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Your question is statistics. It has absolutely nothing to do with calculus.

Nice solution tertre, but I think you meant to say $a^2+b^2=49?$

An easier way is to split the regular hexagon into six equilateral triangles. There are three more in the triangle $ABC$, so $\frac{6}{9}=\boxed{\frac{2}{3}}.$

$\text{Just eyeballing it you've got a regular hexagon with side length equal to}\\ \text{one third the side of the triangle. Let that triangle side length be }s\\ A_t = \dfrac{\sqrt{3}}{4}s^2\\ A_h = \dfrac{3\sqrt{3}}{2}\left(\dfrac{s}{3}\right)^2 = \dfrac{\sqrt{3}}{6}s^2\\ \dfrac{A_h}{A_t} = \dfrac{\frac{\sqrt{3}}{6}}{\frac{\sqrt{3}}{4}}=\dfrac 2 3$

Third one:  Look at your POSITIVE z-score table     25 ounces is   1oz/1.3oz/S.D =  0.7692 ~~.77 Standard deviations ABOVE the mean. This corresponds to

  .7794  z-score       SO  77.94 percent are LESS than this      77.94% x 1200 boxes =  935.28  ~~  935 boxes are 25 oz or less

another way of doing (1)

$h(x)=f(g(x))\\ h^{-1}(x) = g^{-1}(f^{-1}(x))$

$f^{-1}(x) = \dfrac{x-5}{7}\\ g^{-1}(x) = x+1\\ g^{-1}(f^{-1}(x))=\dfrac{x-5}{7}+1 = \dfrac{x+2}{7}$

Here are the first 10 rows of Pascal's Triangle. Just count the EVEN integers:

1   = R 0
1   1   = R 1
1   2   1   = R 2
1   3   3   1   = R 3
1   4   6   4   1   = R 4
1   5   10   10   5   1   = R 5
1   6   15   20   15   6   1   = R 6
1   7   21   35   35   21   7   1   = R 7
1   8   28   56   70   56   28   8   1   = R 8
1   9   36   84   126   126   84   36   9   1   = R 9

Zscore -0.5   standard deviations from mean is   -0.5 (1.8) + 12.5 = 11.25 ft

Look at your NEGATIVE z-score table to find .20    (1 - 80% = .20)       and find approximately  -.84 standard deviations below the mean = .2005  (which means 1-.2005 are ABOVE this).....        

55 gm - .84 x 4.2   = 51.47 gm    80% are greater than this

It has no constant because f(0)=0

c_3x^3+c_2x^2+c_1x

-c_3+c_2-c_1=15

c_3+c_2+c_1=-5

8c_3+4c_2+2c_1=12

4c_3+2c_2+c_1=6

-2c_3+2c_2-2c_1=30

*2

6c_3+6c_2+6c_1=-30

*6

add together

4c_3+8c_2+4c_1=0

divide by 4

c_3+2c_2+c_1=0

compare for x intercepts

c_3x^3+c_2x^2+c_1x=c_3+2c_2+c_1

3*17=51

https://web2.0calc.com/questions/need-help-p

1.) Let f(x)=7x+5 and g(x)=x-1.  If h(x)=f(g(x)), then what is the inverse of h(x)?

2.) If the point (3,6) is on the graph of y=g(x), and h(x)=(g(x))^2 for all x, then there is one point that must be on the graph of y=h(x). What is the sum of the coordinates of that point?

16.

you know that in triangle abc, a+b is more than c, c-b < a, and from there you can get that the largest possible side is 39, yielding 16 as the smallest.

😊😊

$p(q(-1))=4(-1)+7 = 3\\ \text{linear functions are a bijection, i.e. }\\ p(x)=a \text{ and } p(y)=a \Rightarrow x = y\\ p(2)=3 \text{ and } p(q(-1))=3 \Rightarrow q(-1)=2$

Thanks!

Number of term =7
First term =15
Common difference =3
Sum =N/2 * [2*F + (N - 1) * D], where N=Number of terms, F=First term, D = Common difference.
Sum = 7/2 * [2*15 + (7 - 1) * 3]
         = 3.5 * [30    +  (6 * 3)    ]
         = 3.5 * [30 + 18]
         = 3.5 * 48
         =168 - Total number of logs in the stack. 

Hi ThorganHarzard

Welcome to the   Web2.0calc   forum       

That is a very impressive answer but you lost me at the ''=2R"   bit.    What has it to do with the circumscribed circle?

I am not questioning your knowledge, I just don't understand.

hi

No you can't .