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masya89 [10]
3 years ago
13

When point J (-14,6) is rotated 180 degrees around the origin, what is the resulting point J'? a (14,-6) b (-14,6) c (14,6) d (-

14,-6)
Mathematics
1 answer:
elena55 [62]3 years ago
6 0

Answer:

J' = (14,-6)

Step-by-step explanation:

Given

J = (-14,6)

R_0 = 180^\circ

Required

Determine J'

When a point (x,y) is rotated 180 degrees around the origin, the new point is (-x, -y).

So:

J = (-14,6)

J' = (14,-6)

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The length of a rectangle is five times it’s width if the area of the rectangle is 320m^2 find its perimeter
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Answer:

Perimeter of the rectangle = 96 meters.

Step-by-step explanation:

Let the width of the rectangle = K

So, the length of the rectangle = 5K

Area  = 320 sq meters

Now, Area of a Rectangle  = Length x Width

⇒  320  = K x  5K

or,5K^{2} = 320\\or, K^{2}  = \frac{320}{5}   = 64

So, the vlue of K = 8 or, K = -8

But as K represents a length hence, K ≠ -8

So, the <u>width of the rectangle = 8 meters</u>

and the<u> length  of the rectangle = 8 x 5 = 40 meters</u>

Now, Perimeter = 2(length + width)

                            = 2(8 + 40) = 96 meters.

So, the perimeter of the rectangle = 96 meters.

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3 years ago
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Step-by-step explanation:

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Can someone please help me with this math problem.
mr Goodwill [35]
PART A:
Find the rate of change between 1980 and 1989
d for P₁ = 80 - 60
d for P₁ = 20

d for P₂ = 76 - 82
d for P₂ = -6

The rate of change in P₁ is 20 hundred per year. The rate of change in P₂ is -6 hundred per year.

PART B:
Find the rate of change between 1989 and 1996
d for P₁ = 100 - 80
d for P₁ = 20

d for P₂ = 70 - 76
d for P₂ = -6

The rate of change in P₁ is 20 hundred per year. The rate of change in P₂ is -6 hundred per year.

PART C:
Find the rate of change between 1980 and 1996
d for P₁ = 100 - 60
d for P₁ = 40

d for P₂ = 70 - 82
d for P₂ = -12

The rate of change in P₁ is 40 hundred per year. The rate of change in P₂ is -12 hundred per year.
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