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vagabundo [1.1K]
2 years ago
8

Un tren pasa delante de un poste en 10 s y cruza un puente en 15 s. ¿En cuánto tiempo el tren cruzaría el puente si este tuviera

el triple de su longitud?
Mathematics
1 answer:
emmainna [20.7K]2 years ago
6 0

Answer:

45 segundos.

Step-by-step explanation:

Un tren pasa por delante de un puente en 15 segundos; si el puente tuviera el doble de longitud, le tomaría el doble de tiempo en cruzarlo, si tuviera el triple de longitud, le tomaría el triple de tiempo y así sucesivamente.

En este caso, la pregunta es ¿En cuánto tiempo cruzaría el puente si tuviera el triple de su longitud? Por lo tanto, si cruza el puente en 15 segundos, teniendo el triple de longitud le tomaría 3 (15) = 45 segundos en cruzarlo.

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A hotel manager records the number of bookings made at various prices during July.What type of correlation does the graph show?
Usimov [2.4K]

Answer:

below

Step-by-step explanation:

if positive increases

if negative decreases

3 0
2 years ago
salina is one-quarter of her aunts age if a represents her aunts age which expression represents her age
katovenus [111]

Answer:

<em>s</em> = .25<em>a</em>

Step-by-step explanation:

if <em>s</em> is Salina's age, and we know that she is 1/4 (.25) her aunt's age, then ONE QUARTER (.25) times AUNT'S AGE (<em>a</em>) = SALINA'S AGE (<em>s</em>)

7 0
2 years ago
7 times as much as the sum of 1/3 and 4/5
Vlada [557]
7( \frac{1}{3} +  \frac{4}{5} )

=7( \frac{1(5)}{3(5)} +  \frac{4(3)}{5(3)} )

=7( \frac{5}{15} +  \frac{12}{15} )

=7( \frac{17}{15})

=\frac{7}{1} ( \frac{17}{15})

=\frac{119}{15}

= 7\frac{14}{15} ≈ 7.93
4 0
2 years ago
Read 2 more answers
Find the number a such that the line x = a bisects the area under the curve y = 1/x2 for 1 ≤ x ≤ 4. 8 5​ (b) find the number b s
IceJOKER [234]

a. We're looking for a such that

\displaystyle\int_1^a\frac{\mathrm dx}{x^2}=\int_a^4\frac{\mathrm dx}{x^2}

-\dfrac1x\bigg|_{x=1}^{x=a}=-\dfrac1x\bigg|_{x=a}^{x=4}

1-\dfrac1a=\dfrac1a-\dfrac14

\dfrac54=\dfrac2a\implies\boxed{a=\dfrac85}

b. Integrating with respect to y will make things easier.

y=\dfrac1{x^2}\implies x=\dfrac1{\sqrt y}

(where we take the positive square root because we know x>0)

Now we want to find b such that

\displaystyle\int_0^b\frac{\mathrm dy}{\sqrt y}=\int_b^1\frac{\mathrm dy}{\sqrt y}

2\sqrt y\bigg|_{y=0}^{y=b}=2\sqrt y\bigg|_{y=b}^{b=1}

2\sqrt b=2-2\sqrt b

4\sqrt b=2\implies\boxed{b=\dfrac14}

3 0
2 years ago
Milton is floating in an inner tube in a wave pool. He is 1.5 m from the bottom of the pool when he is at the trough of a wave.
posledela
This is the concept of sinusoidal, to solve the question we proceed as follows;
Using the formula;
g(t)=offset+A*sin[(2πt)/T+Delay]
From sinusoidal theory, the time from trough to crest is normally half the period of the wave form. Such that T=2.5
The pick magnitude is given by:
Trough-Crest=
2.1-1.5=0.6 m
amplitude=1/2(Trough-Crest)
=1/2*0.6
=0.3
The offset to the center of the circle is 0.3+1.5=1.8
Since the delay is at -π/2 the wave will start at the trough at [time,t=0]
substituting the above in our formula we get:
g(t)=1.8+(0.3)sin[(2*π*t)/2.5]-π/2]
g(t)=1.8+0.3sin[(0.8πt)/T-π/2]


3 0
2 years ago
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