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morpeh [17]
3 years ago
14

What is the perimeter?

Mathematics
1 answer:
seraphim [82]3 years ago
4 0

Answer:

59.

Step-by-step explanation:

I´m pretty sure the answer is 59. If I´m wrong then sorry.

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Graph the system of inequalities. y is greater than 3x-4 4y-x is less than or equal to 8
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Answer:

no;no because it is going to be negative 12.

Step-by-step explanation:

first you do 3x-4=-12 and so if you get -12 it is not even close to 8.

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Which tools were used by ancient mathematicians to make geometric constructions?
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The tools that were used by ancient mathematicians to make geometric constructions are compasses and straightedges. A compass is a drawing instrument used to draw circles and arcs. The straightedges on the other hand is a tool to check on the accuracy of a straight line.

The answer is first choice.
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Find both possible pairs of numbers if two numbers have a product of 14 and a difference of 5
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tt

Step-by-step explanation:

ttt

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In the relation defined by the equation y = 3x − 4, for all x > 0, y is a function of x because
soldier1979 [14.2K]
A because the starting values is not a negative
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3 years ago
. The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to nd
Blababa [14]

The question is:

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation

x²y'' - 7xy' + 16y = 0; y1 = x^4

Answer:

The second solution y2 is

A(x^4)lnx

Step-by-step explanation:

Given the homogeneous differential equation

x²y'' - 7xy' + 16y = 0

And a solution: y1 = x^4

We need to find a second solution y2 using the method of reduction of order.

Let y2 = uy1

=> y2 = ux^4

Since y2 is also a solution to the differential equation, it also satisfies it.

Differentiate y2 twice in succession with respect to x, to obtain y2' and y2'' and substitute the resulting values into the original differential equation.

y2' = u'. x^4 + u. 4x³

y2'' = u''. x^4 + u'. 4x³ + u'. 4x³ + u. 12x²

= u''. x^4 + u'. 8x³ + u. 12x²

Now, using these values in the original equation,

x²(u''. x^4 + u'. 8x³ + u. 12x²) - 7x(u'. x^4 + u. 4x³)+ 16(ux^4) = 0

x^6u'' + 8x^5u' + 12x^4u - 7x^5u' - 28x^4u + 16x^4u = 0

x^6u'' + x^5u' = 0

xu'' = -u'

Let w = u'

Then w' = u''

So

xw' = -w

w'/w = -1/x

Integrating both sides

lnw = -lnx + C

w = Ae^(-lnx) (where A = e^C)

w = A/x

But w = u'

So,

u' = A/x

Integrating this

u = Alnx

Since

y2 = uy1

We have

y2 = (Alnx)x^4 = (Ax^4)lnx

Therefore, the second solution y2 is

A(x^4)lnx

6 0
3 years ago
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