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Like this:

I'll leave you to finish it.

Hopefully all my errors are now fixed.

7 + 23 = 30 sum of the ratios

90 / 30 = 3 constant ratio

[90 - (7 *3) ]=69 - complementary angle "A"

[90 - (23*3 ]=21 - complementary angle "B"

Ratio of A to B = 69 / 21 = 23 / 7

[5 * 5]  -  [4 * 4.25] = 25 - 17 = 8 - was removed

576,000 * 4 * 48 = 110,592,000

110,592,000 = [2^15 * 3^3 * 5^3]^(1/3)

2^(15/3) * 3 * 5 =2^5 * 3 * 5 =32 * 15 =480

[576,000 * 4 * 48]^(1/3) = 480

Where is the diagram, lazy ?!

Let the number of small boxes = S

Let the number of large boxes = L

S  +  L = 6..................................(1)

10S  +  30L =100......................(2), solve for S, L

S=4 - number of small boxes

L= 2 - number of large boxes.

First, we can eliminate the percent of students who brought both: 100-10=90

Then, we can solve for the percentage of the students bringing an umbrella only

We can see that number of umbrellas compared to raincoats is 2:1

So, we can turn the amount of raincoats into a fraction then solve the problem!

The fraction is: 2/3 because we do 2+1=3 for how we are going to divide the rest 90% into

Finally, we do 90x2/3=60% for the final answer!

It's \(y + \lfloor y \rfloor\)

Thank you!

Is it supposed to be \(y - \lfloor y \rfloor \) or \(y + \lfloor y \rfloor \)?

Yes I can take it from here! Thank you so much for the answer and hint!

Alright. Notice that the red-shaded region is a quarter circle with a radius of 1. This means that its area is \((1^2 \pi) \div 4 = {\pi \over 4}\)

The area of the square is just 1 x 1 = 1, so the probability is just \(\color{brown}\boxed{ \pi \over 4}\)

First, move everything to the left-hand side: \(4t^2 - 22t + 7 \leq 0\)

Now, let's assume that we want to solve \(4t^2 - 22t + 7 = 0\). Using the quadratic formula, this would be \(t = {22 \pm \sqrt{22^2-4 \times 7 \times 4} \over 2 \times 4} ={ 22 \pm 2\sqrt{93} \over 8} = {11 \pm \sqrt{93} \over 4}\)

But, we haven't solved the problem yet! We need to find the values that satisfy the inequality. Note that all the points in between the interval of our zeros (\({11 - \sqrt {93} \over 4}\) and \({11 +\sqrt {93} \over 4}\)) will either be positive of negative. 

So, let's plug in \(t = 1\) because it's in the interval. This gives us \(4(1)^2 -22 (1) + 7 = -11\), which is less than 0. This means that the value of quadratic must be less than 0 within this interval. 

In interval notation, this would be \(\color{brown}\boxed{{11 - \sqrt {93} \over 4}, {11 + \sqrt{93} \over 4}}\)

Note: There's probably a better, easier way of solving this that I can't think of.

I need the whole explanation.  This is due soon.

Can you give the complete explanation?

And solve this one too: https://web2.0calc.com/questions/please-help-please-explain-your-answer

First, simplify the expression to \(-2t^2 + 20t - 10\)

Now, recall that the vertex occurs at \(t = -{b \over 2a} = -{20 \over -4} = 5\)

This means that if we substitute \(t = 5\) into the expression, we get the vertex. 

Want to try it?

Note that the equation is a circle centered at (0, 0) with a radius of 1. 

Using this, we can draw a diagram representing the outcomes:

Basically, the shaded region (red) is the area of the success and the black square (side length 1) is the area of the total region

Can you take it from here?

Thank you for trying but it says that it is incorrect.

The distance is 2*sqrt(13).

Sqrt{[7 * 6 * 5 * 4 * 3 * 2 * 1] + [6 * 5 * 4 * 3 * 2 * 1]}

Sqrt{[ 5040] + [720]}

Sqrt{ 5,760}= 2^7 * 3^2 * 5 = 2^6 * 3^2

Sqrt{2^6 * 3^2 * 2 * 5}

2^3 * 3 sqrt{2 * 5}

24sqrt{10} - final simplification.

56 * 360==20,160

20160 = 2^6 * 3^2 * 5 * 7

n==2^2 * 3^2 * 5^3 * 7^3 ==1,543,500

Check: 20,160  *  1,543,500==[31,116,960,000]^(1/4) ==420

The smallest distance is 2*sqrt(15)/3.

2. Completing the square, we get y = 2(x - 4)^2 - 25, so k = -25.