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4 years ago

To show that ΔFGH ≅ ΔJKL by SAS, what additional information is needed? Check all that apply.

Mathematics

2 answers:

Ugo [173]4 years ago

7 0

Answer:

1 and 2

Step-by-step explanation:

100% correct

k0ka [10]4 years ago

3 0

Answer:

2 and 3 I think

Step-by-step explanation:

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I need help on it, I don’t know what I’m supposed to do

bogdanovich [222]

Answer:

The length of river frontage for each lot are 96.55 ft. 98.85 ft, 101.15 ft and 103.45 ft.

Step-by-step explanation:

See the attached diagram.

The river frontage of 400 ft will be divided into 84 : 86 : 88 : 90 for each lot as AP, BQ, CR, DS and ET all are parallel.

Therefore, PQ : QR : RS : ST = 84 : 86 : 88 : 90 = 42 : 43 : 44 : 45

Let, PQ = 42x, QR = 43x, RS = 44x and ST = 45x

So, (42x + 43x + 44x + 45x) = 400

⇒ 175x = 400

⇒ x = 2.2988.

So, PQ = 42x = 96.55 ft.

QR = 43x = 98.85 ft.

RS = 44x = 101.15 ft and

ST = 45x = 103.45 ft

(Answer)

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cdisplaystyle%5Crm%5Cint%20%5Climits_%7B0%7D%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D%20%5

umka2103 [35]

Replace $x\mapsto \tan^{-1}(x)$ :

$\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \int_0^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx$

Split the integral at x = 1, and consider the latter one over [1, ∞) in which we replace $x\mapsto\frac1x$ :

$\displaystyle \int_1^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx = \int_0^1 \frac{\ln\left(\frac1x\right)}{\sqrt[3]{x} \left(1+\frac1{x^2}\right)} \frac{dx}{x^2} = - \int_0^1 \frac{\ln(x)}{\sqrt[3]{x} (1+x^2)} \, dx$

Then the original integral is equivalent to

$\displaystyle \int_0^1 \frac{\ln(x)}{1+x^2} \left(\sqrt[3]{x} - \frac1{\sqrt[3]{x}}\right) \, dx$

Recall that for |x| < 1,

$\displaystyle \sum_{n=0}^\infty x^n = \frac1{1-x}$

so that we can expand the integrand, then interchange the sum and integral to get

$\displaystyle \sum_{n=0}^\infty (-1)^n \int_0^1 \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \ln(x) \, dx$

Integrate by parts, with

$u = \ln(x) \implies du = \dfrac{dx}x$

$du = \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \, dx \implies u = \dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac23}}{2n+\frac23}$

$\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \int_0^1 \left(\dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac13}}{2n-\frac13}\right) \, dx \\\\ = \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{\left(2n+\frac43\right)^2} - \frac1{\left(2n+\frac23\right)^2}\right) \\\\ = \frac94 \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)$

Recall the Fourier series we used in an earlier question [27217075]; if $f(x)=\left(x-\frac12\right)^2$ where 0 ≤ x ≤ 1 is a periodic function, then

$\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi n x)}{n^2}$

$\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(2\pi(3n+1)x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(2\pi(3n+2)x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(2\pi(3n)x)}{(3n)^2}\right)$

$\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(6\pi n x + 2\pi x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(6\pi n x + 4\pi x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(6\pi n x)}{(3n)^2}\right)$

Evaluate f and its Fourier expansion at x = 1/2 :

$\displaystyle 0 = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(3n+1)^2} + \sum_{n=0}^\infty \frac{(-1)^n}{(3n+2)^2} + \sum_{n=1}^\infty \frac{(-1)^n}{(3n)^2}\right)$

$\implies \displaystyle -\frac{\pi^2}{12} - \frac19 \underbrace{\sum_{n=1}^\infty \frac{(-1)^n}{n^2}}_{-\frac{\pi^2}{12}} = - \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)$

$\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right) = \frac{2\pi^2}{27}$

So, we conclude that

$\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \frac94 \times \frac{2\pi^2}{27} = \boxed{\frac{\pi^2}6}$

3 0

2 years ago

The times for three swimmers are 101.85 seconds, 101.6 seconds, and 59.625 seconds. Which is the winning time? how many seconds

bija089 [108]

Hello there! the winning time is 59.625 seconds.

This is the person who completed the lowest amount of time, meaning they finished first. The others too longer so they finished later, meaning they didn't win.

I hope this was helpful and have a great day! :)

3 0

3 years ago

There are two similar toys

ValentinkaMS [17]

????? I’m confused what toys

8 0

3 years ago

Persamaan garis lurus yang melalui titik potong garis dengan persamaan 2x+5y=1 dan x-3y=-5 serta tegak lurus garis dengan persam

Novosadov [1.4K]

pertama menemukan tempat di mana dua persamaan mencegat
2x+5y=1
x-3y=-5

kalikan persamaan kedua ( x - 3y = -5 ) oleh -2
-2x+6y=10
tambahkan dua persamaan bersama-sama
2x+5=1
-2x+6y=10 +
0x+11y=11

11y=11
membagi kedua sisi dengan 11
y=1

subsitute y = 1 untuk y dalam semua persamaan untuk memecahkan x
x-3y=-5
x-3(1)=-5
x-3=5
x=8

x=8
y=1
(x,y)
titik persimpangan adalah ( 8,1 )


untuk membuat menemukan garis tegak lurus lebih mudah , mengkonversi persamaan terakhir ke bentuk lereng - intercept
2x-y+5=0
2x+5=y
y=2x+5

garis tegak lurus memiliki kemiringan yang , bila dikalikan dengan kemiringan garis lainnya , memberikan -1
y=mx+b
m=lereng

y=2x+5
2 dikalikan x=-1
x=-1/2
y=-1/2x+b
subsitute ( 8,1 ) ke dalam persamaan dan memecahkan untuk b
x=8
y=1
1=-1/2(8)+b
1=-4+b
tambahkan 4 untuk kedua belah pihak
5=b
persamaan adalah y=-1/2+5

( Catatan : Saya menggunakan google translate )

7 0

3 years ago