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horsena [70]
3 years ago
14

Polygon y is a scaled copy of polygon x using a scal factor of 1/3. Polygon y area is what fraction of polygon x area

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
4 0

9514 1404 393

Answer:

  1/9

Step-by-step explanation:

The ratio of areas is the square of the scale factor for linear dimensions.

  Area(y)/Area(x) = (1/3)^2 = 1/9

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Sarah's rectangular garden has a perimeter of 327 +327 +218 +28 cm.
hodyreva [135]

Answer:

Sarah's rectangular garden has a perimeter of 900cm

5 0
3 years ago
Read 2 more answers
What is m∠HLK? thanks
Katen [24]
The answer is 80°.

If there is a line segment in a triangle which cuts two sides in half or if there is a line segment between two midpoints of a triangle's sides, that line segment must be parallel to the remaining side.

In this case, point L is the midpoint of FH, and point K is the midpoint of HG. So, LK is parallel to FG.

When a line cuts two parallel lines, there are angle pairs which are equal. In the problem above, the line segment FH cuts KL and GF.

The angles KLH and GFH are what we called "Corresponding Angles" and they are equal.

So if HFG is 80°, HLK must be equal to 80°
3 0
3 years ago
1. Mr. Hudson bought some shirts for the new members of his band. The cost for the number of shirts, including $3.99 shipping, w
nignag [31]

Answer:

6x + 4 - 1/100 = 77.5

Step-by-step explanation:

Inc shipping all shirts 77.49

Shirt individual cost = 12.25

We find the division first = 77.49-3.99 = 73.5  = 73 1/2

Then we divide 73.5 / 12.25 = 6

Then we have our equation

6x + 4 - 1/100 = 77.5

7 0
3 years ago
The set of all Real numbers between 4 and 7 including 4
erik [133]
4,5,6,7 is your answer
8 0
2 years ago
Find the work done by the force field F(x, y) = xi + (y + 5)j in moving an object along an arch of the cycloid r(t) = (t − sin(t
Nimfa-mama [501]

Integrate the force field along the given path (call it <em>C</em>):

W=\displaystyle\int_C\mathbf F(x,y)\cdot\mathrm d\mathbf r=\int_0^{2\pi}\mathbf F(x(t),y(t))\cdot\frac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt

=\displaystyle\int_0^{2\pi}\bigg((t-\sin t)\,\mathbf i+(6-\cos t)\,\mathbf j\bigg)\cdot\bigg((1-\cos t)\,\mathbf i+\sin t\,\mathbf j\bigg)\,\mathrm dt

=\displaystyle\int_0^{2\pi}(t-t\cos t+5\sin t)\,\mathrm dt=\boxed{2\pi^2}

5 0
3 years ago
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