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3 years ago

(the pic of a graph goes here and is part of the question)

Mathematics

2 answers:

fiasKO [112]3 years ago

8 0

<h3>Answer: { } or empty set.</h3>

Step-by-step explanation:

We are given two lines on the graph.

Line l and line m.

Lines l and m are parallel to each other.

Parallel lines never intersect each other at any point.

<em>Note: Solution of a system or a set of lines on the graph are the point(s) where lines intersect each other.</em>

As we know parallel lines would not intersect each other at any point, there would not be any solution of the system.

<h3>So, the solution set would be { } or empty set.</h3>

Anna35 [415]3 years ago

6 0

Ur lines are parallel...they never intersect...means there is no solution...
ur answer { } or empty set

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The slope is the change in y over the change in x.

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$\frac{3-(-3)}{-4-8} =\frac{6}{-12} =-\frac{1}{2}$

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3 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

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

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3 years ago

Could I have some help please

Contact [7]

Sure what’s the question

6 0

3 years ago

Which linear inequality is represented by the graph

Arlecino [84]

Answer: $\Large\boxed{First~choice.~y\leq \frac{1}{2}x+2 }$

Step-by-step explanation:

<h2>Determine the equation form of the inequality</h2>

<u>Given function form</u>

Point-Slope form : y = mx + b

m = slope
b = y-intercept

<u>Given the information from the graph</u>

Since the graph crosses (0, 2), b = 2

<u>Find the slope</u>

<em>Choose any two points from the graph</em>

(x₁, y₁) = (0, 2)

(x₂, y₂) = (2, 3)

Slope = (y₂ - y₁) / (x₂ - x₁)

Slope = (3 - 2) / (2 - 0)

Slope = 1 / 2

m = 1 / 2

Therefore, the equation form of the inequality is <u>y = (1/2)x + 2</u>

<h2>Determine whether it is ≤ or ≥</h2>

Since it is a <u>solid line</u>, the values on the line are also included. Thus, it must be either greater than or equal to, or less than or equal to.

The shaded region is below the solid line, therefore, it is<u> ≤</u>

Therefore, the inequality is $\Large\boxed{y\leq \frac{1}{2}x+2 }$

To double-check the sign, we can substitute (0, 0) into the inequality to see whether it is true.

y ≤ (1/2) x + 2

0 ≤ (1/2) 0 + 2

0 ≤ 2 TRUE

Hope this helps!! :)

Please let me know if you have any questions

5 0

1 year ago