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Arturiano [62]
3 years ago
12

Oman said that it is impossible to raise a number to the power of 2 and get a value less than the original number.Do you agree w

ith oman? Justify your reasoning
Mathematics
1 answer:
disa [49]3 years ago
3 0
That’s false, if you raise a number that’s less than 1 to the second power, you’d get a lesser number.
For example, if you raise 1/2 to the second power, you’d get 1/4
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3 years ago
The radius of a circle is 58 centimeters. enter the circumference of the circle
Zanzabum

Answer:

C = 364.24

Step-by-step explanation:

Comment

The circumference and the radius are related to one another. The constant relating them is pi which is a fixed number of 3.14 The formula for the circumference is C = 2*pi * r

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5 0
2 years ago
I am lost on what to do
Neko [114]
\bf sin({{ \alpha}})sin({{ \beta}})=\cfrac{1}{2}[cos({{ \alpha}}-{{ \beta}})\quad -\quad cos({{ \alpha}}+{{ \beta}})]
\\\\\\
cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}\\\\
-----------------------------\\\\
\lim\limits_{x\to 0}\ \cfrac{sin(11x)}{cot(5x)}\\\\
-----------------------------\\\\
\cfrac{sin(11x)}{\frac{cos(5x)}{sin(5x)}}\implies \cfrac{sin(11x)}{1}\cdot \cfrac{sin(5x)}{cos(5x)}\implies \cfrac{sin(11x)sin(5x)}{cos(5x)}

\bf \cfrac{\frac{cos(11x-5x)-cos(11x+5x)}{2}}{cos(5x)}\implies \cfrac{\frac{cos(6x)-cos(16x)}{2}}{cos(5x)}
\\\\\\
\cfrac{cos(6x)-cos(16x)}{2}\cdot \cfrac{1}{cos(5x)}\implies \cfrac{cos(6x)-cos(16x)}{2cos(5x)}
\\\\\\
\lim\limits_{x\to 0}\ \cfrac{cos(6x)-cos(16x)}{2cos(5x)}\implies \cfrac{1-1}{2\cdot 1}\implies \cfrac{0}{2}\implies 0
4 0
4 years ago
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