = (-9)^2 - 4(-1)(10) = 121

Delta is positive so there are two solutions

t1 = [9 + sqrt(121)] / 20 = 1
t2 = [9 - sqrt(121)] / 20 = -1/10

Note: Here is how the discriminant works. Let's say the equation is ax^2 + bx + c = 0
delta = b^2 - 4*a*c

When it is positive then there is two solutions
x1 = [-b + sqrt(delta)] / 2a
x2 = [-b - sqrt(delta)] / 2a

When it's negative then there are no solutions.
And when it's equal to 0 then
x1 = x2 = -b / 2a

Answer 5

10t^2 - 9t -1 =0
Discrimiant = 9^2 -4(-1)(10) = 81 +40 =121 = 11^2

t1 = (9 +11)/20 = 1
t2 = (9 -11)/20 = -2/20 =-1/10

Answer 6

So this is a basic algebra problem where you need to set your function equal to zero and factor it to solve for the value of t.

10t^2-9t-1=0

(10t+1)(t-1)=0

so:
10t+1=0
10t=-1
t= -1/10

and t-1=0
t=1

it's that simple!!

Answer 7

Well, I figure the speed of the boat minus the speed of the current would result in the speed of the boat going upriver, which was 16 km/hr. Boat speed plus current speed is the speed going downriver, 24 km/hr. With a velocity difference of 8 km/hr. The speed of the boat in still water would be midway between the 16 and the 24 because you have an added vector in one case and a subtracted vector (same magnitude) in the other. 8 km/hr divided by two is 4 km/hr (CURRENT SPEED). The midway of the different velocities, 20 km/hr, would be the BOAT SPEED.