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Alenkasestr [34]
3 years ago
5

Can someone help me??

Mathematics
1 answer:
makkiz [27]3 years ago
6 0

Answer:

58cm squared

Step-by-step explanation:

Ok so what we first want to do is find out the total area of that rectangle.

The way we do this is by multiplying the width x height which in this case is 10 cm and 7 cm.

This gives us 70 cm as the total area of the rectangle if the circular hole wasn't there. Now we need to find the area of the circle.

The way we do this is by taking the circle and finding the radius. The radius is shown as 2cm.

Now the formula to find the area of a circle is \pi r^{2} which means it is about 3.14 multiplied by r squared.

R squared in this problem is going to be 2 x 2 which is 4.

Now we just do 3.14 times 4 which gives us about 12.

What we do next is subtract 12 from the total area of the rectangle which is 70.

This gives us 58cm squared as an answer.

Hope this helped ! !

<3

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ki77a [65]

Explanation:

The cubic ...

  f(x) = ax³ +bx² +cx +d

has derivatives ...

  f'(x) = 3ax² +2bx +c

  f''(x) = 6ax +2b

<h3>a)</h3>

By definition, there will be a point of inflection where the second derivative is zero (changes sign). The second derivative is a linear equation in x, so can only have one zero. Since it is given that a≠0, we are assured that the line described by f''(x) will cross the x-axis at ...

  f''(x) = 0 = 6ax +2b   ⇒   x = -b/(3a)

The single point of inflection is at x = -b/(3a).

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<h3>b)</h3>

The cubic will have a local extreme where the first derivative is zero and the second derivative is not zero. These will only occur when the discriminant of the first derivative quadratic is positive. Their location can be found by applying the quadratic formula to the first derivative.

  x=\dfrac{-2b\pm\sqrt{(2b)^2-4(3a)(c)}}{2(3a)} = \dfrac{-2b\pm\sqrt{4b^2-12ac}}{6a}\\\\x=\dfrac{-b\pm\sqrt{b^2-3ac}}{3a}\qquad\text{extreme point locations when $b^2>3ac$}

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<h3>c)</h3>

Part A tells you the point of inflection is at x= -b/(3a).

Part B tells you the midpoint of the local extremes is x = -b/(3a). (This is half the sum of the x-values of the extreme points.) You will notice these are the same point.

The extreme points are located symmetrically about their midpoint, so are located symmetrically about the point of inflection.

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Additional comment

There are other interesting features of cubics with two local extremes. The points where the horizontal tangents meet the graph, together with the point of inflection, have equally-spaced x-coordinates. The point of inflection is the midpoint, both horizontally and vertically, between the local extreme points.

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▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

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