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Mathematics

1 answer:

zavuch27 [327]3 years ago

4 0

Answer:

7) G

8) F & G

9) H

i think this is right, hope i helped :)

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Pleaseeee help me. I'm stuck on a math question.

lozanna [386]

The slope of the function for pronghorn antelope is 60.78 which infers that the rate of speed of the pronghorn is 60.78 miles per hour.

7) The given function that represents the speed of the pronghorn is

y = 60.78x - 5.4

Comparing this function with the general equation of a straight line

y = mx + c we can conclude that the slope of the function is 60.78 .

So the Pronghorn's rate of speed is 60.78 miles per hour.

8) Now the speed of the cheetah is given in the form of a table.

Let us take any two points on the graph

(0.5,21.85) and (2,118.60)

Slope of the line passing through these two points

= (118.6-21.85)/(2-0.5)

=64.5

So the slope of the graph is 64.5 and the average rate of speed of the Cheetah is 64.5 miles per hour.

9) From the above two slopes and the rate of speed we can conclude that the speed of the cheetah is 64.5 mph which is greater than that of the pronghorn 's speed of 60.78 miles per hour.

To learn more about slope visit:

brainly.com/question/13281781

#SPJ1

6 0

2 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$