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3 years ago

Hey can you please help me!

Mathematics

1 answer:

melisa1 [442]3 years ago

6 0

Answer:

Step-by-step explanation:

1) Tangent makes 90 degrees with the radius. Hence

a) Angle ABD = 90-angle BDC = 90-61.3 = 28.7

b) Triangle BCD is isosceles since tangents have equal length.

Hence angle BCD = 180-(61.3+61.3) = 57.4

c) Angle BDC = 57.4 (since triangle BCD is isosceles)

d) Angle BAC and Angle BCD are supplementary. Hence angle BAC = 122,6

-------------------------------------------

2) Arc length of minor arc BC = $\frac{79.3}{360} (24) = 5.287$

b) ARc length of minor arc = circumference -minor arc

= 18.713

---------------------------------------------

3) circumference = pi d = $3.14(9) = 28.26$

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Can someone help me with this? I don’t understand it.

horrorfan [7]

Answer:

x+y=z

4+9=13

Step-by-step explanation:

x=4

2(4)+1=y=9

4+9=13

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3 years ago

A plane descends to sea level from 36 comma 400 feet after being airborne for 1 and one half hours. The entire flight time is 3

andrew-mc [135]

Answer:

350 feet per minute

Step-by-step explanation:

The amount of time it took the plane to descend is 3 hours 14 minutes minus 1 hour 30 minutes.

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3 years ago

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Solve the equation for theta (0°≤theta<360°)<br><br>2 sin theta = 1

Oduvanchick [21]

Answer:

30 degrees

Step-by-step explanation:

To figure out theta is using a bit of simple trigonometry.

2sin(theta) = 1

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4 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

3 years ago

Ill give brainliest, please

Flura [38]

Answer:

1st One goes to the third one

2nd one goes to the 1st one

3rd one goes to the 4th one

and the 4th one goes to the 2nd one

7 0

3 years ago