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Angle BAC is 20 degrees.

f(x)=sec(x/2)+7 and f(x)=7sec(x/2) have asymptotes that occur every 2π units.

f(x)=sec(x)+7 has a functional value at on (-7, 7).

Therefore your answer is:

f(x)=7sec(x)

Let the angle C be 60º, and the angle D be 30º

The height of the trapezoid we may denote with an h

AB = 5           CD = 11           EG = 1.5

h / tan(60º) + h / tan(30º) = 6                  ( 11 - 5 = 6 )

h / 1.732 + h / 0.57735 = 6       ( simplify )

0.57735 * h + 1.732 * h = 6

2.3094 * h = 6

h = 2.598077423

EF = sqrt ( h² + 1.5² ) = 3   

Let the the 3rd pile =T
The 2nd pile =T + 2
First pile =[T - 1]
Fourth pile =2[T + 2]
T + T + 2 + T - 1 + 2T + 4 =20, solve for T
5T  + 5 = 20
5T = 15
T =3 quarters - 3rd pile
3+2 =5 quarters- 2nd pile
3-1 =2 quarters- 1st pile.
5 x 2=10 quarters - 4th pile

The radius of the circle is sqrt(7), so the area of the square is (2*sqrt(7))^2 = 28.

1. The maximum value of x is 5 + sqrt(3).

2. Using the angle bisector theorem, I found that the slope of the line in the middle is 7/5.

There is a difference of a factor of 4 between the two triangles. So, the legs of the 2nd triangle would be:

4 x 7 = 28 units(the shorter leg)  and  4 x 24 = 96 units(the longer leg).

So that: 100^2 = 28^2  +  96^2

Any number above 2^3 makes the two expressions relatively prime. Therefore, their greatest GCD = 1

6 x 1/2 = 6/2 = 3

This:  logn(2, 3)* logn(3, 4)* logn(4, 5) *logn(5, 6)*logn(6, 7)*logn(7, 8)*logn(8,9)*logn(9,10*.......*log(2008, 2009) is equivalent to: log(2009) / log(2).

So, a = 2  and   b=2009

Wouldn't 3 and 4 work?

John should give David 3 quarters =0.75 cents and should get 4 dimes =0.75 - 0.40 =0.35 cents that David owes John. So: m + n = 3 + 4 =7

Oh, OK. Thanks for that. I misread the question.

How did you make the text white lmao. I can only see it by highlighting over.

When x= 9, f(g(x)= f(3)= -6.

Answer is -6.

Thanks

n =1, 8, 15, 22, 29.........etc. Every 7th number gives a maximum GCD of 7.

No geno3141 is correct, the problem states to compute the expression when n = 2020, not when the expression equals 2020.

To find the number of terms in an arithmetic sequence we can use formula

\[n = \frac{a_n - a_1}{d}\].

In this case, n is the number of terms in the sequence so \(a_n\) is the last term of the sequence.

\(d\) is the common difference between each term.

Using this formula we plug in the terms we already know \(d = 7, a_n =2008, a_1 = 13 \).

\[n = \frac{2008-13}{7} +1= \frac{1995}{7} +1 = 285 +1 = \boxed{286}\]

geno3141 misread the question !! I think what the poster wants is this:

n · (n + 1) · (n + 2) / 6 =2020, solve for n

n = ~ 22. But then with 22, the total =2024 NOT 2020.

Yes.

$3.20 average per pound, so $0.20 per ounce.

The average price per ounce in cents for her total purchases is 20 cents.

The numbers that you have are:  (1/2)(2/3)(3/4)(4/5) ... (2017/2018)(2018/2019)(2019/2020)

Notice that the denominator of one term cancels the numerator of the next term, just leaving the

first numerator and the last denominator:  1/2020

hi other guest!

first of all, thank you for your given solution (however it was incorrect.) i did the same thing as you before with a and b and they were 3 and 1, but how do i find n?

This series creates a series of triangular numbers:

1  =  1

1 + 2  =  3

1 + 2 + 3  =  6

1 + 2 + 3 + 4  =  10

...

1 + 2 + 3 + ... + 2020  =     < which I don't need to calculate, because I won't need it >

The formula for the sum of n triangular numbers is:  Sum  =  n · (n + 1) · (n + 2) / 6

Sum  =  2020 · (2020 + 1) · (2020 + 2) / 6  =  1 375 775 540