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morpeh [17]
2 years ago
15

Consider in the figure below.

Mathematics
2 answers:
ELEN [110]2 years ago
6 0

⚘ <em>R</em><em>e</em><em>f</em><em>e</em><em>r</em><em> </em><em>t</em><em>o</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>a</em><em>t</em><em>t</em><em>a</em><em>c</em><em>h</em><em>m</em><em>e</em><em>n</em><em>t</em><em>s</em><em>!</em><em>!</em><em>~</em>

marusya05 [52]2 years ago
5 0

Let's solve

in ∆GNK

  • GN=130
  • GJ=94
  • GK=GJ/2=47

Apply pythagorean theorem

\\ \sf\longmapsto NK^2=GN^2-GK^2=130^2-47^2=16900-2209=14691

\\ \sf\longmapsto NK=\sqrt{14691}

\\ \sf\longmapsto NK=121(Approx)

If you observe the traingle

  • GN=NJ=130

\\ \sf\longmapsto LJ^2=JN^2-LN^2

\\ \sf\longmapsto LJ^2=130^2-78^2

\\ \sf\longmapsto LJ^2=16900-6084

\\ \sf\longmapsto LJ^2=10816

\\ \sf\longmapsto LJ=\sqrt{10816}

\\ \sf\longmapsto LJ=104

  • LH=LJ =104

So

in ∆HNL

\\ \sf\longmapsto HN^2=104^2+78^2

\\ \sf\longmapsto HN^2=10816+6084

\\ \sf\longmapsto HN^2=16900

\\ \sf\longmapsto HN=\sqrt{16900}

\\ \sf\longmapsto HN=130

Done .

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Solve 9(8 - x) = 3x.
bulgar [2K]

Hello - let's do this step by step. Show your work.

You're asking: 9(8−x)=3x.

Now, simplify both sides of the equation.

9(8−x)=3x

(9)(8)+(9)(−x)=3x (To do that, distribute)

72+−9x=3x

−9x+72=3x

Now,  subtract 3x from both sides.

−9x+72−3x=3x−3x

Simplify,

−12x+72=0

After, subtract 72 from both sides.

−12x+72−72=0−72

−12x=−72

Last thing, divide both sides by -12.

-12x/12 = -72/-12

Divide.

Therefore, you have <u>x = 6</u> as your final answer.

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3 years ago
There are 8 action, 3 comedy, and 5 drama DVDs on a shelf. Suppose three DVDs are selected at random from the shelf. Find each p
nordsb [41]
8 action, 3 comedy, 5 drama....total = 16 dvd's

P(2 action, then a comedy)...without replacement

8/16 * 7/15 * 3/14 = (8 * 7 * 3) / (16 * 15 * 14) = 168/3360 = 1/20
4 0
3 years ago
C=1/21.22.23+1/22.23.24+................+1/200.201.202<br><br> . = là dấu nhân
Aneli [31]

It looks like you have to find the value of the sum,

C = \displaystyle \frac1{21\times22\times23} + \frac1{22\times23\times24} + \cdots + \frac1{200\times201\times202}

so that the <em>n</em>-th term in the sum is

\dfrac1{(21+(n-1))\times(21+n)\times(21+(n+1))} = \dfrac1{(n+20)(n+21)(n+22)}

for 1 ≤ <em>n</em> ≤ 180.

We can then write the sum as

\displaystyle C = \sum_{n=1}^{180} \frac1{(n+20)(n+21)(n+22)}

Break up the summand into partial fractions:

\dfrac1{(n+20)(n+21)(n+22)} = \dfrac a{n+20} + \dfrac b{n+21} + \dfrac c{n+22}

Combine the fractions into one with a common denominator and set the numerators equal to one another:

1 = a(n+21)(n+22) + b(n+20)(n+22) + c(n+20)(n+21)

Expand the right side and collect terms with the same power of <em>n</em> :

1 = a(n^2+43n+462)+b(n^2+42n+440) + c(n^2+41n + 420) \\\\ 1 = (a+b+c)n^2 + (43a+42b+41c)n + 462a+440b+420c

Then

<em>a</em> + <em>b</em> + <em>c</em> = 0

43<em>a</em> + 42<em>b</em> + 41<em>c</em> = 0

462<em>a</em> + 440<em>b</em> + 420<em>c</em> = 1

==>   <em>a</em> = 1/2, <em>b</em> = -1, <em>c</em> = 1/2

Now our sum is

\displaystyle C = \sum_{n=1}^{180} \left(\frac1{2(n+20)}-\frac1{n+21}+\frac1{2(n+22)}\right)

which is a telescoping sum. If we write out the first and last few terms, we have

<em>C</em> = 1/(2×21) - 1/22 <u>+ 1/(2×23)</u>

… … + 1/(2×22) - 1/23 <u>+ 1/(2×24)</u>

… … <u>+ 1/(2×23)</u> - 1/24 <u>+ 1/(2×25)</u>

… … <u>+ 1/(2×24)</u> - 1/25 <u>+ 1/(2×26)</u>

… … + … - … + …

… … <u>+ 1/(2×198)</u> - 1/199 <u>+ 1/(2×200)</u>

… … <u>+ 1/(2×199)</u> - 1/200 + 1/(2×201)

… … <u>+ 1/(2×200)</u> - 1/201 + 1/(2×202)

Notice the diagonal pattern of underlined and bolded terms that add up to zero (e.g. 1/(2×23) - 1/23 + 1/(2×23) = 1/23 - 1/23 = 0). So, like a telescope, the sum collapses down to a simple sum of just six terms,

<em>C</em> = 1/(2×21) - 1/22 + 1/(2×22) + 1/(2×201) - 1/201 + 1/(2×202)

which we simplify further to

<em>C</em> = 1/42 - 1/44 - 1/402 + 1/404

<em>C</em> = 1,115/1,042,118 ≈ 0.00106994

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If the domain of the coordinate transformation f(x,y)=(y+2,-x-4) is (1,-4),(3,-2), (0,-1)
Mama L [17]

Answer:

The Fundamental Graphing Principle for Functions says that for a point (a, b) to be on the graph,

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(a, b+ 2) is on the graph o

main role x (x, f(x)) f(x) g(x) = f(x) + 2 (x, g(x))

0 (0, 1) 1 3 (0, 3)

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Step-by-step explanation:

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3 years ago
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