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jolli1 [7]
2 years ago
14

\Rectangle ABCDABCDA, B, C, D is graphed in the coordinate plane. The following are the vertices of the rectangle: A(5, 1),A(5,1

),A, left parenthesis, 5, comma, 1, right parenthesis, comma B(7,1)B(7,1)B, left parenthesis, 7, comma, 1, right parenthesis, C(7, 6)C(7,6)C, left parenthesis, 7, comma, 6, right parenthesis, and D(5, 6)D(5,6)D, left parenthesis, 5, comma, 6, right parenthesis.
Mathematics
2 answers:
Zolol [24]2 years ago
7 0

Answer:

hi

this question is already answered so im type gibbrish

Step-by-step explanation:

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dusya [7]2 years ago
6 0

Answer:

3

Step-by-step explanation:

Points A and B lie on the horizontal line y = 1. The length of segment AB is then the difference of the x-coordinates of the points:

 5 -2 = 3

You might be interested in
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
10 months ago
Given the conditional statement:
pogonyaev
Seodosieodoososoeoer dodostvxdhffcoxoddoodododdododod
6 0
3 years ago
Which expressionis equal to (-7x+5.1y)-(-9x+5.3y)
Furkat [3]

Answer:

The simplified expression for the given expression will be:

c. 2x-0.2y

Step-by-step explanation:

Given expression:

(-7x+5.1y)-(-9x+5.3y)

To simplify the expression.

Solution:

In order to simplify the expression, we will first remove the parenthesis by reversing the signs of the terms inside the parenthesis which lies after a negative sign out side the parenthesis.

<em>This is because negative multiplies to a negative to give a positive and negative multiplies to a positive to give a negative.</em>

So, we have:

⇒ -7x+5.1y+9x-5.3y

Combining like terms

⇒ -7x+9x+5.1y-5.3y

<em>The like terms can be evaluated as</em>

-7x+9x=2x

5.1y-5.3y=-0.2y

Thus, the simplified expression will be:

⇒  2x-0.2y

7 0
2 years ago
Help please due now I only have 10 mins..Help please!!! I only have 10 mins!!!​
fomenos

Answer:

a is equilateral and right, b is scalene and acute, c is equilateral and acute, d is isoseles and right, e is scalene and either acute or right, f is equilateral and acute, g is scalene and acute, and h is scalene and obtuse

Step-by-step explanation:

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Help?????????????????
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Answer:

please make the image less blurry

Step-by-step explanation:

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