Question

PLEASE 25 POINTS!!!!!!

Answer 1

-4.9t²+7.5t+1.8=2.1
-4.9t²+7.5t-0.3=0

t=-7.5±√(56.25-4*-4.9*-.3)/-2*4.9

t=-7.5±√(44.49/-9.8

t=(-7.5±6.67)/-9.8

t=-.83/-9.8 = .085 or t= -14.17/-9.8 = 1.446

Answer 2

Answer:

The equation in standard form is: $-4.9t^{2} +7.5t-0.3=0$

The solutions for this equation are:

t1=0.0411

t2=1.4895

Step-by-step explanation:

The initial equation is: $-4.9t^{2} +7.5t+1.8=2.1$.

The standard form of a quadratic equation is:

$a*t^{2} +b*t+c=0$.

Where a, b and c are constants, to transform the initial equation into the standard form we should subtract from both sides of the equality the number 2.1 and simplify:

$-4.9t^{2} +7.5t+1.8=2.1$

$-4.9t^{2} +7.5t+1.8 - 2.1 = 2.1 - 2.1$

$-4.9t^{2} +7.5t+1.8 - 2.1 = 0$

$-4.9t^{2} +7.5t - 0.3 = 0$

From this equation: a=-4.9, b=7.5 and c=-0.3

The solutions of a quadratic equation is given by the expressions:

$t1=\frac{-b+\sqrt[2]{b^{2}- 4*a*c} }{2*a}$

$t2=\frac{-b-\sqrt[2]{b^{2}- 4*a*c} }{2*a}$

Replacing the values of a, b and c and solving, we get:

$t1=\frac{-7.5+\sqrt[2]{7.5^{2}- 4*(-4.9)*(-0.3)} }{2*(-4.9)}$

t1=0.04110

$t2=\frac{-7.5-\sqrt[2]{7.5^{2}- 4*(-4.9)*(-0.3)} }{2*(-4.9)}$

t2=1.4895

Finally, the solutions of the quadratic equation are t1=0.04110 and t2=1.4895