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Consider the equation as two separate equations: \(5x + 3 = 4x - 1\) and \(-(5x + 3) = 4x - 1\)

Now, solve both equations for x as you normally would. Be careful of the negative in the second equation. 

The only solution is x = -4.

See the answer here:  https://web2.0calc.com/questions/counting_34169

Let the perimeter ==P

Then the side length of the square==P/4

Area of the square==(P/4)^2==P^2 / 16

Side length of the equilateral triangle==P/3

Area of the triangle ==1/2 * (P/3)^2 * sin(60)

                                    ==1/2 * P^2/9 * (sqrt(3/2)

                                    ==P^2 / [12 sqrt(3)]

Area A / B ==[P^2 / 16] / [P^2 / (12sqrt(3))]

                    ==3 * sqrt(3) / 4

The radian measure is pi/2.

Let j be the number of juniors and s be the number of seniors. We know that 3/5j=2/3(6/7s), so 3/5j=4/7s. Multiplying both sides by 35/12, we get 21j=20s, so j=20s/21. Thus, the fraction of the group that passed the test is 3/5j+6/7s = 3/5(20s/21)+6/7s = 44/35​.

Would that be 1.5 in a straight line or around the arc of the circumference?  

We can compute the areas of the circumcircle and incircle of an equilateral triangle with side length "a" using formulas involving their radii.

First, let's find the circumradius (R) and the inradius (r) of the equilateral triangle.

For an equilateral triangle with side length "a", the circumradius R is given by the formula:

R = a / (2 * sin(π / 3))

Since sin(π / 3) = √3 / 2, we have:

R = a / (2 * (√3 / 2))
R = a / √3

Now, let's find the inradius. In an equilateral triangle, the inradius r is given by the formula:

r = a / (2 * √3)

Now that we have the radii, we can compute the areas of the circumcircle (x) and the incircle (y) using the formulas:

x = π * R^2
y = π * r^2

Substituting the expressions for R and r, we have:

x = π * (a / √3)^2
x = π * (a^2 / 3)

y = π * (a / (2 * √3))^2
y = π * (a^2 / 12)

Now, compute x - y:

x - y = π * (a^2 / 3) - π * (a^2 / 12)
x - y = (4 * π * a^2 / 12) - (π * a^2 / 12)
x - y = (3 * π * a^2 / 12)
x - y = (π * a^2 / 4)

So the difference between the areas of the circumcircle and the incircle of an equilateral triangle with side length "a" is (π * a^2 / 4).

Thank you. Your example didn't work for me but I managed to fix it, which was basically the same thing!

if AB is a diameter then angle ACB is 90 degrees. The bisector of angle ACB is also a diameter. Hence a right angled triangle and we just need to find the hypotenuse of the isosceles 90 degree triangle whose side is length 8. sqr of 8^2 + 8^2 = sqr128

p = 17, q = 20.

sin angle PAQ = 2/7.

The area of triangle ABC is 14*sqrt(5).

Let D be the vertex of the cube that lies on face ABC and E, F, G be the vertices of the cube adjacent to D.

Since AOB is a right angle, we have AD^2 + BD^2 = AB^2. Similarly, we have AD^2 + CD^2 = AC^2 and BD^2 + CD^2 = BC^2.

Also, since D is a vertex of the cube, we have DE = DF = DG. Let x be the length of the side of the cube. Then we have AD = AE = AF = a - x, BD = BE = BG = b - x, and CD = CF = CG = c - x.

Using the Pythagorean theorem in triangles ADE, BDE, and CDE, we have

(x - a)^2 + DE^2 = AD^2 (x - b)^2 + DE^2 = BD^2 (x - c)^2 + DE^2 = CD^2

Substituting DE = DF = DG = x/√2 and simplifying, we get

(x - a)^2 + (x^2/2) = (a - x)^2 (x - b)^2 + (x^2/2) = (b - x)^2 (x - c)^2 + (x^2/2) = (c - x)^2

Expanding the squares and simplifying, we get

3x^2 = a^2 + b^2 + c^2 - 2ab - 2ac - 2bc

Solving for x, we get

x = √(2abc(a + b + c))/(2(ab + ac + bc))

Substituting this value of x in the expressions for AD, BD, and CD, we get the side lengths of the cube:

AD = a - √(2abc(a + b + c))/(2(ab + ac + bc)) BD = b - √(2abc(a + b + c))/(2(ab + ac + bc)) CD = c - √(2abc(a + b + c))/(2(ab + ac + bc))

The length of the side of the cube is the minimum of these three lengths, so we have

x = min{AD, BD, CD} = a - √(2abc(a + b + c))/(2(ab + ac + bc))

Therefore, the side length of the cube is

abc/(ab + ac + bc).

Q.E.D.

what is 36 over 360 times 30  

This is pretty elementary stuff to be asking.  Anyway....  

           36           30                1            30  

        –––––  x  –––––   =    –––––  x  –––––   =   3  

          360           1                10            1  

         OR  

               36                      1                    1  

        –––––––––   =   –––––––––   =   –––––  

          360 x 30             10 x 30              300  

Find the constant term in the expansion of (2z - 1/(z*sqrt(z)) + 1)^4.

Even though this expression looks like a monster, the constant term simplifies to one.

Because 1/(z*sqrt(z))=1/(z^(3/2)=z^(2/3), the fraction cannot simplify in only the fourth power. Therefore, the only constant term is 1*1*1*1, which is 1.

I hope this helps!

It is actually 8

I checked and that is incorrect

You made a mistake, it was 10x*srqt(21x)

You said all twice