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aleksandr82 [10.1K]
3 years ago
12

I need help is it due tomorrow

Mathematics
1 answer:
Anettt [7]3 years ago
7 0

Answer:

witch one is the selection

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lys-0071 [83]
Your answer would be -38.9 hope that helped
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3 years ago
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Which explanation about figures is correct? * A. All rhombuses are parallelograms. Parallelograms have 2 pairs of parallel sides
bezimeni [28]

Answer:

The correct option is;

A. All rhombuses are parallelograms. Parallelograms have 2 pairs of parallel sides. Therefore, all rhombuses have 2 pairs of parallel sides

Step-by-step explanation:

A rhombus is a quadrilateral that has all 4 sides, it has equal opposite angles and perpendicular diagonals that bisect one another  as well as having a pair of opposite parallel sides making it a parallelogram

A rhombus is similar to a parallelogram which also has equal opposite and parallel sided and equal opposite angles and the diagonals of a parallelogram also bisect each other.

4 0
3 years ago
Please calculate this limit <br>please help me​
Tasya [4]

Answer:

We want to find:

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}

Here we can use Stirling's approximation, which says that for large values of n, we get:

n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} =  \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}

Now we can just simplify this, so we get:

\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\

And we can rewrite it as:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}

7 0
3 years ago
PLEASE I NEED HELP ON THIS QUESTION..
Olenka [21]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the circles.

It's is universally proven that ratio of circumference and diameter is always π

hence the decimal value is π = 3.14

===> answer is option B.) 3.14

4 0
3 years ago
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Negativite eleven is less than negativte one
brilliants [131]
False -11 is furter on the number line then -1
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