All Answers 358443 Answers

In a store window, there was a box of berries having a total weight of 200 kg.  The berries were 99% water, by weight.  After two days in the sun, the water content of the berries was only 90%, by weight.  What was the total weight of the berries after two days, in kg?  

You're starting with 200 kg of berries, and 99% of that is water.  

Since they're 99% water, that means that there's 1% of solid stuff.  

1% of 200 kg means that there are 2 kg of solid stuff.   

Some of the water evaporates in the sun, but the solid stuff doesn't.  

So there still remains 2 kg of solid stuff.  

Since the new water content is 90%, that means that now the solid stuff is 10%.  

So 2 kg (the solid stuff) is 10% of the total weight.  

That means the new total weight is 20 kg.  

_.

16 x sqrt(3) is 27.7

8. The area of triangle GBC is 18.

9. The area of triangle ABC is 54.

Assume the original price before any discounts ==$100

100%  -  40% ==60% first discount

$100  x  0.60 ==$60  - what you pay after first discount.

100%  -  45% ==55% secound discount

$60  x  0.55 ==$33 - what you pay after the second discount

$33 / $100 ==33% - percentage you end up paying after the two discounts.

The degree of g(x) is 3.

We are given that tetrahedron ABCO has angle AOB = AOC = BOC = 90. This means that triangle OAB, triangle OAC, and triangle OBC are all right triangles.

We are also given that a = OA, b = OB, and c = OC. This means that the sides of triangle OAB are a, b, and c.

A cube is inscribed in the tetrahedron so that one of its vertices is at O, and the opposite vertex lies on face ABC. This means that the cube is tangent to face ABC.

The side length of the cube is the distance from O to the opposite vertex. This distance is also the length of the altitude from O to face ABC.

The altitude from O to face ABC is the perpendicular bisector of edge BC. This means that the altitude from O to face ABC is also the median of triangle ABC.

The median of triangle ABC is half the length of side BC. This means that the altitude from O to face ABC is half the length of side BC.

The length of side BC is ab + ac + bc. This is because the length of side BC is the sum of the lengths of sides AB, AC, and BC.

Therefore, the side length of the cube is abc/(ab + ac + bc).

Guest, I don't think the problem is complete. I believe the LaTeX did not copy over (e.g. if she instead gave away [BLANK] of the original pile), could you please update the problem?

let the number of juniors be j and the number of seniors be s.

j/s = 3/2, so j = 3s/2

(j+6)/(s-1) = 2/3

using cross-multiplication, 3j+18 = 2s-2.

substituting j for 3s/2, 9s/2+18 = 2s-2.

9s+36=4s-2

5s=-38

s=-38/5   <----- Doesn't make any sense (???)

Guest, are you sure the problem is correct? Intuitively, if there are more juniors and less seniors, the new ratio would have been with more juniors and less seniors. For example, in this similar problem: https://web2.0calc.com/questions/help-pls_10673, the ratio increases.

Did I make a mistake in my calculations? If so, please point it out. Thank you!

This also helped alot with my promblem too,

Thanks.

It's the opposite if this promblem

The promblem is \(\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}}\)

Using the same method, that works out to 2.

\(\mbox{draw[blue, very thick] (0,0) rectangle (3,2)}\)Thanks, now I get it. 

Well the next promblem is

  \(\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}}\)

Thanks for the help!

Hi, in this type of questions we always follow these steps:

Step 1: let the a variable be equal to this expression.

For instance,

\(x=\sqrt{12+\sqrt{12+...}}\)

Step 2: Square both sides:

\(x^2=12+\sqrt{12+...}\)

But, notice, x was defined to be these "squareroots" right?

So: \(x^2=12+x\)

Step 3: We solve this equation.

\(x^2=12+x \\ \iff x^2-x-12=0 \\ \iff (x-4)(x+3)=0\\ \iff x=4 \\ \text{Note: We rejected x=-3 as we know x>0 (as the squareroot of any number is non-negative)}\)

Hi xud33,

We just substitute a=4 and b=y to get:

\(4+\sqrt{y+\sqrt{y+\sqrt{y+...}}} =10\)

Thus, 

\(\sqrt{y+\sqrt{y+\sqrt{y+...}}}=6\)

Next, the idea is to square both sides:

\(y+\sqrt{y+\sqrt{y+...}}=36\)

But notice, these nested squareroots are equal to 6 from the previous line!
Hence,

 \(y+6=36 \\ \iff y=30\)

I hope this helps!

I have no Idea of what to do.

I need help fast!!!

Good try but the answer was 30.

It says this:

There are no values of \(x \) that make the equation true.

No solution

I recommand you try to use mathway to solve this.

Here's the link:

\(20+69=\boxed{89}\)

\(\mbox{XY=} \frac{180}{4} \mbox{=} \mbox{45} \)

Thank you! I tried out your method...it made sense and the answer was actually correct! You're amazing!!

Hi, I hope this sketch helps:

Steps:

Ignore anything in red for now, we are going to find it.

But first, we need to identify what we really want. 
We want the area of Triangle BXA. To find the area of a triangle, it must have a base and a height.

The height is BC (The height is defined to be a vertical line connected the top most point of a triangle to the lower most point).

And the base (I.e. the side where the height is perpendicular on it) is AX.

Thus, the desired area is just: \(\dfrac{1}{2}*AX*BC\).

Now, we need to find both of BC and AX.

Firstly, for triangle ABC: we have AC = 12 and angle B is 60. 
So using trigonometry: \(\tan(60)=\dfrac{12}{BC} \implies BC=4\sqrt{3}\)

(Note: If trigonometry is not allowed, then you have to remember the ratios in the 30-60-90 triangle).

So we have found BC.

Next, consider triangle BXC:

Again, using trigonometry to find XC:

\(tan(30)=\dfrac{XC}{4\sqrt{3}}\implies XC=4\)

But we want AX not XC. Thus,  \(AX=12-4=8\)

Thus the area of the triangle is: \(\dfrac{1}{2}*8*4\sqrt{3}=16\sqrt{3}\)

I hope this helps!