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Yep....I got 5 for M, too.....but....you beat me to it,  hectictar.....!!!!!

What whole numbers when multiplied by another number equal 45?

5, 9, 15, 3

Hmm it looks like O can only into 2 pieces. the letters F, S, H, all tie at 4 pieces. M beats them all with 5 pieces.

This reminds me of several games I have played,

rarinstraw1195

Atomic mass is molar (or 1 mole of an element is this much mass)

 

CaCO3 is the element

 

Think of it as an addition problem: 

 

Ca + C + O + O + O

 

Substitute numbers

 

40+12+16+16+16

 

Simplify

 

52+48

 

Add 52 and 48

 

100

So, the molar weight (molar mass) of the compound CaCO3 is 100 g/mol (grams per mole)

I love you. Thank you so much. :-) 

Atomic mass is molar (or 1 mole of an element is this much mass)

CaCO₃ is the element

Think of it as an addition problem: 

Ca + C + O + O + O

Substitute numbers

40+12+16+16+16

Simplify

52+48

Add 52 and 48

100

So, the molar weight (molar mass) of the compound CaCO₃ is 100 g/mol (grams per mole)

Well let's look at the third one.

They tell us that the radius is 19 yards.

So how can we find the diameter?

Look at the equation for diameter.

diameter = 2 × radius

Oh, but we know what the radius is!

So let's replace the word "radius" with what we know the radius is, 19 yards.

diameter = 2 × 19 yards

Now just grab your calculator and type in, "2 × 19." Whatever you get is the diameter.

diameter = 38 yards

Does that make any sense?

3) Diameter = 19*2= 38

Circumference = 119.32

For the area, use the formula pi r^2

V ...............................No.............i can't Do it.....................It's so hard ;-;  ..............   :|

Ah you can do these! I believe in you!

Just use these equations and plug in what you know:

radius = ½ × diameter

diameter = 2 × radius

circumference = 3.14 × diameter

area = 3.14 × radius²

1)

radius = 11 inches

diameter = 2 × radius = 2 × 11 = 22 inches

circumference = 3.14 × diameter = 3.14 × 22 = 69.08 inches

area = 3.14 × 11² = 3.14 × 121 = 379.94 square inches

2)

radius = ½ × diameter = ½ × 20 = 10 feet

diameter = 20 feet

circumference = 3.14 × diameter = 3.14 × 20 = 62.8 feet

area = 3.14 × radius² = 3.14 × 10² = 3.14 × 100 = 314 square feet

I hope this made sense!

Can you do the third one by yourself? It's almost identical to the first one.

O  F    S     H       M

0   3     0    3      |4|

4+5=9, so 1800/9= 200. Each set is 200. There are 4 sets for the bagels, so that means the answer is 200*4=800. 

I'm guessing that 144 goes under the radical.

\(\sqrt{\frac{45}{144}} \\~\\ \frac{\sqrt{45}}{\sqrt{144}} \\~\\ \frac{\sqrt{(3)(3)(5)}}{12} \\~\\ \frac{3\sqrt{5}}{12} \\~\\ \frac{\sqrt{5}}{4}\)

If 144 is not under the radical:

\(\frac{3\sqrt{5}}{144} = \frac{\sqrt{5}}{48}\)

Thank you so much!

it is the x squared button to square stuff, but if you need a specific power, do X to the Y power, and you can put in your own power

0.0465847495312456187

9¾ + 4¾ + 3/10 + 3/10

\(9\frac34 + 4\frac34 +\frac3{10}+ \frac3{10} \\~\\ = \frac{39}{4} + \frac{19}{4} + \frac3{10}+ \frac3{10} \\~\\ = \frac{390}{40} + \frac{190}{40} + \frac{12}{40}+ \frac{12}{40} \\~\\ = \frac{390 +190+12+12}{40} = \frac{604}{40} = \frac{151}{10} = 15.1\)

Jacob wants to know how many students in his school enjoy watching space travel programs on TV. He poses this question to all 20 students in his history class and finds that 80% of his classmates enjoy watching space travel programs on TV. He claims that 80% of the school's student population would be expected to enjoy watching space travel programs on TV. Is Jacob making a valid inference about this population?

 No, it is not a valid inference because his classmates do not make up a random sample of the students in the school 

No, it is not a valid inference because he asked all 20 students in his history class instead of taking a sample from his math class

 Yes, it is a valid inference because his classmates make up a random sample of the students in the school 

Yes, it is a valid inference because he asked all 20 students in his history class

I believe this related to something known as  Gauss's Circle Problem....however...I think Gauss was only concerned with the lattice points inside the circle...

The problem actually boils down to finding how many integer pairs (m,n) exist such that:

m^2 + n^2 ≤ r^2

Anyway,  using "brute force," there are 35  lattice points above the x axis.....so by symmetry, there are also 35 below

And 11 lattice points lie on the x axis....so we have.....

[ 35 + 35 + 11 ]   = 81    {If I counted correctly....!!!  }

Note that, if we assume that each square unit of the circle contains one lattice point....then we should have about 25*pi  ≈ 79 points....so....81 is pretty close to this.....

P.S.  - Maybe heureka or Alan can enlighten us with a presentation of the exact formula for this....I don't know enough higher math to carry it out ....!!!!

550.0214294005214294

Nobody knows. It will always be a mystery.

\(\frac{7.9 \times 10-4}{8} = \frac{79-4}{8} = \frac{75}{8} = 9\frac{3}{8} = 9.375\)

500 * (1.08) = 540

540 / 16 = 33.75