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Alchen [17]
3 years ago
13

Greg is designing a clock face. Using the center of the clock face as the origin, he keeps its diameter at 10 units. Match the p

ositions of the hours on the clock face to their corresponding coordinates. Tiles 3 o'clock (0, -5) 6 o'clock (5, 0) 9 o'clock (0, 5) 12 o'clock (-5, 0) Pairs arrowBoth arrowBoth arrowBoth arrowBoth
Mathematics
2 answers:
Alex17521 [72]3 years ago
7 0
3o'clock - directly to the right of the origin; on the x axis - (5,0)
6o'clock - directly below the origin; on the y-axis - (0,-5)
9o'clock - directly to the left of the origin; on the x-axis, (-5,0)
12o'clock - directly above the origin; on the y-axis, - (5,0)
kozerog [31]3 years ago
7 0

Answer:

Given: Center of the clock face is at origin that is ( 0 , 0 )

To find : Position of the hours on the clock faces 3 o'clock , 6 o'clock , 9

o'clock and 12 o'clock.

We know that 3 o'clock on watch is on the straight right side of the center.

So, here 3 o'clock lies on right side of x-axis or positive side of x-axis.

Coordinate of 3 o'clock = ( 5 , 0 )

We know that 6 o'clock on watch is on straight bottom of the center.

So, here 6 o'clock lies on downside of y-axis or negative side of y-axis.

Coordinate of 6 o'clock = ( 0 , -5 )

We know that 9 o'clock on watch is on the straight left side of the center.

So, here 9 o'clock lies on left side of x-axis or negative side of x-axis.

Coordinate of 9 o'clock = ( -5 , 0 )

We know that 12 o'clock on watch is on the straight upside of the center.

So, here 12 o'clock lies on upside of y-axis or positive side of y-axis.

Coordinate of 12 o'clock = ( 0 , 5 )

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Answer:

The area of the region is 25,351 units^2.

Step-by-step explanation:

The Fundamental Theorem of Calculus:<em> if </em>f<em> is a continuous function on </em>[a,b]<em>, then</em>

                                   \int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) |  {_a^b}

where F is an antiderivative of f.

A function F is an antiderivative of the function f if

                                                    F^{'}(x)=f(x)

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x + 15 and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx

\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2

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