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baherus [9]
3 years ago
13

What is the value of the function at x = 3?

Mathematics
1 answer:
BARSIC [14]3 years ago
6 0
The value is 3 because x=3 right there! it's not even a question, just an answer!
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Can someone help me these 2 questions? PLS and ty!
adelina 88 [10]

Answer:

y=60° b=60°

Step-by-step explanation:

the sum of all degrees of interior triangle is 180

in the first one, 30° and 90° have been given.(that small square represent 90°)

so 180°-90°-30°=60°

in the second one, the 120° is on the other side of a line, a line has a degree of 180, therefore the interior angle is 60°, the same rule 180°-2*60°=60°

5 0
2 years ago
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Levart [38]
(9y+y) + (6+3)
The answer is 10y+9

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2 years ago
Explain how to find the upper quartile of a box-and-whisker plot.
kirza4 [7]
B. The upper quartile is the median of the upper half of the data.

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How do I subtract 3 - 1 2/7
inna [77]

Answer:

5/7

Step-by-step explanation:

3-1 2/7

3=21/7 (by multiplying denominator)

1 2/7=16/7 (by multiply 1 and 7. Then add to 2)

21/7 - 16/7 = 5/7

5 0
2 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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