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If we eliminate y from the first two equations, we get x+3z=10−2x, or x=5−3z. Substituting this into the third equation, we get [kx + z = 3,]so k(5−3z)+z=3. This simplifies to 6z−5k=2. For this to have no solutions, we must have 6z−5k=0, or k=6​.

Well, first of all, we can simplify the bottom equation:

6a + 2b = k + 3a + mb

into 

3a + 2b = k + mb

Because the top equation is also 3a+2b, and the bottom is also 3a+2b, and there are infinitely many solutions, that means:

k+mb = 2

And because b can be anything, that means m has to be 0 to work. that means k = 2

k = 2

m = 0

In triangle ACD, using the Law of Cosines:

Angle CAD =34.772 degrees

Angle BAC=90 - 34.772 ==55.228 degrees

Sin of BAC ==55.228 ==0.821428

Angle BDC=180 - (45 + 60) = 75 degrees

Angle ADC=180 - (45 + 30) =105 degrees

In triangle ADC, using the Law of Cosines:

CD = Sqrt(2)

DB = Sqrt(3)

The answer is 90%

The new weight is 150 kg.

BD = sqrt(2).

Let the total number of people be P. 7/12P = 280. Dividing both sides by 12, we get 1/12P = 40. The number of people not bored is P - 7/12P = 5/12P. 5/12P = 40 x 5 = 200. Hope it helps!

According to my calculations, 2+22=24

Please inform me if I'm wrong. 

Uh...... 24 obviously. 

Since the ratio of red paint to blue paint to white paint is 3:2:1, the greatest number of cans of mixed paint Rosencrantz and Guildenstern can make is the least common multiple of 3, 2, and 15, or 30 cans. To make the greatest number of cans, Rosencrantz and Guildenstern will use 30 cans of red paint, 20 cans of blue paint, and 10 cans of white paint. The amount of red, blue, and white paint left over is 48−30=18 cans, 40−20=20 cans, and 15−10=5 cans, respectively. Therefore, Rosencrantz and Guildenstern will have 18 cans of red paint, 20 cans of blue paint, and 5 cans of white paint left over.

To see why 30 is the greatest number of cans of mixed paint Rosencrantz and Guildenstern can make, notice that if they make 29 cans of mixed paint, they will have 1 can of red paint left over, and if they make 31 cans of mixed paint, they will have 2 cans of white paint left over.

RT = (2/5)SQ TS = (3/5)SQ

The solution is 51/8 < x < 73/9.

By the Pascal Identity, we have

C(a + b, 2) = C(a, 2) + C(a - 1, 1) C(b, 1) + C(a - 2, 2)

Substituting this into the expression, we get

C(a + b, 2) - C(a, 2) - C(b, 2) = C(a - 1, 1) C(b, 1) + C(a - 2, 2)

Since a≥2 and b≥2, we know that a−1≥1 and a−2≥0. Therefore, C(a−1,1)≥1 and C(a−2,2)≥0. This means that the expression is always nonnegative.

We can also simplify the expression by expanding the terms. We get

C(a + b, 2) - C(a, 2) - C(b, 2) = \frac{(a + b)(a + b - 1)}{2} - \frac{a(a - 1)}{2} - \frac{b(b - 1)}{2}

This simplifies to

(ab + a + b)/2.

This expression is always nonnegative, and it is equal to zero if and only if a=b.

We can think of Marvin's path as a sequence of 6 steps, 3 of which are rightward and 3 of which are upward. We can divide the set of all paths into two cases: those that go through (1,4) and those that do not.

If Marvin's path goes through (1,4), then he must take 3 rightward steps and 3 upward steps before he gets there. There are C(6,3)=20 ways to choose which 3 of the 6 steps are rightward.

If Marvin's path does not go through (1,4), then he must take 2 rightward steps and 4 upward steps before he gets there. There are C(6,2)=15 ways to choose which 2 of the 6 steps are rightward.

Therefore, there are 20+15=35​ ways for Marvin to reach his home.

\(3a+4b=5a-6b+24-3a\)

\(-2a=-10b+24-3a\)

\(a=24-10b\)

(a) No, 1/4 is not in the set S.

The number 1/4 can be represented as a geometric series with first term 1/4 and common ratio -1/3. However, the terms of the geometric series in the set S are all either 1/3 or -1/3. Therefore, 1/4 cannot be represented as a sum of terms in the set S.

(b) Yes, 1/7 is in the set S.

The number 1/7 can be represented as a geometric series with first term 1/7 and common ratio -1/3. The terms of the geometric series in the set S are all either 1/3 or -1/3. Therefore, 1/7 can be represented as a sum of terms in the set S.

Unfortunately, the answer was 5, but thank you for the effort, friend.

We have that 6n+1, 6n+2, and 6n+3 are all multiples of 3, so 6n+k must be odd. Therefore, k must be divisible by 2. If k is divisible by 4, then 6n+k is also divisible by 4, so k must be a multiple of 6.

Now, we need to show that if k is a multiple of 6, then 6n+k is relatively prime to each of 6n+3, 6n+2, and 6n+1.

Suppose k=6m for some positive integer m. Then \begin{align*} \gcd(6n+k, 6n+3) &= \gcd(6n+6m, 6n+3) \ &= \gcd(6n+3, 6m) \ &= \gcd(3, 6m) \ &= 3, \end{align*}unless m=0.

If m=0, then k=0, and in this case, 6n+k=6n is relatively prime to each of 6n+3, 6n+2, and 6n+1.

Similarly, \begin{align*} \gcd(6n+k, 6n+2) &= \gcd(6n+6m, 6n+2) \ &= \gcd(2, 6m) \ &= 2, \end{align*}unless m=0.

And \begin{align*} \gcd(6n+k, 6n+1) &= \gcd(6n+6m, 6n+1) \ &= \gcd(1, 6m) \ &= 1, \end{align*}unless m=0.

Therefore, if k is a multiple of 6, then 6n+k is relatively prime to each of 6n+3, 6n+2, and 6n+1.

Hence, the smallest positive integer k such that, for every positive integer n, 6n+k is relatively prime to each of 6n+3, 6n+2, and 6n+1 is 6​

Rosencrantz and Guildenstern can make the greatest number of cans of mixed paint if they use all of the red paint, because it is the most. They can use 48 cans of red paint, which is 12 groups of 4 cans.

To make the same ratio of colors, they need to use 8 groups of 4 cans of blue paint and 3 groups of 4 cans of white paint. This will leave them with 20 cans of blue paint and 12 cans of white paint.

Therefore, the greatest number of cans of mixed paint they can make is 12​.

Can't understand your question with all those $ signs! Translate it into plaintext.

3 + 2 + 1 ==6 sum of all ratios

48 + 40 + 15 ==103 cans of all 3 colors.

103 / 6 ==17+1/6 cans of any one color.

But we have maximum of 15 cans of white paint, therefore:

15  x  6 ==90 cans be mixed up.

15 cans of red paint ==15  x  3 ==45 cans of red paint

15 cans of blue paint==15 x   2==30 cans of blue paint

15 cans of white paint ==15 x 1==15 cans of white paint.

So, we mix: 45 red + 30 blue + 15 white ==90 cans in total

48  - 45 ==3 cans of red paint left over

40 - 30 ==10 cans of blue paint left over

15  - 15 ==0 cans of white paint left over.

We can find the greatest number of cans of mixed paint they can make by finding the greatest common factor of the number of cans of each color. The greatest common factor of 48, 40, and 15 is 12. Therefore, they can make 12​ cans of mixed paint. They will have 48−12=36 cans of red paint left over, 40−12=28 cans of blue paint left over, and 15−12=3 cans of white paint left over.