All Answers 358468 Answers

PM = 2*sqrt(11)

The probability is 4/47.

That answer is wrong, it's actually 144.  But thanks for trying to help

--Qube73

(a,b) = (-1,13).

The answer is 50.

1/a + 1/b = 27.

x can be 3 and 4.

asinus answered one of your questions and you gave him no feed back at all.

You are expected to respond to answerers, especially if they have obviously put time into the answer.

BC = 16/3.

BZ = 34/9

Angle A = 81 degrees.

y = 4/9.

The answer is 3 - 8i

There are 45 possible dinner combinations.

 What is the value of a+b+c?

Hello Guest!

\( x^2 + 15x + 54=0\\ x=-7.5\pm \sqrt{56.25-54}\\ x\in \{-9,-6\}\\ (x+a)(x+b)=(x+6)(x+9)=0\\ a=6\\ b=9\)

\(x^2 - 17x + 72=0\\ x=8.5\pm \sqrt{72.25-72}\\ x\in \{8,9\} \\ (x-b)(x-c)=(x-9)(x-8)\\ b=9\\ c=8\\ \color{blue}a+b+c=6+9+8=23\)

  !

This simplifies to 8*sqrt(10x).

y = 25.

The answer is 22.

Find the center and radius of this circle.

Hello Guest!

\( x^2 - 2x + y^2 = 24\\ -y^2=x^2-2x-24\\ x=1\pm \sqrt{1+24}\\ x\in \{-4,6\}\)

The center of the circle is (1,0).

The radius of the circle is 5.

\(x^2-2x+y^2-24=0\\ (x-1)^2-1 +y^2=24\\ (x-1)^2+y^2=25\)

The center of the circle is (1,0).

The radius of the circle is 5.

  !

\(f(x)=ax^3+bx^2+cx+d\\ a=1\\ f(x)=x^3+bx^2+cx+d\\ -b=r+s+t\\ c=rs+rt+st\\ -d=rst\)

So I am asked to find c

\(f(x)=x^3+bx^2+cx+d\\ f(6)=6^3+b*6^2+c*6+d\\ f(6)=216+36b+6c+d\\~\\ f(-6)=-216+36b-6c+d\\~\\ f(6)-f(-6)=432+12c=7-2\\ 12c=-427\\ c=35\frac{7}{12}\)

You need to check for careless errors.

The side length of the square is 150/13.

The remainder is 6x - 13.