What is the answer to number 4

Need step by step detailed answer

kolezko [41]

Answer:

y = -4x - 4

Step-by-step explanation:

If we have two lines in slope intercept form as

$y = m_1x + c_1 \text{ and}\\y = m_2x + c_2\\$

then the product of the slopes $m_1m_2 = -1$

In other words, $m_2 = -\frac{1}{m_1}$

We have the first line as

$y = \frac{1}{4}x + 5\\\\$

The slope of this line is $\frac{1}{4}$

The slope of a perpendicular line will be the negative of the reciprocal of this line

Reciprocal of 1/4 is 4

So slope of perpendicular line is -4 which implies

y = -4x + c

We have to determine c

Since the line passes through (-3, 8), plug in these values for x and y in the equation and solve for c

8 = (-4)(-3) + c

8 = 12 + c

c = -4

So the equation of the perpendicular line is y = -4x - 4 (Answer)

It is always a good idea to plot these graphs and see if they fit the data provided

The attached plot shows the two graphs and you can see they are perpendicular to each other and the perpendicular line(the answer) passes through point (-3,8)

4 0

2 years ago

A scale drawing of a rectangular prism is 10cm by 8.5 cm. If the actual field has dimension of 40m by 34m, what scale can be use

Bad White [126]

The answer will be D. Hope it help!

5 0

3 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 :

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.