What two numbers add to -17 and multiply to -16

Write one digit on each side of 36 to make a four-digit multiple of 36. How many different solutions does this problem have?

vaieri [72.5K]

9514 1404 393

Answer:

3 solutions: 1368, 5364, 9360

Step-by-step explanation:

In order to be divisible by 36, the number must be divisible by 9 and by 4. The number will be divisible by 9 if the sum of its digits is divisible by 9. 36 already has a digit sum of 9, so the two added digits must sum to 9.

The number will be divisible by 4 if the least-significant digit is a multiple of 4. That is, it must be 0, 4, or 8. Then the corresponding most-significant digits must be 9, 5, or 1.

The three solutions are ...

1368 = 38 · 36

5364 = 149 · 36

9360 = 260 · 36

6 0

3 years ago

so a son and dad went on a fishing trip and the got in to a car accident the dad died and the son needed surgery and the doctor

Kobotan [32]

Answer:

if the condition of the son is very critical

a surgeon will say NO

Because after a successful operation, the son will become like a vegetable

and there's no point of doing so.

It's a torture to the son for the rest of his life.

5 0

3 years ago

Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

lidiya [134]

2x^2 - quadratic - monomial
-2 - constant - monomial
3x - 9 - linear - binomial
-3x^2 - 6x + 9 - quadratic - trinomial

3 0

3 years ago

A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six

kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

$V = Length \times Width \times Height \\\\$

Let x denote the length of the sides of the square as shown in the diagram.

The width of the shaded region is given by

$Width = 3 - 2x \\\\$

The length of the shaded region is given by

$Length = \frac{1}{2} (5 - 3x) \\\\$

So, the volume of the box becomes,

$V = \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V = \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V = \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V = \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\$

In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.

$\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\$

We are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

$a = 18 \\\\b = -38 \\\\c = 15 \\\\$

$x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 + 19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\$