algebra 2 help?

Question

please help with these problems:

solve the inequality 3(1+ x) >1+5x

the equation d= 4.9t^2 gives the distance in meters, an object falls in t seconds. how many seconds would it take for an object to fall 472 meters?

let f(x)= 4-x^2 g(x)= 2-x find f(x) + g(x)

Answer 1

1) 3(1 + x) > 1 + 5x

First thing you have to do is expand the left hand side.

3 + 3x > 1 + 5x

Move everything with a x on the left hand side; everything else goes on the right hand side.

3x - 5x > 1 - 3

Simplify,

-2x > -2

Now, here's the part you have to remember; whenever you multiply or divide by a negative number in an inequality, the sign flips. We're going to divide both sides by -2, flip the sign, and obtain

x < 1

2) d = 4.9t^2

You're given the value of d, which is 472. All you have to do is plug this in and solve for t.

472 = 4.9(t^2)

472/(4.9) = t^2
4720/49 = t^2

To solve for t, you have to take the square root of both sides. Note that you have to add a "plus or minus", or +/-, whenever you take the square root of both sides of an equation.

t = +/- sqrt(4720/49)

But, given that time can never be negative, we can reject the negative answer.

t = sqrt(4720/49)

We can simplify this to

t = sqrt(4720)/sqrt(49) = sqrt(4720)/7

t = (4/7)sqrt(295)

3) f(x) = 4 - x^2, g(x) = 2 - x

f(x) + g(x) = (4 - x^2) + (2 - x)
= 4 - x^2 + 2 - x
= 6 - x^2 - x

Answer 2

3(1+x) > 1+ 5x
3+3x> 1+5x
3x - 5x>1-3
-2x > -2
x < 1


when d=472, the equation becomes:
472 = 4.9 t ^ 2
t^2 =472/4.9
t= √(472/4.9)
t=9.81s



∵f(x)=4-x^2 and g(x)=2-x
∴f(x)+g(x) = (4-x^2) + (2-x)
= 4-x^2 +2-x
= -x^2 -x +6

Answer 3

3+3x >1+5x;
3-1 > 5x-3x;
2 > 2x;
1 > x;


4.9*t^2 = 472; t = abs ( sqrt(472/4.9) ); t = 9.81 s. approx
We only take the positive value because there isn t negative seconds.

f(x) + g(x) = 4-x^2 + 2-x = x^2 - x + 6;

Answer 4

X<1

on the seccond one..you replace d with 472..then divde both sides by 4.9
then you square root both sides..and there you have t