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3 years ago

How many students are there. 28 students is 40% of 100

Mathematics

2 answers:

Amiraneli [1.4K]3 years ago

4 0

28/x = 40/100
Cross multiply
40x = 28(100)
40x = 2800
Divide both sides by 40
x = 70 students

Vladimir [108]3 years ago

3 0

There would be 70 childern

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a pyramid with a square base that is 9 cm on a side has a height of 10 cm the volume of the pyramid is

Zanzabum

$\bf \textit{volume of a pyramid}\\\\
V=\cfrac{1}{3}Bh~~
\begin{cases}
B=area~of\\
\qquad its~base\\
h=height\\[-0.5em]
\hrulefill\\
B=\stackrel{9\times 9}{81}\\
h=10
\end{cases}\implies V=\cfrac{1}{3}(81)(10)\implies V=270$

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3 years ago

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

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Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

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3 years ago

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Step-by-step explanation:

absolute value, if it's a negative number, will be positive. if it's a positive number, it will stay positive.

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A posted speed limit of 55 mph means that: You may drive 55 mph only under favorable conditions.

<h3>What is the speed limit?</h3>

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Now generally, speed limits are normally indicated on traffic signs reflecting the maximum allowable speed in kilometers per hour (km/h) or eevn miles per hour (mph). They are primarily set by the legislative bodies of national or provincial governments and then enforced by law enforcement agencies. This suggests that speed limits varies from road to road and is even absent in some certain areas.

Now, in this question we are told that the posted speed limit is 55 mph. Thus, from our definitions of speed limits above it is clear that A posted speed limit of 55 mph means that: You may drive 55 mph only under favorable conditions.

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