Firebolt

SA = 2(hl+hw+lw)

It doesn't matter which specifically is the height or width or length in this case.

1.8 x 0.5 = 0.9

1.8 x 0.5 = 0.9

0.5 x 0.5 = 0.25

2(0.9 + 0.9 + 0.25) = 2(1.8 + 0.25) = 2(2.05)

2 x 2.05 = 4.1, which is the surface area.

Since it is a square, all four sides are equal. Since the perimeter of the square is 144, 144/4 = 36 is the side length of one side of the big square. One of the sides is divided into 4 equal parts, so 36/4 = 9. This means one rectangle has a length of 36, and a width of 9. 

Perimeter: 36+36+9+9 = 90.

Thank you, all three of you. I appreciated it very much.

Thank you, all three of you, for helping me with this problem.

Thank you both; it really helped me.

If the arithmetic mean of a two numbers is 10 and the harmonic mean is 360. Find the geometric mean of the numbers.

SInce there are two numbers, and the arithmetic mean is 10, then the sum of the two numbers is 20.

The harmonic mean is $2/(1/a+1/b)$ where a and b are the two numbers. Since it equals 360, then 1/a + 1/b must equal 1/180.

We can rewrite 1/a + 1/b to get $(a+b)/ab$  and since we know a+b = 20, we get 20/ab = 1/180. Therefore ab = 3600.

The geometric mean is $\sqrt{ab}$ so we get an answer of 60. 

I hope this helped.

Let a and b be the solutions to $5x^2-11x+4=0$. Find $1/a + 1/b$ .

According to Vieta's Formula, the sum of the two roots are $-b/a$ and the product of the roots is $c/a$ for an equation in the form of$ax^2+bx+c=0$ .

In this case, $a=5, b= -11,$ and $c=4$ .

We can factor $1/a + 1/b$ to become $(a+b)/ab$. We can plug our values for a, b, and c in for the sum and product of the roots, giving us the answer of $11/4$ . We don't need to solve for the roots.

Therefore our answer is 11/4. 

I hope this helped.

In a right triangle, the hypotenuse is 65 and the altitude to the hypotenuse is 300/13. Find the legs of the right triangle.

We first want to find the area of the triangle, which is $(65*300/13)/2 = 750$

Note that we divide by two since this is a triangle. We could set up an algebraic equation to solve for the legs, but by some observation, we can find that this is a 25-60-65 triangle, with the area being 750 and the squares of the bases summing to to the hypotenuse's square.

Thus the legs of the right triangle are 25 and 60. 

I hope this helped.

The 3rd and 6th terms of an arithmetic progression are 94 and 85 respectively. Find the value of the 15th term.

From the 3rd term to the 6th term we have a 3 term skip, with a decrease of 9. Since this is an arithmetic progression, for every next term, we will subtract 3 from the pervious term since 9/3 =3.

15 - 6 = 9 more terms from the 6th term, so 9 x 3 = 27. Since the 6th term is 85, 85 - 27 = 58.

I hope this helped.

$k, a_2, a_3$ and $k, b_2, b_3$ are both nonconstant geometric sequences with different common ratios. We have $a_3 -b_3 =3(a_2 -b_2)$ . Find the sum of the common ratios of the two sequences.

We can rewrite these sequences as $k, ak, a^2k$ and $k, bk, b^2k$ , since they are geometric sequences. 

We now have:

$a^2k-b^2k = 3(ak-bk)$, which we can factor to become:

$k(a^2-b^2)=3k(a-b)$.

We can rewrite:

$k(a+b)(a-b)=3k(a-b)$

Dividing by $k(a-b)$ on both sides, we can get $a+b=3$

Since $a$ is the common ratio of the first sequence, and $b$ is the common ratio of the second, our answer is simply 3 .