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swat32
3 years ago
13

Multiplication of numbers with their uncertainty​

Mathematics
1 answer:
Bess [88]3 years ago
3 0

if you're adding or subtracting quantities with uncertainties, you add the absolute uncertainties. If you're multiplying or dividing, you add the relative uncertainties. If you're multiplying by a constant factor, you multiply absolute uncertainties by the same factor, or do nothing to relative uncertainties

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125% of what number is 80?<br> Oa 36<br> Oь 40<br> Os 64<br> Od 72
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Answer:

d. 72

Step-by-step explanation:

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2 years ago
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Which correctly uses bar notation to represent the repeating decimal
ioda

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2nd one..

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PLEASE HELP!!!! PHOTO IS USED!!! DONT SKIP!!!
sattari [20]
The answer would be a because it says to write two equations to support your answer
5 0
3 years ago
PLEASE HELP
lisabon 2012 [21]

Step-by-step explanation:

The figure below shows a portion of the graph of the function j\left(x\right) \ = \ 4^{x-2}, hence the average rate of change (slope of the blue line) between the x and x+h is

                     \text{Average rate of change} \ = \ \displaystyle\frac{\Delta y}{\Delta x} \\ \\ \rule{3.7cm}{0cm} = \dsiplaystyle\frac{f\left(x+h\right) \ - \ f\left(x\right)}{\left(x \ + \ h \right) \ - \ x} \\ \\ \\  \rule{3.7cm}{0cm} = \displaystyle\frac{f\left(x + h\right) \ - \ f\left(x\right)}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x+h-2} \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x-2+h} \ - \ 4^{x-2}}{h}

                                                            \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h}\right) \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h} \ - \ 1 \right)}{h}

7 0
2 years ago
Jocelyn's swimming pool is 14.75 ft by 7.5 ft. find the Area of the swimming pool
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6 0
3 years ago
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