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dem82 [27]
2 years ago
6

Select the correct answer. A model rocket was launched from the top of a building. The rocket's approximate height can be modele

d by the quadratic equation h = -4.9t2 + 53.9t + 127.4, where h is the rocket's height, in meters, after t seconds. When factored, this equation is h = -4.9(t - 13)(t + 2). What is a reasonable time for it to take the rocket to land on the ground? Answer A 11 seconds Answer B 13 seconds Answer C 2 seconds Answer D 15 seconds
Mathematics
1 answer:
garri49 [273]2 years ago
6 0

Answer:

13 seconds.

Step-by-step explanation:

Given

h = -4.9(t - 13)(t + 2).

Required

Determine a reasonable solution for t

To do this, we equate h to 0.

h = -4.9(t - 13)(t + 2). becomes

0 = -4.9(t - 13)(t + 2).

Divide through by -4.9

0 = (t - 13)(t + 2).

Reorder the expression

(t - 13)(t + 2) = 0.

Split

t-13 = 0 or t + 2 = 0

This gives:

t = 13 or t = -2

But time can't be negative, So:

t = 13

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If Denise scored 3 fewer points, then you would add three to 18 to get the number of points that Carter scored.

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Find the equation of a line that is perpendicular to y = 3x – 5 and passes through the point (1, -3).
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keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5

well then therefore

\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1})\implies y+3=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y=-\cfrac{1}{3}x+\cfrac{1}{3}-3\implies y=-\cfrac{1}{3}x-\cfrac{8}{3}

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Answer:

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Step-by-step explanation:

\frac{2}{5} / 6\\= \frac{6}{15} / 6\\

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vlabodo [156]

The equation of the graphed line 2x + 3y = -6 written in standard form is,\rm y= -\frac{2}{3}x -2 and for line 2x + 3y = 6 will be,\rm y = \frac{-2}{3}x+2

<h3>What is the equation?</h3>

A mathematical statement consisting of an equal symbol between two algebraic expressions with the same value is known as an equation.

a) 2x + 3y = -6

The line in which the standard form is found as;

Ax + By + C = 0

The typical form of a slope intercept is

y = mx+c

where m denotes the slope and b the line's y-intercept.

3y= -6-2x\\\\ y=\frac{-6-2x}{3}\\\\ y= -\frac{2}{3}x -2

To obtain the slope standard form, we rewrite. The obtained equation is in the form of the slope-intercept.

b) 2x + 3y = 6

The line in which the standard form is found as;

Ax + By + C = 0

The typical form of a slope intercept is

y = mx+c

where m denotes the slope and b the line's y-intercept.

\rm 3y= 6-2x \\\\ y=  \frac{6-2x}{3} \\\\ y = \frac{-2}{3}x+2

Hence, The equation of the graphed line 2x + 3y = -6 written in standard form is,\rm y= -\frac{2}{3}x -2

To learn more, about equations, refer;

brainly.com/question/10413253

#SPJ1

What is the equation of the graphed line written in standard form?

2x + 3y = -6, 2x + 3y = 6

Summary:

a. The equation of the graphed line 2x + 3y = -6 written in standard form is y= -2/3 x -2.

b. The equation of the graphed line 2x + 3y = 6 written in standard form is y= -2/3 x +2.

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