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hammer [34]
3 years ago
8

1. Joshua has a ladder that is 17 ft long. He wants to lean the ladder against a vertical wall so that the top of the ladder is

16.5 ft above the ground. For safety reasons, he wants the angle the ladder makes with the ground to be no greater than 70°. Will the ladder be safe at this height? Show your work please.
Mathematics
1 answer:
yan [13]3 years ago
3 0

Answer:

NO, the ladder is too steep.

Step-by-step explanation:

Length of ladder = 17 feet.  This is the hypotenuse.  

 Side a of a right triangle = 16.6    

To answer this, you need one of the trigonometric functions.      Opposite = 16.5    

hypotenuse = 17    

The function you need is the sine function.    

Sin(theta) = opposite / hypotenuse    

sin(theta) = 16.5 / 17    

Sin(theta) = 0.97059    

theta = sin-1(0.07059)    

theta = 76.07  

No the ladder is too steep.

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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 :

a). The level curves of the function :

$C(x,y) = e^{-(x^2+2y^2)/10^4}$

are actually the curves

$e^{-(x^2+2y^2)/10^4}=k$

where k is a positive constant.

The equation is equivalent to

$x^2+2y^2=K$

$\Rightarrow \frac{x^2}{(\sqrt K)^2}+\frac{y^2}{(\sqrt {K/2})^2}=1, \text{ where}\ K = -10^4 \ln k$

which is a family of ellipses.

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

If the shark always swim in the direction of maximum increase of blood concentration, its direction at any point would coincide with the gradient vector.

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

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

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

7 0
2 years ago
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kati45 [8]
I would say the answer is 5, please let me know if it’s right!
4 0
3 years ago
Clearly mark on the coordinate plane the location of the ordered pairs and
ladessa [460]

Answer:

Point J(4,0) lies on x-axis and point M(-2,-3) lies quadrant III.

Step-by-step explanation:

A coordinate plane is a plane that is divided into four quadrants by the coordinate axes.

Here, coordinate axes are x-axis and y-axis.

Let's first mark the points in the coordinate plane.

See the attached figure.

From the figure, it can be observed that point J(4,0) lies on x-axis and point M(-2,-3) lies quadrant III.

3 0
3 years ago
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podryga [215]

13/15-1/4

=52/60-15/60

=52-15

______

60

=37/60

<em>hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em><em>.</em><em>uu</em><em> </em><em>frnd</em>

5 0
3 years ago
Read 2 more answers
(20 PTS)<br> 4x-(3x-2x)=3(2x-1) <br> A. 0<br> B. 1<br> C. Infinitely Many
Mkey [24]

4x-(3x-2x)=3(2x-1)

4x - 3x + 2x = 6x - 3

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Answer

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3 0
3 years ago
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