Write the equation of the line in point-slope form

Mathematics

1 answer:

Nostrana [21]3 years ago

8 0

Answer:

The slope of this line is -1/2

Step-by-step explanation:

In order to find the value of any slope, you use the slope formula.

m (slope) = (y1 - y2)/(x1 - x2)

Then we pick two points in order to plug in. For the purpose of this graph, we'll use (0, -1) and (2, -2)

m (slope) = (y1 - y2)/(x1 - x2)

m (slope) = (-1 - -2)/(0 - 2)

m (slope) = 1/-2

m (slope) = -1/2

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PLEASE HELP! (1/4) - 50 POINTS -

Annette [7]

Answer:

D) 0.35

Step-by-step explanation:

The table gives the area between z=0 and the given magnitude of z. That is, the area between z = 0 and z = -0.6 is 0.23, as found in the 0.6 column of the table. Similarly, the area between z = 0 and z = 0.3 is 0.12, as found in the 0.3 column of the table.

The area between z = -0.6 and z = +0.3 is the sum of these areas:

p(-.6<z<.3) = 0.23 +0.12 = 0.35

5 0

3 years ago

A rocket is fired upward from some initial distance above the ground. Its height in feet, h,above the ground, t seconds after

kirill115 [55]

Check the picture below.

$\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
h(t) = &{{ -16}}t^2&{{ +96}}t&{{ +2560}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

so.. the maximum point is at $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}~~~ft$

and it happened after $\bf -\cfrac{{{ b}}}{2{{ a}}}~~~seconds$

--------------------------------------------------------------------------

now, when the rocket reaches the ground, as you saw on the picture, y = 0, thus

$\bf h(t)=-16t^2+96t+2560\implies 0=-16t^2+96t+2560
\\\\\\
0=-16(t^2-6t-160)\implies 0=t^2-6t-160
\\\\\\
0=(t-16)(t+10)\implies t=
\begin{cases}
\boxed{16}\\
-10
\end{cases}$

"t" is an amount in seconds, for this specific phenomena, it cannot be negative, so -10 is not a feasible value.