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DIA [1.3K]
3 years ago
13

Solve each system of inequalities by graphing. x-2y<3 2x+y>8

Mathematics
1 answer:
salantis [7]3 years ago
6 0
Not easy to graph but

multily first equaiton by -2 remember to flip sign
-2x+4y>-6
add to other equaiton

2x-2x+y+4y>-6+8
5y>2
divide by 5
y>2/5

subtitute
2x+2/5>8
subtract 2/5 from both sdies
2x>7 3/5
divid eby 2
x>38/10=19/5=3 4/5

x>3 4/5
y>2/5
just shade the area in that zone, not including point (0,0)
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For each x and n, find the multiplicative inverse mod n of x. Your answer should be an integer s in the range 0 through n - 1. C
krok68 [10]

Use the Euclidean algorithm to express 1 as a linear combination of x and n.

a. 52^{-1}\equiv40\pmod{77} because

77 = 1*52 + 25

52 = 2*25 + 2

25 = 12*2 + 1

so we can write

1 = 25 - 12*2 = 25*25 - 12*52 = (77 - 52)(77 - 52) - 12*52 = 77^2 - 2*52*77 + 52^2 - 12*52

Taken modulo 77 leaves us with

1\equiv52\cdot52-12\cdot52\equiv40\cdot52\pmod{77}\implies52^{-1}\equiv40\pmod{77}

b. First, 77\equiv25\pmod{52}, so really we're looking for the inverse of 25 mod 52. We've basically done the work in part (a) already:

1 = 25*25 - 12*52

Taken modulo 52, we're left with

1\equiv25\cdot25\pmod{52}\implies25^{-1}\equiv25\pmod{52}

c. The EA gives

71 = 1*53 + 18

53 = 2*18 + 17

18 = 1*17 + 1

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1\equiv(-4)\cdot53\pmod{71}\implies53^{-1}\equiv-4\equiv67\pmod{71}

d. Same process as with (b). First we have 71\equiv18\pmod{53}, and we've already shown that

1 = 3*18 - 53

which means, taken modulo 53, that

1\equiv3\cdot18\pmod{53}\implies71^{-1}\equiv18^{-1}\equiv3\pmod{53}

6 0
3 years ago
A circular mat has a circumference of 220 centimeters. Find the radius rounded to the nearest whole number.
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mezya [45]
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2 years ago
PLEASE HELP!!!!
scZoUnD [109]

Answer:

(D) 1+0.06m

Step-by-step explanation:

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I hope this helps!

3 0
3 years ago
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