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Veseljchak [2.6K]

2 years ago

15

An irregular parallelogram rotates 360° about the midpoint of its diagonal. How many times does the image of the parallelogram c

oincide with its preimage during the rotation?

1 answer:

Helga [31]2 years ago

6 0

2 times it will coincide

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A construction worker needs to put in a rectangular window in the side of a building. He knows from measuring that the top and b

maksim [4K]

Answer:

13

Step-by-step explanation:

The diagonals of a rectangle are congruent.

7 0

2 years ago

A regular pentagon is created using the bases of five congruent isosceles triangles, joined at a common vertex. The total number

Gala2k [10]

It's 54 degrees.

The internet is a wonderful friend. Type in "Pentagon split into triangles" on google images and you will find a diagram that finds it.

4 0

1 year ago

Read 2 more answers

A ball is kicked 4 feet above the ground with an initial vertical velocity of 51 feet per second. The function h(t)=−16t2+51t+4

Katyanochek1 [597]

Check the first picture below.

now, the exercise states to get it from the graph, so we have to graph it first.

Check the second picture below.

recall, the height in feet is on the y-axis, and the time in seconds is on the x-axis.

and rounded up to the nearest tenth will be 0.6.

7 0

1 year ago

Nina's monthly budget is $2,250. Every month she makes the following payments: $175 for insurance, $129 for utilities, $283 for

choli [55]

The total costs that Nina would have to pay is the sum of the cost of insurance, utilities, and taxes.
C = $175 + $129 + $283 = $587
The highest house payment she could afford is then the difference between her budget and the costs,
P = $2,250 - $587 = $1663
Thus, the answer would be letter B. House B because all the rest of the monthly payment are way over her budget.

6 0

1 year ago

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If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe

Natalka [10]

Answer:

They encircle the planet $3.76\times 10^{11}$ times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is $6.022\times 10^{23}$

Thus, the number of dust particles is:

$2.5\times 6.022\times 10^{23}=15.055\times 10^{23}$

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: $10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m$

If we line up the particles the distance they could cover is:

$15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}$

Circumference in meters:

$40,076km\times \frac{1000m}{1km}=40,076,000 m$

Therefore,

$\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}$

Hence, they encircle the planet $3.76\times 10^{11}$ times.

8 0

2 years ago