Owinner

1 Questions
12 Answers

0

4

1

+28

geometry

Four cubes of volumes $1 \text{ cm}^3$, $8 \text{ cm}^3$, $27 \text{ cm}^3$, and $125 \text{ cm}^3$ are glued together at their faces. What is the number of square centimeters in the smallest possible surface area of the resulting solid figure?

Owinner 4 ene 2025

+28

+1

We can plug in the points into the standard form of a parabola to get $a$, $b$, and $c$.

First, notice that $c$ must be equal to $0$ because when you plug in $(0,0)$ in, you get $c=0$. That way, we only have to solve for $2$ variables.

Now we can plug in the rest of the points, and we get the equations $8=4a-2b$ and $8=4a+2b$.

We can set the equations equal to each other and then we get $b=0$ as well.

Plugging $b=0$ into the equations, we get $a=2$. So $a+b+c$ is $2+0+0$ which is $2$

Owinner20 ene 2025

+28

+1

$y = -2x^2 + 8x - 15 - 3x^2 - 14x + 25$

First, let's simplify the equation a bit so we get a nice quadratic equation in standard form: $y = -5x^2-6x+10$.

There are a lot of ways to find the vertex of a parabola, but the easiest way is to use the formula $x=-b/2a$ and then plug that x-value into the equation to solve for $y$.

From the formula, $x=-6/10=-3/5$

Now we can substitute this $x$ value into the original equation: $y=-5*9/25+18/5+10=9/5+10=59/5$

Therefore, the vertex of the equation is $(-3/5,59/5)$. You can even see this if you use a graphing calculator.

Owinner20 ene 2025