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Yes, at least 4 ... maybe more.  I am not sure.

that's some interesting terminology...

A line "clamped" by v?

Does that mean the line passes through v?

it looks like the answer is wrong. Do you know why?

10/11 is correct

14/22 + 10/22 - 4/22 = 20/22 = 10/11

for a girl you have a   8 out of 22 students chance....for an A you have an additional 4 (boys) out of 22 chance

8/22+4/22 = 12/22 = 6/11

for a boy there is a 14 out of 22 chance of being a boy   and an andditional 6 (girls) out of 22 for an A

14/22 + 6/22 = 20/22= 10/11

(Not ABSOLUTELY positive of this....but ....mostly ! )

https://web2.0calc.com/questions/help_36919

Look at the entire thread for the correct answer.....

Hmmmm.....   seems each person has to hug 7 other people....they get two hugs each week ...so it will take 4 weeks to complete this?

amount x interest x years  = amount earned         + original amount is the amount you would have at end of 4 years....

  thei simplifies to   amount x (1 +( interest x years))

                                  400    x (1+(.08 x 4))  =   400 x 1.32 = .......

John's scenario:

                 8437 x (1+(.055 x 1/2)) = .......             .055 is annual interest of 5 1/2%    the 1/2 is because it is only for 1/2 year

No guest-it is 146.

The asking guest has every opportunity to make another post and clarify their question.

It is quite likely that this person asked the question and left. Never even bothering to look at any answer.

So why does it matter?

Take a look at this old post and try to learn from it :)

Yex, \(5^{x+1}=5^x*5^1=30 \longrightarrow 5^x=6.\) Breaking. \(5^{3x+1}=5^{3x}*5^1, (5^x)^3*5^1=(6)^3*5=\boxed{1080}.\)

Hint: If it's easier for you, split the hexagon into six equilateral triangles, and calculate the area of each one using \(\frac{\sqrt{3}}{4}s^2\) . Then, multiply that by six.

To generalize. the area of a hexagon is \(\frac{3\sqrt{3}}{2}s^2\) .

Ok, all of these are relativaly simple.

The formula you will be working with for all of these is a^2+b^2=c^2

Im not going to give you the answers only explain how to do it.

#1. 28 and 21 are a and b. Insert them into the formula to get 28^2+21^2=c^2

Solve for 28^2+21^2 and get 1225

now you are at 1225=c^2

Your answer will be the square root of 1225.

#2. In this problem 6 and 7 are a and b.

Put them in the equation above to get 36+49=c^2

Your answer for this will be the square root of 85

#3. In this problem 2 times the square root of 3 is c and the square root of 3 is a.

Put them into the equation above to get the square root of 3 squared (3) + b^2=2times the square root of 3 squared (2 times the square root of 3.

This gets us to the Square root of 3+ b^2=2 times the square root of 3

Minus Square root of 3 from right side to get b^2 equals the square root of three. so b equals the square root of the square root of three

Have A nice day!

1. 28^2+21^2=c^2,\(c=35.\)

2. 6^2+7^2=c^2, c^2=85, \(c=\sqrt{85}.\)

3. \(c^2+(\sqrt{3})^2=(2\sqrt{3})^2, c^2=9, c=3.\)

Hexagon Area = 3*sqrt3 /2     * a^2      where a = base length (8 inches)

I'm not sure if I understand your question!!. I will give it my interpretation!

11001010100101011 +  110100011000100 = 1111111 0111101111 - this is their sum in base 10 = n

11001010100101011 +  110100011000100 = 1111111 0111101111 - this is their sum in base 2  = m

Then, if we interpret m in base 10, then obviously the difference n - m =0

Mostly, you can treat the < > as an equal sign  EXCEPT Mutiplying or Dividing both sides of the equation by a negative number  REVERSES the < or >

8<5p-2    add 2 to both sides

10 < 5p    Divide both sides by 5

2 < p    or   p > 2

I believe your 2nd equation should read:

bc+c+c=156 NOT 146.

If that is correct, then through simple substitutions, you should get the following values:

a = 20, b = 24, c = 6, d = 14

With above values, ALL your 4 equations balance out.

Hello again, Melody.  Yes, I know the PEMDAS protocol.  But sometimes we must recognize human ... er ... imprecision. 

.

4)

The median and the height of an isosceles trapezoid are equal in length. One diagonal of the trapezoid has length 12. What is the area of the trapezoid?

See the following image  ( not to scale )

Let AD be the diagonal......call the  median GJ  and   the  height AE , M

Let GH  , IJ   =  X

And let  the top  base  = B

Then  GJ  - HI  = GH + IJ  =  2IJ

So....M  - B  = GH + IJ  = 2X  = 2IJ  

And  since triangles BFD and BIJ  are similar...and 2BI  = BF  and 2BJ = BD....then 2IJ = FD

But 2IJ  = 2X

So.....

EF +  FD  = 

B + (2IJ ) =

B + (M - B)  =

 M  =

ED

And  because AED is a right angle.. and AED a right triangle......we have  that  

AE^2 + ED^2  = AD^2

M^2 + M^2  =12^2

2M^2   = 144

M^2  =  72

Note that the  area of the trapezoid is     (1/2) (height ( sum of the bases) = 

(height)  (sum of the bases) / 2 =

And ...(sum of the bases) / 2  =  the median M  ....so....

M  *   M  =  M^2  =    72   = area of the trapezoid

You could also use the log function if you are familiar with that:

5^x+1 =30

log(5^x+1)=log30

x+1 = log30/log5

x+1 = 2.113282752559378-1 = 1.113282752559378

Then  5^3x+1 = 5^4.339848257678134 = 1079.999999999998196774=~~1080