Owinner

1 Questions
12 Answers

0

5

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 Jan 4, 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 $2$

OwinnerJan 20, 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.

OwinnerJan 20, 2025

+28

0

First, let's call the center of the circle, $C$. Essentially, all the sides are equal ($P_1 P_2, P_2 P_3, P_3 P_4$, etc.) since this is a regular polygon so we just need to find one of those sides.

Connect $C$ to $P_1$ and $P_2$, and those lengths are $1$ since the radius of the circle is $1$.

Now, we need to find the central angle between those two sides. Since this is a decagon, the central angle is $360/10$, which is $36^{\circ}$.

To find $P_1 P_2$, we can use Law of Cosines: $(P_1 P_2)^2=1+1-2*1*1*\cos(36^{\circ})$. When you solve for $P_1 P_2$, you get $P_1 P_2 \approx 0.618$.

Since all the sides are equal, the answer is just $10(P_1 P_2)$, so the answer is approximately $6.18$.

OwinnerJan 19, 2025

+28

0

$\frac{3-z}{z+1} \ge 2(z + 4)$

Multiply both sides by $z+1$: $3-z \ge 2(z+4)(z+1)$

When we simplify, we get: $3-z \ge 2z^2+10z+8$

Take everything to one side: $2z^2+11z+5 \le 0$

Now, we can factor the quadratic: $(2z+1)(z+5) \le 0$. Clearly, the critical points are $z=-1/2,-5,$ and $-1$ (because that is what is undefined in the original equation)

We can now test values below $-5$, between $-5$ and $-1$, between $-1$ and $-1/2$ and values greater than $-1/2$, and see which intervals satisfy the inequality.

Substiuting $-6$ in (interval below $-5$): $-9/5 \ge -4$ - True

Substituting $-2$ in (interval between $-5$ and $-1$): $-5 \ge 8$ - Not True

Substituting $-3/4$ in (interval between $-1$ and $-1/2$): $15 \ge 13/2$ - True

Substituting $0$ in (interval greater than $-1/2$): $3 \ge 8$ - Not True

There is two intervals which satisfy the original inequality, so our answer In interval notation is $(-\infty,-5)\cup(-1,-1/2)$

OwinnerJan 19, 2025

+28

+1

$4t^2 \le -9t + 12 - 4t + 23 + (2t - 1)(2t + 1)$

When you combine like terms and simplify, you get $4t^2 \le -13t+35+4t^2-1$ which turns into $13t \le 34$

Divide both sides by 13: $t \le 34/13$

So the answer in interval notation is $(-\infty,34/13)$

OwinnerJan 19, 2025

+28

+1

If one angle of the rhombus is 90 degrees, then the shape is a square, so the area is $12^2$, which is $144$.

OwinnerJan 12, 2025

+28

+1

Since the perimeter of the rectangle is $40$, we can make the equation $2(x+y)=40$, and then divide both sides by $2$ to get $x+y=20$ Then, since the diagonal is $10 \sqrt{2}$, we can make the equation $x^2+y^2=200$. We can substitute $x=20-y$ into the 2nd equation, and we get $400-40y+y^2+y^2=200$ When we simplify everything we get the equation: $y^2-20y+100=0$ -> $(y-10)^2=0$, so $y=10$ Therefore, $x=10$, so the area of the rectangle is $100$.

OwinnerJan 12, 2025

+28

0

Taking everything to the left side, we get $2x^2+20x+18=0$. To simplify the equation more, we can divide both sides by $2$ and get $x^2+10x+9=0$ Then, we can factor the quadratic and get $(x+9)(x+1)=0$, so the solutions to the equation are $x=9,-1$.

OwinnerJan 7, 2025

+28

+1

We can rewrite the angles as $\angle A = 60^\circ$, $\angle B = 60^\circ+x$, and $\angle C = 60^\circ+2x$. Now, we can use the Triangle Sum Theorem to solve for $x$. We can make the equation $60+60+x+60+2x=180$ Solving for $x$ gets $x=0$, so therefore all of the angles are $60^\circ$ so $\angle C = 60^\circ$.

OwinnerJan 7, 2025

+28

+1

We can square both sides of the first given equation to get $(x+y)^2=4$. This can be simplified to $x^2+y^2+2xy=4$. Now, we can substitute xy into the equation and get $x^2+y^2+14-2x^2-2y^2=4$, which when simplified turns into $x^2+y^2=10$, so 10 Is the answer.

OwinnerJan 7, 2025

+28

+1

I'm going to assume you meant angle YXW = 120 because otherwise, that wouldn't make sense Connect an altitude for X to WZ and call the point that intersects WZ, V. By similar triangles, we see that angle XZW is 30 degrees, and since XY and WZ are parallel (property of trapezoids), angle YXZ must be 30 degrees.

OwinnerJan 5, 2025

+28

0

The only possible way P can exist inside square WXYZ so that triangle PWX is a right isosceles is in the center of the square. Therefore, triangle PZY is 1/4th the square, and PY is part of the diagonal, so angle PYZ is 45 degrees.

OwinnerJan 5, 2025