a), 15 ft/sec

b) 25 ft/sec

c) 0 ft/sec

d) 10 ft/sec

Hayley Bernard
2022-07-19
Answered

A ball rolls down an inclined plane with an acceleration of $6\text{}ft/se{c}^{2}$? What initial velocity must be given for the ball to roll 300 feet

a), 15 ft/sec

b) 25 ft/sec

c) 0 ft/sec

d) 10 ft/sec

a), 15 ft/sec

b) 25 ft/sec

c) 0 ft/sec

d) 10 ft/sec

You can still ask an expert for help

Steven Bates

Answered 2022-07-20
Author has **15** answers

Initial velocity= ${V}_{0}$

Velocity after t times $v(t)={v}_{0}+at$

a=acceleration

s=total distance

$s=4t+\frac{1}{2}a{t}^{2}$

$300=10\times u+\frac{1}{2}\times 6\times 100$

300=10 u +300

10u=0

u=0

option (c) correct

Velocity after t times $v(t)={v}_{0}+at$

a=acceleration

s=total distance

$s=4t+\frac{1}{2}a{t}^{2}$

$300=10\times u+\frac{1}{2}\times 6\times 100$

300=10 u +300

10u=0

u=0

option (c) correct

asked 2022-07-10

Big intersection operation

$\mathrm{\varnothing}\ne A\subseteq B\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\Rightarrow \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\bigcap B\subseteq \bigcap A.$

In each case, the proof is straightforward. For example, in the last case, we assume that every member of A is also a member of B. Hence if $x\in \bigcap B$, i.e., if x belongs to every member of B, then a fortiori x belongs to every member of the smaller collection A. And consequently $x\in \bigcap A$.

I wonder if the last part is really a proof. It is okay to assume $x\in \bigcap B$ and arrive at $x\in \bigcap A$. But can I use this reasoning "then a fortiori x belongs to every member of the smaller collection" in a proof exercise? It looks like just using intuition (it does not seem rigorous).

$\mathrm{\varnothing}\ne A\subseteq B\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\Rightarrow \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\bigcap B\subseteq \bigcap A.$

In each case, the proof is straightforward. For example, in the last case, we assume that every member of A is also a member of B. Hence if $x\in \bigcap B$, i.e., if x belongs to every member of B, then a fortiori x belongs to every member of the smaller collection A. And consequently $x\in \bigcap A$.

I wonder if the last part is really a proof. It is okay to assume $x\in \bigcap B$ and arrive at $x\in \bigcap A$. But can I use this reasoning "then a fortiori x belongs to every member of the smaller collection" in a proof exercise? It looks like just using intuition (it does not seem rigorous).

asked 2021-01-27

Two-digit numbers are formed, with replacement, form the digits 0 through 9.

How many two-digit even numbers are possible?

How many two-digit even numbers are possible?

asked 2021-08-14

Discrete Math Question

Negate the following(in English) :

a) If it is snowing, then it is cold.

b) If

c) If

asked 2022-05-11

Prove that $\bigcap _{x\in [-1,0]}[x,1+x]=\{0\}$

Answer:

So my approach is a little bit different than the usual, that $\{0\}\subset \bigcap _{x\in [-1,0]}[x,1+x]$ and that $\bigcap _{x\in [-1,0]}[x,1+x]\subset 0$.

So I tried to do it that we have $a\in [x,1+x],$, $x\in [-1,0]\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}a\in [-1,1]$, and is a real number.

So now if we look at $x=-1+\u03f5$, where $\u03f5>0$ and is a small real number, we have some ${a}_{0}\in [-1+\u03f5,\u03f5]$ and if we look at $x=-\u03f5$, we have ${a}_{1}\in [-\u03f5,1-\u03f5]$.

Ok, so what I want to show is that if ${a}_{1},{a}_{0}\in \bigcap _{x\in [-1,0]}[x,1+x]$ iff ${a}_{1}={a}_{0}$.

Now if we look at the inequalities (we say that $\{a\in \mathbb{R}|\mathrm{\forall}x\in [-1,0]:a\in \bigcap [x,1+x]\}$):

$-1+\u03f5\le a\le \u03f5\phantom{\rule{1em}{0ex}}\wedge \phantom{\rule{1em}{0ex}}-\u03f5\le a\le 1-\u03f5$

$-\u03f5\le a\le \u03f5$

and as $\u03f5\to 0$, we get: $0\le a\le 0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}a=0={a}_{1}={a}_{0}$

Is my proof correct? How would you show it the usual way (with subset going both ways)?

Answer:

So my approach is a little bit different than the usual, that $\{0\}\subset \bigcap _{x\in [-1,0]}[x,1+x]$ and that $\bigcap _{x\in [-1,0]}[x,1+x]\subset 0$.

So I tried to do it that we have $a\in [x,1+x],$, $x\in [-1,0]\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}a\in [-1,1]$, and is a real number.

So now if we look at $x=-1+\u03f5$, where $\u03f5>0$ and is a small real number, we have some ${a}_{0}\in [-1+\u03f5,\u03f5]$ and if we look at $x=-\u03f5$, we have ${a}_{1}\in [-\u03f5,1-\u03f5]$.

Ok, so what I want to show is that if ${a}_{1},{a}_{0}\in \bigcap _{x\in [-1,0]}[x,1+x]$ iff ${a}_{1}={a}_{0}$.

Now if we look at the inequalities (we say that $\{a\in \mathbb{R}|\mathrm{\forall}x\in [-1,0]:a\in \bigcap [x,1+x]\}$):

$-1+\u03f5\le a\le \u03f5\phantom{\rule{1em}{0ex}}\wedge \phantom{\rule{1em}{0ex}}-\u03f5\le a\le 1-\u03f5$

$-\u03f5\le a\le \u03f5$

and as $\u03f5\to 0$, we get: $0\le a\le 0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}a=0={a}_{1}={a}_{0}$

Is my proof correct? How would you show it the usual way (with subset going both ways)?

asked 2022-06-01

Can't solve $(A\setminus B)\setminus C=(A\setminus C)\setminus (B\setminus C)=A\setminus (B\cup C)$ any help?

I have discrete math exam and i can't quite figure out one example.

$(A\setminus B)\setminus C=(A\setminus C)\setminus (B\setminus C)=A\setminus (B\cup C)$

i tried solving it like this and i got stuck: $(A\setminus B)\setminus C=(A\cap \overline{B})\cap \overline{C}$, what next ?

I have discrete math exam and i can't quite figure out one example.

$(A\setminus B)\setminus C=(A\setminus C)\setminus (B\setminus C)=A\setminus (B\cup C)$

i tried solving it like this and i got stuck: $(A\setminus B)\setminus C=(A\cap \overline{B})\cap \overline{C}$, what next ?

asked 2022-03-26

A car travels at 108 kph from A to B for t seconds, applies the brakes for 4 seconds between B and C to give the car a constant deceleration and a speed of 72 kph at C, and finally travels at this speed from C to D, also for t seconds. If the total horizontal distance from A to D is 3100 m, determine the distance between A and B.

asked 2022-06-13

Showing that a relation is an equivalence relation

Let $\in \mathbb{N}$ and ${\sim}_{n}\subseteq \mathbb{R}\times \mathbb{R}$ be defined by: $x{\sim}_{n}y{\iff}_{df}{x}^{n}-{y}^{n}=nx-ny$. Show that ${\sim}_{n}$ is an equivalence relation.

And here is my take on it (I apologize in advance for any notation mistakes, I'm still not quite good at it.):

Proof Reflexivity: Let $x\in \mathbb{R}$ and $n\in \mathbb{N}$; to show: $x{\sim}_{n}x$

$\begin{array}{rl}x{\sim}_{n}x& {\iff}_{df}{x}^{n}-{x}^{n}=nx-nx\\ & \iff 0=0\end{array}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\sim}_{n}$ is reflexive

Proof Symmetry: Let $x\in \mathbb{R}$ and $n\in \mathbb{N}$; to show: $x{\sim}_{n}y\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y{\sim}_{n}x$

$\begin{array}{rl}x{\sim}_{n}y& {\iff}_{df}{x}^{n}-{y}^{n}=nx-ny\\ y{\sim}_{n}x& \iff -({x}^{n}-{y}^{n})=-nx+ny\end{array}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y\nsim x$

So, the assumption made is wrong, that is, the relation is not symmetric, and therefore not an equivalence relation.

However, I'm unsure whether my proof is valid or not, as the wording of the question, and the following question, that requires ~ to be an equivalence relation, makes me believe it is in fact an equivalence relation.

Let $\in \mathbb{N}$ and ${\sim}_{n}\subseteq \mathbb{R}\times \mathbb{R}$ be defined by: $x{\sim}_{n}y{\iff}_{df}{x}^{n}-{y}^{n}=nx-ny$. Show that ${\sim}_{n}$ is an equivalence relation.

And here is my take on it (I apologize in advance for any notation mistakes, I'm still not quite good at it.):

Proof Reflexivity: Let $x\in \mathbb{R}$ and $n\in \mathbb{N}$; to show: $x{\sim}_{n}x$

$\begin{array}{rl}x{\sim}_{n}x& {\iff}_{df}{x}^{n}-{x}^{n}=nx-nx\\ & \iff 0=0\end{array}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\sim}_{n}$ is reflexive

Proof Symmetry: Let $x\in \mathbb{R}$ and $n\in \mathbb{N}$; to show: $x{\sim}_{n}y\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y{\sim}_{n}x$

$\begin{array}{rl}x{\sim}_{n}y& {\iff}_{df}{x}^{n}-{y}^{n}=nx-ny\\ y{\sim}_{n}x& \iff -({x}^{n}-{y}^{n})=-nx+ny\end{array}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y\nsim x$

So, the assumption made is wrong, that is, the relation is not symmetric, and therefore not an equivalence relation.

However, I'm unsure whether my proof is valid or not, as the wording of the question, and the following question, that requires ~ to be an equivalence relation, makes me believe it is in fact an equivalence relation.