proyaop

By alternate interior angles, angleAEF is congruent to angleDAE. 

Since CAE is a straight line, then it has a straight angle. Therefore angleBAC + angleBAD + angleDAE = 180degrees.

angleDAE = angleAEF = 10angleBAC - 22degrees.

Since angleBAC = angleBAD because BA is an angle bisector, then we can simplify the equation to:

angleBAC + angleBAC + 10angleBAC - 22 = 180degrees.

To simplify:

12 * angleBAC = 202degrees.

To simplify:

angleBAC = (101/6)degrees.

Since we are looking for angleDAC, and we also know 2 * angleBAC = angleDAC, then angleDAC = $101\over3$degrees.

If the square root of $(63 - 3\sqrt{x})$ is an integer, then it has to be a perfect square.

Here are perfect squares under 63:

49 = 7^2 

36 = 6^2

25 = 5^2

16 = 4^2

9 = 3^2

4 = 2^2

1 = 1^2

0 = 0^2

Since there are 8 perfect squares under 63, then there are 8 real values of $x$ where the $\sqrt{(63 - 3\sqrt{x})}$ is an integer.

If the angles form an arithmetic sequence, then we can create a variable for the difference in angles.

If the largest angle is 128 degrees, then the second largest angle is 128 - x degrees, the third largest is 128 - 2x degrees, and the smallest angle is 128 - 3x degrees. 

We also know the interior angle sum of a quadrilateral adds up to 360 degrees.

Therefore, 128 + 128 - x + 128 - 2x + 128 - 3x = 360. 

Then we get x = 76/3 

Since we are looking for the second largest angle, it is just 128 - 76/3. 

Thus, the second largest angle is 308/3 degrees or 102.666666 degrees.

nvm Thanks for the reminder melody I am new, also i solved it so yea 

Triangle inequalities declares that the sum of two sides of a triangle must be greater than the third side of the triangle.

Then applying the triangle inequality to the problem, to find the third side 'x', we can write:

$11 + 17 > x$

Then the largest value of x that is under 28 is 27.

The longest possible integer length for the third side of the triangle is 27 units.

Since x is a 3 digit integer, then the equation is:

$100x = 11 + 997k$

The smallest value of k is 11, or else x could not be a 3 digit integer satisfying the restrictions. 

100x must end with two 0's, so 997k must end with 89. 

Since 7 is the only number that can multiply the last digit of 997 to make the new number end with a 9.

After some testing we get k = 37.

Here is our new equation:

100x = 11 + 997 * 37

100x = 36900

x = 369.

First we open parenthesis.

$w^2 + 26w + 169 = (3w + 7)(2w + 5)$

Then we open parenthesis on the right side of the equation.

$w^2 + 26w + 169 = 6w^2 +29w + 35$

Then we have to combine like terms.

$5w^2 + 3w -134 = 0$

Now we can use the quadratic formula: w = $-b {+\over} \sqrt{b^2 - 4ac}\over2a$ where a is the coefficient of $w^2$, b is the coefficient of w, and c is the constant.

Now to plug in the values into the formula: we get

w = $-3 {+\over}\sqrt{2689}\over10$

If you consider the "squares" as rectangles, then we can count the number of rectangle sizes there are.

For the size 1x1 there are 12 squares/rectangles.

For the next size which is a 2 x 1, there are 17 rectangles.

For the next size which is a 3 x 1, there are 10 of those rectangles.

For the next size which is a 4 x 1, there are 3 of those rectangles.

For the next size which is a 2 x 2, there are 6 of those squares/rectangles.

For the next size which is 2 x 3, there are 7 of those rectangles.

For the next size which is 2 x 4, there are 2 of those rectangles.

For the next size which is 3 x 4, there is 1 rectangle.

For the last size which is 3 x 3, there are 2 of those squares/rectangles.

Altogether there are 40 rectangles and 20 squares. (You might count squares as rectangles.)

This logic question has been answered here already:

First we simplify the radicals in the parenthesis. 

$\sqrt{5} = \sqrt{5}$

$\sqrt{45} = 3\sqrt{5}$

$\sqrt{125} = 5\sqrt{5}$

$\sqrt{80}=4\sqrt{5}$

Since we are adding up the radicals, our new expression is:

$(\sqrt{5} + 3\sqrt{5} +5\sqrt{5}+ 4\sqrt{5})^2$

Simplifying we get:

$(13\sqrt{5})^2$

This simplifies to:

845 which is the answer.