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hoa [83]
3 years ago
12

URGENT!!!!!!!!! will give 30 points and brainlist and if its right i might pay 20$$$$ dollars no cap!!!

Mathematics
1 answer:
7nadin3 [17]3 years ago
7 0
Hey buddy so basically you have to find the unknown sides length and it leads to the escape so here’s the outcome:
A>F>C>H>ESCAPE
Hope this helps also I don’t need the 20$ dollars :)
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3(18-7)+2<br><br> evaluate each expression using the order of operations
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it is 35

Step-by-step explanation:

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If f(a) =11, then use the table above to find f(a-2)​
KatRina [158]

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<h2>7</h2>

Step-by-step explanation:

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5 0
3 years ago
Using the method of mathematical induction to prove that equalities are true for values ​​of n indicated:
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2^2+4^2+6^2+...(2n)^2=\frac{2n(n+1)(2n+1)}{3};\ n\geq1\\\\chek\ for\ n=1:\\L=2^2=4;\ R=\frac{2\cdot1(1+1)(2\cdot1+1)}{3}=\frac{2\cdot2\cdot3}{3}=4\\L=R\\-----------------------\\&#10;assumption\ for\ n=k\\2^2+4^2+6^2+...+(2k)^2=\frac{2k(k+1)(2k+1)}{3}\\-----------------------\\thesis\ for\ n=k+1\\2^2+4^2+6^2+...+(2k)^2+[2(k+1)]^2=\frac{2(k+1)(k+1+1)[2(k+1)+1]}{3}\\-----------------------
proff:\\L=2^2+4^2+6^2+...+(2k)^2+(2k+2)^2=\frac{2k(k+1)(2k+1)}{3}+(2k+2)^2\\\\=\frac{(2k^2+2k)(2k+1)}{3}+\frac{3(2k+2)^2}{3}=\frac{4k^3+2k^2+4k^2+2k+3(4k^2+8k+4)}{3}\\\\=\frac{4k^3+6k^2+2k+12k^2+24k+12}{3}=\boxed{\frac{4k^3+18k^2+26k+12}{3}}\\\\R=\frac{2(k+1)(k+1+1)[2(k+1)+1]}{3}=\frac{(2k+2)(k+2)(2k+2+1)}{3}\\\\=\frac{(2k^2+4k+2k+4)(2k+3)}{3}=\frac{(2k^2+6k+4)(2k+3)}{3}=\frac{4k^3+6k^2+12k^2+18k+8k+12}{3}\\\\=\boxed{\frac{4k^3+18k^2+26k+12}{3}}\\\\L=R
4 0
3 years ago
Sydney wrote the equation y+1=12(x−3).Explain her error and then write the correct equation in point-slope form. x -1 2 y 3 9
valentinak56 [21]

if the equation is y+=1/2 (x-3) the slope is wrong.  the slope is actually 6. that was the error that Sydney made. i hope this help some of my people out:)

Step-by-step explanation:

now you got this go kill that essay:))))

8 0
3 years ago
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