What is the degree measure of f?. A) 56° B) 58° C) 60° D) 64°

Graph y < x2 + 4x. Click on the graph until the correct graph appears.

jonny [76]

Answer: The correct answer is:

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The given "graph" in the bottom right, lowest corner

Step-by-step explanation:

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Note: When there is only one (1) equation give for a graph;

and/or: only one (1) "inequality given";

we look for the symbol.

If the symbol is "not" an "equals" symbol (i.e. not an: = symbol) ;

we check for the type of "inequality" symbol.

If there is a: "less than" (<) ; or a "greater than" (>) symbol; the graph of the "inequality" will have "dashed lines" (since there will be a "boundary").

If there is an "inequality" that is a: "less than or equal to" (≤) ;

or a: "greater than or equal to" (≥) ;

→ then there will be not be a dashed line when graphed;

but rather—a "solid line" ; since "less than or equal to" ;

or "greater than or equal to" —is similar to:

"up to AND including"; or: "lesser/fewer than AND including".).

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Note: We are given the "inequality" :

→ " y < x² + 4x " .

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Note that we have a "less than" symbol (< ) ; so the graph will have a:

"solid line" [and not a "dotted line".].

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Note that all of the graphs among our 4 answer choices have "dotted lines".

Not that all values (all x and y coordinated) within the "shaded portion" of the corresponding graph are considered part of the graph.

As such, given any point within the shaded part, the x and y coordinates must match the inequality (i.e. the given inequality must be true when one puts in the "x-coordinate" and "y-coordinate" into the "given inequality" :

→ " y < x² + 4x " .

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Likewise, we can take any point within the "white, unshaded" portion of any of the graph, and take the "x-coordinate" and "y-coordinate" of that point, and the inequality: → " y < x² + 4x " ; will not hold true when the "x-coordinate" and "y-coordinate" values of that point— are substituted into the "inequality".

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{Note: Answer is continued on images attached.}.

Wishing you the best!

6 0

3 years ago

A small rocket is fired from a launch pad 10 m above the ground with an initial velocity left angle 250 comma 450 comma 500 righ

jonny [76]

Let $\vec r(t),\vec v(t),\vec a(t)$ denote the rocket's position, velocity, and acceleration vectors at time $t$ .

We're given its initial position

$\vec r(0)=\langle0,0,10\rangle\,\mathrm m$

and velocity

$\vec v(0)=\langle250,450,500\rangle\dfrac{\rm m}{\rm s}$

Immediately after launch, the rocket is subject to gravity, so its acceleration is

$\vec a(t)=\langle0,2.5,-g\rangle\dfrac{\rm m}{\mathrm s^2}$

where $g=9.8\frac{\rm m}{\mathrm s^2}$ .

a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,

$\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}$