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BARSIC [14]
2 years ago
9

Can I get some help with question 12? I need to solve it while making a proof. I’ve asked at least 5 tutors for help and none of

them know what to do.

Mathematics
1 answer:
Oliga [24]2 years ago
3 0

Answer:

Step-by-step explanation:

Angle ABD is congruent to angle BDC by alternate interior angles of the parallel sides AB and DC.

Angle GEB is a right angle by given.

Angle DFC is a right angle by given.

The remaining angles for each triangle are congruent by sum of angles of a triangle is 180 degrees.

The two triangles are similar by  angle-angle-angle.

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The coordinates of point D are (7, 4) and the coordinates of point E are (1, −3) . What is the slope of the line that is perpend
Katarina [22]
The slope of a line is given by the change in y coordinates over the change in x coordinate.
The slope of line DE will be (4 - -3)/ (7-1) = 7/6
If M1 and M2 are slopes of two perpendicular lines at a point then
 M1× M2 = -1
 Thus, the slope of a perpendicular to line DE will be -6/7.
7 0
3 years ago
In the dig ram below line l is parallel to line m. Which angles are congruent to angle 1
mixas84 [53]

Answer:

3, 8, 6

Step-by-step explanation:

<u>The angles that are congruent to angle 1 are:  3, 8, 6</u>

<u />

Answer:  Congruent angles are 3, 8, 6

4 0
3 years ago
Find the total surface area of the following
xeze [42]

Step-by-step explanation:

The volume of the cone is:

1/3 *π* r² *h =

1/3 *π *4* 2square root of 3=

(8*square root of 3*π)/3

7 0
3 years ago
The answer for this lol thanks
Olenka [21]

Answer:

x times (x+1).

If it's wrong, then I'm an idiot.

Step-by-step explanation:

5 0
2 years ago
Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (
tekilochka [14]

Answer:

Radius of convergence of power series is \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n        n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n       n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

Power series centered at x = a is:

\sum_{n=1}^{\infty}c_{n}(x-a)^{n}

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]

Applying the ratio test:

\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}

\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

Applying n → ∞

\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

The numerator as well denominator of \frac{a_{n}}{a_{n+1}} are polynomials of fifth degree with leading coefficients:

(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}

4 0
2 years ago
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