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10 months ago

See attached questionthe last line is N (not visible line)

Mathematics

1 answer:

algol1310 months ago

8 0

The figure has reflective symmetry about a line if that line divides the figure in two similar parts, which are reflections of one another.

Notice that the two figures formed when the shape is divided by line k are not similar, since one has 7 vertices and the other just 4.

Notice that the two figures formed when the shape is divided by line m are not similar, since one has 7 vertices and the other just 4.

The line l divides the figure in two shape, one is a trapezoid and the other isn't.

Only line n divides the figure in two similar shapes.

Therefore, the correct choice is B) n only.

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Write the numbers in the order from least to greatest. 4.74, 4.72, 4.98

LUCKY_DIMON [66]

These numbers from least to greatest would be: 4.72, 4.74, 4.98

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2 years ago

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The radius of a circle is 6 in. Find the circumference to the nearest tenth.

ycow [4]

Answer:

37.7

Step-by-step explanation:

The orginal answer is 37.699

But you want to round the tenth

So 9 to the 6 you want to rasie the score

6 turns into a 7 and the numbers behind it (9 and the other 9) turn into a zero

37.7 or 37.700

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2 years ago

Please see attachment

Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

Step-by-step explanation:

Step(i):-

Given function

$f(x) = \frac{-x}{2x^{2} +1}$ ...(i)

Differentiating equation (i) with respective to 'x'

$f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$ ...(ii)

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$2 x^{2}-1 = 0$

$2 x^{2} = 1$

$x^{2} = \frac{1}{2}$

$x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }$

Step(ii):-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

$x = \frac{1}{\sqrt{2} }$

$f^{ll} (x) > 0$

The absolute minimum value at $x = \frac{1}{\sqrt{2} }$

Step(iii):-

The value of absolute minimum value

$f(x) = \frac{-x}{2x^{2} +1}$

$f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

on calculation we get

The value of absolute minimum value = - 0.3536

Final answer:-

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

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2 years ago

The outside edge of a circular saw whose dimensions are measured in inches can be defined by the equation x2 + y2 = 33.0625. A h

Eduardwww [97]

Answer:

In the same place, And 4.0

Step-by-step explanation:

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