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For N to satisfy all the choices given, we must find the LCM of 6, 7, and 8. 

We have that \(LCM[6,7,8]=168\)

Since they are all containing remainder 2, we have the equation

\(168x+2 \leq 1000\)

Since we want the max number, we set the equations to equal each other and solve for x. We get

\(168x+2=1000\\ 168x=998\\ x=\frac{499}{84} \approx 5.940\)

We round down, so our final answer is 5. 

Thanks! :)

First, let's use the FOIL method and combine two of the terms together. 

Combining two at a time, we get

\(n\left(n^2+16n-17\right)\left(n^2-54n+693\right)\)

Then, once again, combining the trinomials, we get

\(n\left(n^4-38n^3-188n^2+12006n-11781\right)\)

Distributing in the n. we finally find that

The final answwer is \(n^{5}-38n^{4}-188n^{3}+12006n^{2}-11781n\)

Thanksd! :)

You get n^5 - 38n^4 + 108n^3 - 336n^2 + 7220n + 148060

The remainder is 21.

(5 x 6 x 7 x 8) + 1  =  1680 + 1  =  1681 apples          

_.

Analyzing the Problem

We are given a triangle ABC with points P, Q, and R on sides AB, AB, and AC respectively. Specific ratios are provided for segment lengths related to these points. We need to find the ratio of the areas of triangles QBC and CRQ.

Solution Strategy

Relate Segment Lengths: Use the given ratios to express the lengths of AP, PQ, QB, AR, and RC in terms of a single variable.

Express Triangle Areas: Formulate the areas of triangles QBC and CRQ using the segment lengths obtained in step 1.

Find the Ratio of Areas: Divide the area of triangle QBC by the area of triangle CRQ and simplify the expression.

Steps to Solve

Relating Segment Lengths:

We are given: AP = PQ / 3 = QB / 2 and AR / 2 = RC / 3

Let x be the length of AP (or PQ / 3 or QB / 2).

Then, PQ = 3x and QB = 2x.

Similarly, let y be the length of AR (or 2 * RC / 3).

Then, RC = 3y / 2.

Expressing Triangle Areas:

Area of triangle QBC = (1/2) * Base * Height = (1/2) * QB * RC = (1/2) * 2x * (3y / 2) = 3xy

Area of triangle CRQ = (1/2) * Base * Height = (1/2) * CR * PQ = (1/2) * (3y / 2) * 3x = 4.5xy

Ratio of Areas:

[QBC] / [CRQ] = (3xy) / (4.5xy) = 2 / 3

Answer

Therefore, the ratio of the areas of triangles QBC and CRQ is 2:3.

Let's work from bottom up. Starting with the smallest, we have \(\frac{1}{\frac{17}{4}}=\frac{4}{17}\)

Then, we have \(4+\frac{4}{17}=\frac{72}{17}\)

\(4+\frac{1}{4+\frac{1}{4}}=\frac{72}{17}\)

Thus, we have

\(\frac{1}{4+\frac{1}{4+\frac{1}{4}}}=4+\frac{17}{72}=\frac{305}{72}\)

The final answer is \(\frac{305}{72}\)

Let's focus on the two smallest numbers first. 15 and 24. 

Since N leaves a remainder, we can elmiinate the factors. There are

Thus, N cannot be \( 3, 5, 2, 4, 6, 8, 12\)

This leaves 7, 9, 10, 11, 13, and 14.   Now, let's test out all the other numbers

15/7 = 2 R 1      24/7 = 3 R 3      so it isn't 7     

15/9 = 1 R 6      24/9 = 2 R 6      so 9 looks hopeful, we'll try another   

1457/9 = 161 R 8                        so 9 is eliminated   

15/10 = 1 R 5     24/10 = 2 R 4    so it isn't 10   

15/11 = 1 R 4     24/11 = 2 R 2     so it isn't 11   

15/13 = 1 R 2     24/13 = 1 R 11   so it isn't 13   

15/14 = 1 R 1     24/14 = 1 R 10   so it isn't 14   

mmm...I don't think there is a solution

Thanks! :)

Let's use a handy trick to solve this problem. 

First off, let's set a variable. Let's set \(X=2.\overline{57}\)

For that value of X, then we have that

\(100 X = 257.\overline{57}\)

Now, we subtract x from 100x. Note that the repeated part of the decimal cancels out. Thus, we have

\(\eqalign{100 X &= &\hfill257.57...\cr X &= &\hfill2.57...\cr \hline 99X &= &255\cr}\)

We end up witht he equation

\(99 X = 255\)

Simplifying and isolating x, we get 

\(X = \frac{255}{99}\)

Thus,

The answer is \(2.\overline{57} = 2 \frac{19}{33}\)

Thanks! :)

First, let's turn the decimal into a fraction, unsimplified. We do the process

\(\frac{0.11012}{1}\times \frac{100000}{100000}= \frac{11012}{100000}\)

Now, we must find common factors. 

The GCD of the two numbers are 4, so dividing by 4, we get

\(\frac{11012 \div 4}{100000 \div 4}= \frac{2753}{25000}\)

From here, we can no longer simplify, so

The answer is \( \frac{2753}{25000}\)

Thanks! :)

I didn't follow an algorithm. I legit tried every single possibilitility until I found 1. 

When wee have A, B, C = 1, 0, 1, we get the number \(10011010011101110\)

Converting this into base 10, we have

\((10011010011101110)_2 = (1 × 2^{16}) + (0 × 2^{15}) + (0 × 2^{14}) + (1 × 2^{13}) + \\(1 × 2^{12}) + (0 × 2^{11}) + (1 × 2^{10}) + (0 × 2^9) + (0 × 2^8) + \\(1 × 2^7) + (1 × 2^6) + (1 × 2^5) + (0 × 2^4) + (1 × 2^3) + \\(1 × 2^2) + (1 × 2^1) + (0 × 2^0) = (79086)_{10}\)

Dividing by 7, we get \(79086/7=11298\)

THUS, 

The answer is \((A, B, C) = (1, 0, 1)\)

Thanks! :)

*I'm so tired

It is impossible. There is no way every scientist can sit next to a mathematician. 

Look at the imahe. 

The value of N is \(124\)

The value of N is \(212\)

You are missing values. 

How many replublicans are there?

How many democrats are there?

The value of N is \(37\)

First, let's combine all like terms and reaarange so we have the shape of a quadratic formula. 

We get

\(x^2 + 15x + k \)

Having a double root means that the descriminant is greater than 0, so we have the equation

\(15^2 - 4 (1)(k) = 0 \\ 225 - 4k = 0 \\ 225 = 4k \\ k = 225 / 4\)

thus, The value of k is \(k = 225 / 4 \)

Thanks! :)

First, let's multiply by the LCM of the denominators. This will get rid of all the denominators and leave us with a nice clean sheet. We get

\(10\left(x+4\right)=5\left(x-3\right)\left(x+5\right)+2\left(x+7\right)\left(x+5\right)\)

Doing the painful process of multiplying out and combining all like terms, we get

\(10x+40=7x^2+34x-5\)

Moving all terms to one side and combining all like terms, we get

\(7x^2+24x-45=0\)

Now, applying the quadratic equation, we find that

\(x_{1,\:2}=\frac{-24\pm \:\sqrt{24^2-4\cdot \:7\left(-45\right)}}{2\cdot \:7}\)

The 2 values of x are \(x=\frac{3\left(\sqrt{51}-4\right)}{7},\:x=-\frac{3\left(4+\sqrt{51}\right)}{7}\)

Thanks! :)

There are actually multiply parabolas that satsify the conditions given.

The simplest one is \(y=2x^2\)

It has a vertex at (0, 0) and faces upwards. 

If this the case, the answer is \(2+0+0 =2\)

Thanks! :)

ABJelly, I believe you were referring to this box in the problem

I understand you don't know how to add an image.