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Ivahew [28]
3 years ago
5

The graph of a linear equation contains the points (6,7) and (-4,13) find another point that lies on the graph

Mathematics
2 answers:
Feliz [49]3 years ago
7 0

Answer:

(10,4.6)

Step-by-step explanation:

We need to find the equation for the line first

Find the slope:

m = y2-y1/x2-x1

m = 13 - 7 / -4 - 6

m = 6 / -10

Now we need the y intercept (b)

subtitute one of the points into the equation

y = -6/10x + b

7 = -6/10(6) + b

7 = -3.6 + b

10.6 = b

Now we have the full equation for the line

y = -6/10x + 10.6

and we can input any value into x to find the y of the point

y = -6/10(10) + 10.6

y = -6 + 10.6

y = 4.6

MAVERICK [17]3 years ago
5 0

Answer:

(11, 4)

Step-by-step explanation:

Find the slope = rise/ run  or (change in y values)/(change in x-values)

Using points (6, 7) and (-4, 13) ; Slope = (13 - 7)/(-4 - 6) = 6/-10 = -3/5

From this point if the slope is -3/5 this means the rise is -3 and the run 5.

remember that a negative rise you will move down on the rise from the point and a positive run means you will move across from the point.

Using point (6, 7) .

The run is the x movement on a graph so start at 6 and move to the right 5 places,  you in up at  x= 11.

The rise is the y movement on the graph so start at 7 and move down 3 places, you are at y = 4

Point (x, y) on the graph is (11, 4)

to check the problem

you from the slope = -3/5 . If you solve for "b"  in y = mx + b. The equation of the line in slope intercept form is y = -3/5x + 53/5.  Graph  the line and check the point or substitute any value for x and solve for y.

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Step-by-step explanation:

We are given the dimensions of the box and the wrap paper:

Box:

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Warp paper:

(26 in)(17 in)

Now we need to find the surface area of the box and the area of the wrap paper:

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surface_{1}=(56 in)(8 in)(2)=896 in^{2}

surface_{2}=(8 in)(36 in)(2)=576 in^{2}

surface_{3}=(56 in)(36 in)(2)=4032 in^{2}

surface-area=896 in^{2}+576 in^{2}+4032 in^{2}=5504 in^{2}

Warp paper:

area=(26 in)(17 in)=442 in^{2}

Dividing the area of the box by the area of the paper:

\frac{surface-area}{area}=\frac{5504 in^{2}}{442 in^{2}}=12.45 \approx 13

This means Angel's dad needs to purchase 13 sheets of wrapping paper.

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

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Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

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Therefore, the solution is:

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As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

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And

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Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

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