30) (19, 15), (5,11)<br> find the slope of the line through each pair points

First, let's rearrange the given equation into something more recognizable. If we add 13 to both sides, we now have the polynomial $x^{2}-10x+13$ . We can now use the quadratic formula to solve.

Remember that the quadratic formula is

$\frac{-b+/-\sqrt{b^{2}-4ac } }{2a}$

Substitute the numbers from the equation into the formula.

$\frac{-(-10)+/-\sqrt{(-10)^{2}-4(1)(13) } }{2(1)}$

Simplify:

$\frac{10+/-\sqrt{100-52} }{2}$

$\frac{10+/-\sqrt{48} }{2}$

Here, I'm going to assume that there was a mistype in option B because if we divide out the 2 we end up with $5+/-\sqrt{48}$ .

Hope this helps!

6 0

3 years ago

you drop a ball from a height of 98 feet. at the same time, your friend throws a ball upward. the polynomials represent the heig

Ostrovityanka [42]

a) $h_0 -u_y t$

b) See interpretation below

Step-by-step explanation:

a)

The motion of both balls is a free-fall motion: it means that the ball is acted upon the force of gravity only.

Therefore, this means that the motion of the ball is a uniformly accelerated motion, with constant acceleration equal to the acceleration of gravity:

$g=32 ft/s^2$

in the downward direction.

For the ball dropped from the initial height of $h_0 = 98 ft$ , the height at time t is given by

$h(t) = h_0 -\frac{1}{2}gt^2$ (1)

The ball which is thrown upward from the ground instead is fired with an initial vertical velocity $u_y$ , and its starting height is zero, so its position at time t is given by

$h'(t)=u_y t - \frac{1}{2}gt^2$ (2)

Therefore, the polynomial that represents the distance between the two balls is:

$h(t)-h'(t)=h_0 - \frac{1}{2}gt^2 - (u_y t - \frac{1}{2}gt^2) = h_0 -u_y t$

b)

Now we interpret this polynomial, which is:

$\Delta h(t) = h_0 -u_y t$

which represents the distance between the two balls at time t.

The interpretation of the two terms is the following:

- The constant term, $h_0$ , is the initial distance between the two balls, at time t=0 (in fact, the first ball is still at the top of the building, while the second ball is on the ground). For this problem, $h_0 = 98 ft$

- The coefficient of the linear term, $u_y$ , is the initial velocity of the second ball; this terms tells us that the distance between the two balls decreases every second by $u_y$ feet.

5 0

3 years ago

30) (19, 15), (5,11) find the slope of the line through each pair points