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So we know that a unit square is just a 1 by 1 square, so the area of a unit square is 1.

Since rectangle EFGH has 1/5 of the area of the unit square, the area of EFGH is 1/5.

For simplicity, area will be denoted by [] below. (For example, the sentence above translates to [EFGH] = 1/5.)

\([CGF] = \dfrac12 [BCGF]\) (imagine the rectangle BCGF is a piece of sandwich and you want to cut it in halves.)

Similarly, \([AEH] = \dfrac12 [ADHE] \).

Then, we do some algebra.

\(\quad [AFCH]\\ = [CGF] + [EFGH] + [AEH]\\ = [EFGH] + \dfrac12 [BCGF] + \dfrac12 [ADHE]\\ = \dfrac15 + \dfrac12 \left([BCGF] + [ADHE]\right)\)

Then notice that [BCGF] + [ADHE] is just the remaining area when [EFGH] is taken away from the unit square, so that is 4/5.

\([AFCH] = \dfrac15 + \dfrac12 \cdot \dfrac45 = \text{(you do the remaining calculation)}\) 

Whenever (x', y') is a point on y = f(x), it is always true that y' = f(x').

Therefore \(7 = f(9)\).

Now consider the graph of y/2 = f(x) - 3. What happens when we substitute x = 9 on the right-hand side?

\(\dfrac y2 = f(9) - 3 = 7 - 3 = 4\)

Then y = 8.

Therefore, when x = 9, the point where y = 8 is the point lying on the graph of y/2 = f(x) - 3.

The required point is (9, 8). You can proceed to calculate the sum of the coordinates of that point.

If f(i) = i + 1, the only divisors of i are 1 and i. That means i is a prime. You can calculate the number of primes in the given range for your answer.

Whenever you see a circle in the form \(Ax^2 + Ay^2 + Bx + Cy + D = 0\), complete the squares to get something like \((x - h)^2 + (y - k)^2 = r^2\), then the center will just be (h, k) and the radius will just be r.

So, for this one, you move every term to the left-hand side first.

\(x^2 + y^2 + 4x + 16y - 18 = 0\)

Now, complete the square.

\((x^2 + 4x + 4) - 4 + (y^2 + 16y + 64) - 64 - 18 = 0\\ (x + 2)^2 - 4 + (y + 8)^2 - 64 - 18 = 0\\ (x + 2)^2 + (y + 8)^2 = 86\)

Can you take it from here to identify h and k?

Hint: You can write \(x + a\) as \(x - (-a)\) at all times.

Simplify the equation, and you will get \(x^2 + y^2 = 0\). 

This represents the point (0, 0) only, as no other points (x, y) satisfy the equation.

Now it is very obvious what the y-intercept is.

When you move every term to the left-hand side, you get \(x^2 + y^2 - 14x - 8y - 13 \leq 0\).

Complete the squares and you will get \((x - 7)^2 + (y - 4)^2 \leq 78\).

So this is actually just the area enclosed by the circle centered at (7, 4) and with radius \(\sqrt{78}\). Can you take it from here?

Determine whether each number is rational or not, so you can determine its function values.

f(-8) + f(-pi) + f(50) + f(9/2) = -8 + pi^2 + 50 + 9/2 = 93/2 + pi^2

You have d = 1/2, and the first equation just says a = (5 - 8d)/2. So you can substitute your answer into the first equation to get a.

You have to do a little equation solving here.

If you want a value of x to be equal to 4, you take the function -2x+2 and make it equal to 4 so you have -2x+2=4.

Your turn to solve it.

I believe I did everything right

What do you think it is?

Subsitute x = 0, x = 1, x = 2, and x = 3.

See what you get, and determine which graph makes more sense for the blue line. 

The answer is 10*sqrt(5)/5.

There are 24 ways of arranging the balls.

The number of sparse subset is 108.

The answer is 4.

To find the circumcenter, we first need to know the radius of the circle. 

We can rewrite the circle equation to: \(x^2-4x+y^2+8y=-11\)

We can now complete the square, and we get: \((x-2)^2+(y+4)^2=4+16-11\)

The left hand is equal to 9, meaning the radius is 3. 

We now know the circumcenter is 3. 

The equation for the circumcenter is: \(\large{{abc} \over {4[ABC]}}\)

But, we have an equilateral triangle, so \(a = b= c\). This means we can substitute both b and c for a. 

This gives us the equation: \(\large{{a^3} \over 4[ABC] } = 3\)

The formula for the area of an equilateral triangle is \({\sqrt 3 \over 4} s^2\)

We can substitute this in, and we get: \({{a^3} \over {{4 \sqrt3 \over 4} a^2}} = 3\)

We can simplify and divide by \(a^2\) to get: \(\large{a \over \sqrt3} = 3\)

This means that the side of the triangle is \(3 \sqrt 3 \). 

Can you find the area from here?

Hint: Use the formula for the area of the triangle

I googled the result so the responder can compare their result. I wish him fun and patience with this incredibly complicated task!

  !

57 is a multiple of 3

33 - 1 =32 integers =1,608 - their sum

I did a similar question yesterday.  ...

what have you done towards solving this question.

Originally there are  B bags in shop B   and  B+180 bage in shop A

then after transfer

 there will be 

  \(B+\frac{B+180}{5}\;\;bags\;\ in\; shop\;\;B\)

and

\(\frac{4}{5}(B+180)\)    bags in shop A

You can take it from there.

There are 19 integers between 1 and 100 with at leat one 6.

19  x  7 ==133 between 1 and 800

there are 100 numbers between 600 and 699

Total ==133 + 100 ==233 integers that have at least one 6

(2z - 1/sqrt(z) + z^3)^9.

\((2z - \frac{1}{\sqrt z}+ z^3)^9\)

Is this your intended question?

See https://web2.0calc.com/questions/product_3#r2

Here's one way:

C to Y is 37 and  C to L is 51  so the minimum distance possible between Y and L is 14

Now M to Y is 6 and  and M to L is 8 .   

 I already know that YL is a minimum of 14

So that means that   MYL must be collinear points  ( all on one line.

this means that C, Y, M, L are all  collinear.

You can finish it.

Here is a pic.