If the state legislature cannot agree on a state budget, what would MOST LIKELY happen?

Stacy hits the jackpot one day at the gumball machine. She puts in a quarter and gets 444 gumballs rather than 111. The radius o

inn [45]

Answer: 3617.28 cubic millimeters.

Step-by-step explanation: I used the formula of a sphere since a gumball is a shape of a sphere and plugged the radius into the formula.

I got that the volume of a single gumball is 288π cubic millimeters.

Since there are 4 gumballs, i multiplied the volume by 4.

The total volume of the four gumballs add up to be 1152π .

Lastly, I multiplied 1152 by pi

and got the answer to be 3617.28 cubic millimeters

8 0

3 years ago

Two landscapers must mow a rectangular lawn that measures 100 feet by 200 feet. Each wants to mow no more than half of the lawn.

Citrus2011 [14]

The total area of the complete lawn is (100-ft x 200-ft) = 20,000 ft².
One half of the lawn is 10,000 ft². That's the limit that the first man
must be careful not to exceed, lest he blindly mow a couple of blades
more than his partner does, and become the laughing stock of the whole
company when the word gets around. 10,000 ft² ... no mas !

When you think about it ... massage it and roll it around in your
mind's eye, and then soon give up and make yourself a sketch ...
you realize that if he starts along the length of the field, then with
a 2-ft cut, the lengths of the strips he cuts will line up like this:

First lap:
   (200 - 0) = 200
   (100 - 2) = 98
   (200 - 2) = 198
   (100 - 4) = 96

Second lap:
   (200 - 4) = 196
   (100 - 6) = 94
   (200 - 6) = 194
   (100 - 8) = 92

Third lap:
   (200 - 8) = 192
   (100 - 10) = 90
   (200 - 10) = 190
   (100 - 12) = 88

These are the lengths of each strip. They're 2-ft wide, so the area
of each one is (2 x the length).

I expected to be able to see a pattern developing, but my brain cells
are too fatigued and I don't see it. So I'll just keep going for another
lap, then add up all the areas and see how close he is:

Fourth lap:
   (200 - 12) = 188
   (100 - 14) = 86
   (200 - 14) = 186
   (100 - 16) = 84

So far, after four laps around the yard, the 16 lengths add up to
2,272-ft, for a total area of 4,544-ft². If I kept this up, I'd need to do
at least four more laps ... probably more, because they're getting smaller
all the time, so each lap contributes less area than the last one did.

Hey ! Maybe that's the key to the approximate pattern !

Each lap around the yard mows a 2-ft strip along the length ... twice ...
and a 2-ft strip along the width ... twice. (Approximately.) So the area
that gets mowed around each lap is (2-ft) x (the perimeter of the rectangle),
(approximately), and then the NEXT lap is a rectangle with 4-ft less length
and 4-ft less width.

So now we have rectangles measuring

   (200 x 100), (196 x 96), (192 x 92), (188 x 88), (184 x 84) ... etc.

and the areas of their rectangular strips are
   1200-ft², 1168-ft², 1136-ft², 1104-ft², 1072-ft² ... etc.

==> I see that the areas are decreasing by 32-ft² each lap.
   So the next few laps are
   1040-ft², 1008-ft², 976-ft², 944-ft², 912-ft² ... etc.

How much area do we have now:

   After 9 laps,    Area =   9,648-ft²
   After 10 laps, Area = 10,560-ft².

And there you are ... Somewhere during the 10th lap, he'll need to
stop and call the company surveyor, to come out, measure up, walk
in front of the mower, and put down a yellow chalk-line exactly where
the total becomes 10,000-ft².

There must still be an easier way to do it. For now, however, I'll leave it
there, and go with my answer of: During the 10th lap.

5 0

3 years ago

1. If a scale factor is applied to a figure and all dimensions are changed proportionally, what is the effect on the perimeter o

tester [92]

Answer:

Part 1) The perimeter of the new figure must be equal to the perimeter of the original figure multiplied by the scale factor (see the explanation)

Part 2) The area of the new figure must be equal to the area of the original figure multiplied by the scale factor squared

Part 3) The new figure and the original figure are not similar figures (see the explanation)

Step-by-step explanation:

Part 1) If a scale factor is applied to a figure and all dimensions are changed proportionally, what is the effect on the perimeter of the figure?

we know that

If all dimensions are changed proportionally, then the new figure and the original figure are similar

When two figures are similar, the ratio of its perimeters is equal to the scale factor

so

The perimeter of the new figure must be equal to the perimeter of the original figure multiplied by the scale factor

Part 2) If a scale factor is applied to a figure and all dimensions are changed proportionally, what is the effect on the area of the figure?

we know that

If all dimensions are changed proportionally, then the new figure and the original figure are similar

When two figures are similar, the ratio of its areas is equal to the scale factor squared

so

The area of the new figure must be equal to the area of the original figure multiplied by the scale factor squared

Part 3) What would happen to the perimeter and area of a figure if the dimensions were changed NON-proportionally? For example, if the length of a rectangle was tripled, but the width did not change? Or if the length was tripled and the width was decreased by a factor of 1/4?

we know that

If the dimensions were changed NON-proportionally, then the ratio of the corresponding sides of the new figure and the original figure are not proportional

That means

The new figure and the original figure are not similar figures

therefore

Corresponding sides are not proportional and corresponding angles are not congruent

so

A) If the length of a rectangle was tripled, but the width did not change?

Perimeter

The original perimeter is P=2L+2W

The new perimeter would be P=2(3L)+2W ----> P=6L+2W

The perimeter of the new figure is greater than the perimeter of the original figure but are not proportionals

Area

The original area is A=LW

The new area would be A=(3L)(W) ----> A=3LW

The area of the new figure is three times the area of the original figure but its ratio is not equal to the scale factor squared, because there is no single scale factor

B) If the length was tripled and the width was decreased by a factor of 1/4?

Perimeter

The original perimeter is P=2L+2W

The new perimeter would be P=2(3L)+2(W/4) ----> P=6L+W/2

The perimeter of the new figure and the perimeter of the original figure are not proportionals

Area

The original area is A=LW

The new area would be A=(3L)(W/4) ----> A=(3/4)LW

The area of the new figure is three-fourth times the area of the original figure but its ratio is not equal to the scale factor squared, because there is no single scale factor

5 0

3 years ago

In science class, the girl to boy ratio is 3 to 7. If there are 50 students in the class, how many are boys

Julli [10]

Answer:

there are 35 boys in the class

Step-by-step explanation:

love the anime character!

3 0

3 years ago

Read 2 more answers

How do u do this? I have a quiz tomorrow and this is so confusing. Any easier ways to do it?

ehidna [41]

Answer:

sure there is

Step-by-step explanation:

SSS means side side side so if 3 of the sides on each triangle are the same then its congruent Any time you see an A it stands for angle so you only have to learn the leters and then everything else is pretty straight forward

8 0

3 years ago