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3 years ago

14

During the early 20th century, in what way was the radio better than the phonograph in encouraging the concept of crosspollinati

on?
Select one:
a. Phonographs were mostly owned by the wealthy
b. There were not enough records available
c. Phonographs were inferior in sound quality

1 answer:

ololo11 [35]3 years ago

6 0

C. Phonograph were inferior in sound quality.

Step-by-step explanation:

Researches have shown that plants respond to the sounds of their pollinator.

Some plants begin to produce a greater amount of nectar and some begin to vibrate to attract the attention of the pollinators towards them.

Greater the intensity of the sound of the pollinator, the faster the reaction will be.

Phonographs have a sound of lower intensity when compared to the intensity of sound produced by the radio.

Thus the radio was better than the phonograph in encouraging the concept of cross-pollination.

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2 years ago

Marine biologists have determined that when a shark detectsthe presence of blood in the water, it will swim in the directionin w

siniylev [52]

Solution :

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where k is a positive constant.

The equation is equivalent to

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which is a family of ellipses.

We sketch the level curves for K =1,2,3 and 4.

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Then we know the shark's path is perpendicular to the level curves it intersects.

b). We have :

$\triangledown C= \frac{\partial C}{\partial x}i+\frac{\partial C}{\partial y}j$

$\Rightarrow \triangledown C =-\frac{2}{10^4}e^{-(x^2+2y^2)/10^4}(xi+2yj),$ and

$\triangledown C$ points in the direction of most rapid increase in concentration, which means $\triangledown C$ is tangent to the most rapid increase curve.

$r(t)=x(t)i+y(t)j$ is a parametrization of the most $\text{rapid increase curve}$ , then

$\frac{dx}{dt}=\frac{dx}{dt}i+\frac{dy}{dt}j$ is a tangent to the curve.

So then we have that $\frac{dr}{dt}=\lambda \triangledown C$

$\Rightarrow \frac{dx}{dt}=-\frac{2\lambda x}{10^4}e^{-(x^2+2y^2)/10^4}, \frac{dy}{dt}=-\frac{4\lambda y}{10^4}e^{-(x^2+2y^2)/10^4}$

∴ $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2y}{x}$

Using separation of variables,

$\frac{dy}{y}=2\frac{dx}{x}$

$\int\frac{dy}{y}=2\int \frac{dx}{x}$

$\ln y=2 \ln x$

⇒ y = kx^2 for some constant k

but we know that $y(x_0)=y_0$

$\Rightarrow kx_0^2=y_0$

$\Rightarrow k =\frac{y_0}{x_0^2}$

∴ The path of the shark will follow is along the parabola

$y=\frac{y_0}{x_0^2}x^2$

$y=y_0\left(\frac{x}{x_0}\right)^2$

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2 years ago

The length of a rectangle is 6 more than twice the width. if the area is 40 cm^2, find the length and breadth of the rectangle

ra1l [238]

Answer: 3.217 & 12.434

Step-by-step explanation:

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Hence, the length is $2w+6$ .

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$A=w(2w+6)$

We can also substitute 40 in for A as it's given in the question.

$40 = w(2w+6)$

Distributing <em>w</em> by multiplying it by both terms in the parentheses, we get

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We can make the equation simpler by dividing both sides by 2.

$20 = w^2+3w$

Subtracting both sides by 20 will make the left-hand side 0.

$0=w^2+3w-20$

Now that we have put this <em>quadratic equation</em> into standard form (ax²+bx+c), we can find its solutions using the quadratic formula.

For reference, the quadratic formula is

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a is 1, b is 3, and c is -20.

Substituting, we get

$w=\frac{-3\pm\sqrt{3^2-4(1)(-20)}}{2(1)}$

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Since the second solution results in a negative number, it cannot be the length of w.

$w=\frac{-3+\sqrt{89}}{2}\approx3.217$

The width/breadth of the rectangle is 3.217 cm.

To calculate the length, let's substitute the width into the expression for the length:

$l=2(3.217)+6$

$l=12.434$

The length of this rectangle is 12.434 cm.

6 0

1 year ago