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Here's how to find the number of ordered pairs (a, b) of integers:

Multiply both sides by the common denominator (a + 5) in the first term:

a + 2 = b * (a + 5) / 4

Simplify and expand the expression on the right:

4 * (a + 2) = b * (a + 5)

4a + 8 = ab + 5b

Move all terms containing a to one side:

4a - ab = 5b - 8

a(4 - b) = 5b - 8

Factor out the greatest common factor (GCF) in both sides:

a(b - 4) = 5(b - 8/5)

Analyze the equation:

This equation implies that a and (b - 4) are factors of 5.

Since a and b are integers, the possible factors of 5 are 1, -1, 5, and -5.

Consider each factor combination:

Case 1: a = 1, (b - 4) = 5 - b = 9

Case 2: a = -1, (b - 4) = -5 - b = -1

Case 3: a = 5, (b - 4) = 1 - b = 5

Case 4: a = -5, (b - 4) = -1 - b = -3

Check for solutions:

All four cases lead to valid integer solutions for a and b: (1, 9), (-1, -1), (5, 5), and (-5, -3).

Therefore, there are 4 ordered pairs (a, b) of integers that satisfy the given equation.

Let's analyze the situation step-by-step:

Initial Weight:

Total weight = 200 kg

Water content = 98%

Non-water content = 2% (100% - 98%)

Weight of non-water content = 200 kg * 2% = 4 kg

Water Evaporation:

After 2 days, water content decreases to 95%.

Water loss = 3% (98% - 95%)

Weight of water lost = 200 kg * 3% = 6 kg

Final Weight:

All non-water content remains (it doesn't evaporate).

Final weight = Weight of non-water content + Remaining water

Final weight = 4 kg + (200 kg - 6 kg) = 4 kg + 194 kg = 198 kg

Therefore, the total weight of the berries after two days in the sun is 198 kg.

Here's how to find the solution for z:

Multiply each equation by its denominator:

xy = x + y (Equation 1)

xz = 2(x + z) (Equation 2)

yz = 4(y + z) (Equation 3)

Rearrange equations to isolate z:

From Equation 2: z = (xz - 2x) / 2 (Divide both sides by x)

Substitute this expression for z in Equation 3: yz = 4(y + (xz - 2x) / 2)

Simplify and solve for y:

yz = 2y + 2xz - 4x

Group terms with y: yz - 2y = 2xz - 4x

Factor out y: y(z - 2) = 2x(z - 2)

Divide both sides by (z - 2) and solve for y: y = 2x

Substitute y back into Equation 1:

xy = x + 2x

Combine like terms: xy = 3x

Divide both sides by x: y = 3

Combine information about y from steps 3 and 4:

We know y = 3 and y = 2x.

Therefore, 3 = 2x.

Solve for x: x = 3/2.

Substitute x and y back into the expression for z from step 2:

z = (x(3/2) - 2(3/2)) / 2

Simplify: z = (3/2 - 3) / 2

Therefore, z = -3/4.

Therefore, in the solution where the system of equations has exactly one solution, z is -3/4.

In general, in a geometric series, three terms, say \(a{r}^{k}, a{r}^{k+1}, a{r}^{k+2}\), then \(a{r}^{k}*a{r}^{k+2}=({a{r}^{k+1}})^{2}\), or, otherwise, in three terms of a geometric series, the middle term squared, equals the product of the two next to it.

Using this, we get the equation

\((60+x)(140+x)={(120+x)}^{2}\), where x is the added number.

\(8400+200x+{x}^{2}=14400+240x+{x}^{2}\). Cancelling out the x²:

\(8400+200x=14400+240x\)

\(40x=-6000\)

\(x=-150\).

The resulting series will be:

\(-90, -30, -10\).

The common ratio is \(\frac{-30}{-90}=\frac{1}{3}\). So, the resulting sequence has a common ratio of 1/3.

\(\begin{cases} a+2b=-3-7a+b\\4a+2b=k-5a-mb \end{cases}\)

\(\begin{cases} 8a+b=-3 \\ 9a+(2+m)b=k \end{cases}\)

To have infinitely, these two equations must be exactly equal, or in other words, one equation is simplifiable to the other. (Another way to visualize it, is that these two lines, when plotted are the same graph).

So, by \(8a*\frac{9}{8}=9a\), the second equation is the first equation, both sides multiplied by \(\frac{9}{8}\).

For the b terms,\(\frac{9}{8}*1=2+m\)

For the constant term, \(-3*\frac{9}{8}=k\)

Solving, we get \(m=\frac{-7}{8}, k=\frac{-27}{8}\). 

\(\sqrt{5}<\pi\), plug into the first case.

\(f(f(f(\sqrt{5})))=f(f(4\sqrt{5}))\)

\(4\sqrt{5}>5.2\), plug into the third case.

\(f(f(4\sqrt{5}))=f(70)\).

\(70>5.2\), plug into third case.

\(f(70)=4893\).

So, \(\underline{\bf{f(f(f(\sqrt{5})))=4893}}\).

The common difference between terms is \(\frac{\frac{1}{4}}{16}=\frac{1}{64}\).

The 30th term is 30-24=6 terms away, meaning 6 common differences from 24.

So, the 30th term is \(\frac{1}{4}*{(\frac{1}{64})}^6\), \(\frac{1}{{2}^{2}}*\frac{1}{{2}^{36}}=\frac{1}{{2}^{38}}\), is our final answer.

"I wasn't really sure if you could separate the magnitude of z+w to the sum of two magnitudes."

You can't in general!

Connent the adjecent points together to form a chord. In a cricle the radius is a perpendicular bisector of a chord, therefore:
\(A=(-3,-4)\\ B=(3,-4)\\ C=(3,4)\\ \text{we need to find: } O=(x,y)\)

Now perp bisector of 

\(\overline {AB} \text{ is } x=0\\ \overline{BC}\text{ is } y=0\)
\(x=0 \text{ and } y=0 \text{ meet at:}\\ \boxed{(0,0)}\)

Thank you! It makes more sense now!

Thank you for your help! Could you explain how you got that? I wasn't really sure if you could separate the magnitude of z+w to the sum of two magnitudes.

Here are the steps for problem 2:

Find the equation of the perpendicular line passing through A: The slope of the given line is 21​, and its negative reciprocal is −2.

Thus, the equation of the perpendicular line passing through A will be of the form y=−2x+b. We can plug A's coordinates into this equation to solve for b: $1=−2(7)+b⟹b=15.$

Therefore, the equation of the perpendicular line is y=−2x+15.

Find the intersection point of the two lines: The intersection point of the two lines will be the midpoint of the segment connecting A and its reflection B.

To find it, we can solve the system of equations: ${y=21​x+4y=−2x+15​$

Substituting the first equation into the second equation, we get: $21​x+4=−2x+15.$

Solving for x, we find x=35​. Substituting this value back into either equation, we find y=317​.

Therefore, the intersection point is (35​,317​).

Find the reflection point B: Since the reflection point B is the same distance from A as the intersection point is from A, we can find B by calculating the vector from A to the intersection point and multiplying it by 2.

The vector from A to the intersection point is (35​−7,317​−1)=(−316​,314​).

Multiplying this vector by 2, we get (−332​,328​). Finally, adding this vector to A's coordinates gives us the coordinates of B: (7,1) + (-32/3, 28/3) = (1/3, 31/3).

Therefore, B = (1/3, 31/3).

This also has the same logic. We let \(a_1=1\) and let everyother term be 0. This satisfies the condition abaove of:
\(|a_1 - a_2| + 2 |a_2 - a_3| + 3 |a_3 - a_4| + 4 |a_4 - a_5 | + \dots + 9 |a_9 - a_{10}| + 10 |a_{10} - a_1| = 1\)(Note we can let a1 be one greater than a2 but to make the eq 0, everything else must be equal to a2. So a1 can be 10000000000, and if a2 is 9999999999, a1-a6 is still 1)

\(\boxed{\max{|a_1-a_6|}=1}\)

\(|a_1 - a_2| + 2 |a_2 - a_3| + 3 |a_3 - a_1| = 1\) All the terms are positive or zero so each term is between 0 and 1. Therefore to maximize \(|a_1-a_2|\) we set it equal to the max value which is \(\boxed{1}\)

\(f(x) = \left\lfloor\frac{2 - 3x}{x + 3 + 2x}\right\rfloor\)

f(1) = floor(-1/6) = -1

f(2) = floor(-4/9) = -1

f(3) = floor(-7/12) = -1

f(4) = floor(-10/15) = -1

...

f(10) = floor(-28/33) = -1

...

f(100) = floor(-298/303) = -1

... f(x) = floor((-3x + 2) / (3x + 3)), which the numerator will never be able to be -3x + 3, if it did it would be -1, so the floor'd always be -1.

Hence, f(1) + f(2) + ... f(1000) = -1 * 1000 = -1000

\({a}^{3}-\frac{1}{{a}^{3}}=(a-\frac{1}{a})({a}^{2}+1+\frac{1}{{a}^{2}})=(a-\frac{1}{a})({(a-\frac{1}{a})}^{2}+3)\), which is what we want to find.

We see \(a-\frac{1}{a}\) will me good to know.

We have:

\({a}^{2}-a-1=0\), divide both sides by a.

\(a-1-\frac{1}{a}=0\), \(a-\frac{1}{a}=1\).

Plugging in, \({a}^{3}-\frac{1}{{a}^{3}}=(1)({1}^{2}+3)=4\).

So, \(\underline{{a}^{3}-\frac{1}{{a}^{3}}}\) is 4.

\(a_{11}=a_{10}+a_{9}\), so following this logic.

\(4=1+a_{9}\)

\(a_{9}=3\)

Similarly,

\(1=3+a_{8}\)

\(a_{8}=-2\)

\(3=-2+a_{7}\)

\(a_{7}=5\)

\(-2=5+a_{6}\)

\(a_{6}=-7\).

a₆ is -7.