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skad [1K]
1 year ago
11

Please help! math question for college algebra mth 107!

Mathematics
1 answer:
kherson [118]1 year ago
8 0

Answer:

Local Minimum: (-2.5, -2), (0, -2)

Local Maximum: (1, 2)

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Write 7:30 in two ways
Aloiza [94]
Three ways are: 

<u>In Fraction Form:</u>  \frac{7}{30}  
<u>In Ratio its</u>: What you said  7:30

And

In a Percent Form Its:  0.23- %
(I got that from dividing 7 by 30.)

Hope IT helped You. :D
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3 years ago
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Which expression is equivalent to (2x2 +3)(x+4)
FrozenT [24]
The answer should be 7x+28
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5 0
3 years ago
Find two power series solutions of the given differential equation about the ordinary point x = 0. y'' + xy = 0
nalin [4]

Answer:

First we write y and its derivatives as power series:

y=∑n=0∞anxn⟹y′=∑n=1∞nanxn−1⟹y′′=∑n=2∞n(n−1)anxn−2

Next, plug into differential equation:

(x+2)y′′+xy′−y=0

(x+2)∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

x∑n=2∞n(n−1)anxn−2+2∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

Move constants inside of summations:

∑n=2∞x⋅n(n−1)anxn−2+∑n=2∞2⋅n(n−1)anxn−2+∑n=1∞x⋅nanxn−1−∑n=0∞anxn=0

∑n=2∞n(n−1)anxn−1+∑n=2∞2n(n−1)anxn−2+∑n=1∞nanxn−∑n=0∞anxn=0

Change limits so that the exponents for  x  are the same in each summation:

∑n=1∞(n+1)nan+1xn+∑n=0∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−∑n=0∞anxn=0

Pull out any terms from sums, so that each sum starts at same lower limit  (n=1)

∑n=1∞(n+1)nan+1xn+4a2+∑n=1∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−a0−∑n=1∞anxn=0

Combine all sums into a single sum:

4a2−a0+∑n=1∞(2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an)xn=0

Now we must set each coefficient, including constant term  =0 :

4a2−a0=0⟹4a2=a0

2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an=0

We would usually let  a0  and  a1  be arbitrary constants. Then all other constants can be expressed in terms of these two constants, giving us two linearly independent solutions. However, since  a0=4a2 , I’ll choose  a1  and  a2  as the two arbitrary constants. We can still express all other constants in terms of  a1  and/or  a2 .

an+2=−(n+1)nan+1+(n−1)an2(n+2)(n+1)

a3=−(2⋅1)a2+0a12(3⋅2)=−16a2=−13!a2

a4=−(3⋅2)a3+1a22(4⋅3)=0=04!a2

a5=−(4⋅3)a4+2a32(5⋅4)=15!a2

a6=−(5⋅4)a5+3a42(6⋅5)=−26!a2

We see a pattern emerging here:

an=(−1)(n+1)n−4n!a2

This can be proven by mathematical induction. In fact, this is true for all  n≥0 , except for  n=1 , since  a1  is an arbitrary constant independent of  a0  (and therefore independent of  a2 ).

Plugging back into original power series for  y , we get:

y=a0+a1x+a2x2+a3x3+a4x4+a5x5+⋯

y=4a2+a1x+a2x2−13!a2x3+04!a2x4+15!a2x5−⋯

y=a1x+a2(4+x2−13!x3+04!x4+15!x5−⋯)

Notice that the expression following constant  a2  is  =4+  a power series (starting at  n=2 ). However, if we had the appropriate  x -term, we would have a power series starting at  n=0 . Since the other independent solution is simply  y1=x,  then we can let  a1=c1−3c2,   a2=c2 , and we get:

y=(c1−3c2)x+c2(4+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(4−3x+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(−0−40!+0−31!x−2−42!x2+3−43!x3−4−44!x4+5−45!x5−⋯)

y=c1x+c2∑n=0∞(−1)n+1n−4n!xn

Learn more about constants here:

brainly.com/question/11443401

#SPJ4

6 0
1 year ago
Approximately 10% of all people are left-handed. Consider a grouping of fifteen people.
vaieri [72.5K]
1.5 but you can’t get half a person
8 0
3 years ago
Answer of a ²-10a+16-6b-b^2​
german

\quad a^2-10a+16-6b-b^2

$=(a^2-10a)-(b^2+6b) +16$

$=[(a^2-2(5)a+25)-25]-[(b^2+2(3)b+9)-9]+16$

$=(a-5)^2-25-(b+3)^2+9+16$

$=(a-5)^2-(b+3)^2$

5 0
3 years ago
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