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Luba_88 [7]
3 years ago
5

Which of the following intervals represents the set (1,4) u [2,6]?

Mathematics
2 answers:
fenix001 [56]3 years ago
8 0
Set is a sum of two intervals :(1,4) u [2,6] 
In the first interval it is open on both sides so 1 and 4 don't belong to this interval, the second is closed so both 2 and 6 belong to interval and numbers between 2 and 6
We can write it as 1<x\leq 6
summing both sets we have one set: (1,6]
so
a) represent our set
b) also represent because all numbers between 4-6 and 6 are in our set
c)also represent, all numbers are in main set
d)also represent

garri49 [273]3 years ago
4 0
(1;\ 4)\ \cup\ [2;\ 6]=(1;\ 6]\\\\Answer:A.

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Which expression is equivalent to 15p^-4q^-6/-20p^-12q^-3 in simplified from? assume p=/ 0,q=/0
enyata [817]
(15p⁺⁴.q⁻⁶) /(-20p⁻¹².q⁻³). 

Remember that  a⁻ⁿ = 1/aⁿ  and 1/a⁻ⁿ = aⁿ


(-15/4).(p⁻⁴.q⁻⁶)(p⁺¹².q⁺³). 

(-15/20).(p⁻⁴.p¹².q⁻⁶.q³)

Remember aⁿ.aˣ = aⁿ⁺ˣ

(-15/20).(p⁸.q⁻³)

-3/5(p⁸.q⁻³)


3 0
3 years ago
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Choose 3/8 written as a decimal
Solnce55 [7]

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Step-by-step explanation:

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5 0
3 years ago
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Help please i am confused
Lena [83]
-2 ≥ u - 11 > -3

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5 0
3 years ago
Sum of money is placed at simple interest for 3 years at 10% per annum and then the amount is invested for 2 years at the same r
JulsSmile [24]

Answer:

sum was 300000

Step-by-step explanation:

let sum of money is placed  is x rs

Term T = 3 years, rate R = 10%

Then simple interest for for 3 years is

SI= \frac{PTR}{100}\\ SI=\frac{x\times 3\times 10}{100} \\SI = 0.3x

Sum of x will become x+0.3x = 1.3x after 3 years. now this 1.3x is kept for compound interest for two years. ie

A =P[1+R]^t\\471900=1.3x[1+\frac{10}{100} ]^2\\x=\frac{471900}{1.573} \\x=300000

Sum was 300000

3 0
3 years ago
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