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Multiply by (z + 1)^2 on both sides. Then,

\((3 - z)(z + 1) \geq 2(z + 4)(z + 1)^2\text{ and }z + 1 \neq 0\\ (z + 1)(2(z + 4)(z + 1) -(3-z)) \leq 0\text{ and }z + 1 \neq 0\\ (z + 1)(2z^2 +11z+5)\leq 0 \text{ and }z + 1 \neq 0\\ (z+1)(2z+1)(z+5)\leq 0 \text{ and }z + 1 \neq 0\\ z \leq -5 \text{ or }-1 < z \leq -\dfrac12\)

In interval notation, \(z \in (-\infty, -5] \cup \left(-1, -\dfrac12\right]\)

Similar to the one you posted before, so I will only give partial work:

Comparison between before and after drinking coffee gives

\((t - 3)(4p + 5) = 2tp\\ 4tp - 12p + 5t - 15 = 2tp\\ 2tp - 12p + 5t - 15 =0 \\ 2p(t - 6) + 5(t - 6) = -15\\ (2p + 5)(6 - t) = 15\)

Note that since p is a positive integer, \(p \geq 1\). That means \(2p + 5 \geq 7\), and 2p + 5 is also a divisor of 15. 

What is the only divisor of 15 that is greater than 7?

Please continue on your own from this point. If you are stuck, look at my answer to your previous similar problem. I gave full solution there.

Comparison between before and after drinking coffee gives:

\((t - 11)(4p + 2) = 2tp\\ 4tp - 44p + 2t - 22 = 2tp\\ 2tp + 2t - 44p - 22 = 0\\ 2t(p + 1) - 44(p + 1) = -22\\ (22 - t)(p + 1) = 11\)

How is that possible? The only possibilities are \(\begin{cases} 22-t = 1\\ p+1=11 \end{cases} \text{ or } \begin{cases} 22-t=11\\ p+1=1 \end{cases}\). But p + 1 can't be 1 since p would be 0, which is not positive.

So, the solution is (t, p) = (21, 10), which means the mathematician solved 4p + 2 = 42 problems that day.

The second equation gives r = 1 so the third equation says p + 1 = 9 + p, which has no solution for p.

Hence, the system has no solutions.

\(\begin{cases} x = y^4\\ x + y^2 = 1 \end{cases}\)

Substituting,

\(y^4 + y^2 - 1= 0\\ y^2 = \dfrac{-1 \pm \sqrt 5}2\text{(reject negative root)}\\ y^2 = \dfrac{\sqrt 5 - 1}2\\ y = \pm\sqrt{\dfrac{\sqrt 5 - 1}2}\)

Since \(x + y^2 = 1\),

\(x = 1 - \dfrac{\sqrt 5 - 1}2 = \dfrac{3 -\sqrt 5}2\)

So the intersection points are \(\left(\dfrac{3 - \sqrt 5}2, \pm\sqrt{\dfrac{\sqrt 5 - 1}2}\right)\).

The distance is \(2 \sqrt {\dfrac{\sqrt 5 - 1}2} = \sqrt{2(\sqrt 5 - 1)} = \sqrt{-2 + 2 \sqrt 5}\).

Then \((u,v) = (-2, 2)\).

\(\begin{cases} x^2 + 4y^2 = 4\\ x^2 - m(y + 2)^2 = 1 \end{cases}\)

Substituting, 

\(4 - 4y^2 - m(y + 2)^2 = 1\\ m(y + 2)^2 + 4y^2 - 3 = 0\\ (m + 4)y^2 + 4my + 4m - 3 = 0\)

Take \(\Delta = 0\), we have 

\((4m)^2 - 4(m + 4)(4m - 3) = 0\\ 16m^2 - 4(4m^2 +13m -12) = 0\\ -52m+48 = 0\\ m = \dfrac{12}{13}\)

Suppose the legs are a and b. Then by the definition of cool trianlges,

\(\dfrac{ab}2 = 2(a + b)\\ ab = 4a + 4b\\ ab - 4a - 4b + 16 = 16\\ (a - 4)(b - 4) = 16\)

Assume without loss of generality that \(a \leq b\). Then

\(\begin{cases} a-4=1\\ b-4=16 \end{cases}\text{ or }\begin{cases} a-4=2\\ b-4=8 \end{cases}\text{ or }\begin{cases} a-4=4\\ b-4=4 \end{cases} \)

This gives all three cool triangles, the one with legs 5 and 20, the one with legs 6 and 12, and the one with both legs equal to 8.

The sum of all possible areas is \(\dfrac{5 \times 20 + 6 \times 12 + 8 \times 8}2 = 118\).

To solve this problem, let's break it down step by step:

1. **Finding the probability of drawing an even number on the first draw:**

   There are 7 numbers on each color, and 3 of them are even (2, 4, 6). So, the probability of drawing an even number on the first draw is \(\frac{3}{7}\).

2. **Finding the probability of drawing a multiple of 3 on the second draw:**

   There are still 7 numbers on each color, and 2 of them are multiples of 3 (3, 6). After the first draw, there are 13 cards remaining. Among these, there are 4 cards labeled with multiples of 3. So, the probability of drawing a multiple of 3 on the second draw is \(\frac{4}{13}\).

3. **Multiplying the probabilities:**

   The probability of drawing an even number on the first draw and a multiple of 3 on the second draw is the product of the individual probabilities:
   \(\frac{3}{7} \times \frac{4}{13}\)

4. **Calculating the final probability:**

   Multiply the fractions:

   \(\frac{3}{7} \times \frac{4}{13} = \frac{12}{91}\)

So, the probability that the first card Grok draws is labeled with an even number and the second card Grok draws is labeled with a multiple of 3 is \(\frac{12}{91}\).

The axis of symmetry, x = -1, tells us that the vertex of the parabola is located at the point (-1, y-value). In any parabola, the axis of symmetry passes exactly through the vertex.

Since we are given that a = 2, we can rewrite the equation as:

y = 2x^2 + bx + c

We know that the x-coordinate of the vertex is -1. When we plug this value for x in the equation, we should get the y-coordinate of the vertex.

However, we are not given the specific y-value of the vertex. This doesn't prevent us from finding b though!

Let's substitute x = -1 into the equation:

y = 2(-1)^2 + b(-1) + c

This simplifies to:

y = 2 - b + c

Key Point: Regardless of the specific value of c (which determines the vertical positioning of the parabola), the axis of symmetry will still pass through the vertex. This means that even though we don't know the exact y-value of the vertex, we know it must be the same on both sides of the axis of symmetry (x = -1).

Therefore, the equation when we plug in x = 1 (on the other side of the axis of symmetry) must also equal the expression we obtained above.

In other words:

y = 2(1)^2 + b(1) + c must also equal y = 2 - b + c (from above)

Simplifying the left side of the new equation:

y = 2 + b + c

Equating both sides:

2 + b + c = 2 - b + c since they represent the same y-value

Solving for b, we get:

2b = 0

Therefore, b = 0.

So, even without knowing the specific y-value of the vertex, we were able to find that b = 0.

Absolutely, I’ve been improving my problem-solving abilities in simplifying polynomials. Let's find the vertex of the graph of the equation: y=−2x2+8x+15−3x2+7x−25

We can find the vertex by rewriting the equation in vertex form, which is of the form y=A(x−h)2+k

where (h,k) is the vertex.

Steps to solve: 1. Combine terms: y=(−2−3)x2+(8+7)x+(15−25) y=−5x2+15x−10

2. Factor the expression: y=−5(x2−3x+2) y=−5(x−2)(x−1)

3. The equation is already in vertex form: The coefficient of the x2 term is −5, which is negative, so the parabola opens downward. The vertex is at the point where x=2 and y=−5(2−2)(2−1)=0.

Answer: The vertex of the parabola is (2,0).

The sum of an arithmetic series is the number of terms in the series times the average of the first and last term. In this case, the number of terms is 80 and the average of the first and last term is (1+80)/2=40.5. Therefore, the sum of the series is 80⋅40.5=3240.

We can now factor 3240. We see that 3240=2⋅2⋅2⋅2⋅3⋅5⋅7. The greatest prime factor is 7​.

Interest rate per day is 15% / 100 = 0.15

The amount Mark owes increases by 0.15 each day.

We want to find the number of days needed to reach an amount that is at least twice what he borrowed (which is 10 x 2 = $20).

Let's set up a table to track the daily increase:

DayAmount Owed

1$10 + $0.15 = $10.15

2$10.15 + $0.15 = $10.30

3$10.30 + $0.15 = $10.45

4$10.45 + $0.15 = $10.60

5$10.60 + $0.15 = $10.75

We can see that after 5 days, the amount owed is $10.75, which is still less than $20.

One more day of interest will bring the total to $10.75 + $0.15 = $10.90.

Therefore, it will take Mark at least 5​ days to pay back at least twice the amount he borrowed.

The product of two positive integers minus their sum is 39. The integers are relatively prime, and each is less than 20. What is the sum of the two integers?  

                                11 x 5 = 55     This is their product.  

                                11 + 5 = 16     This is their sum.  

                                 55 – 16 = 39    

I can't give an explanation.  I just tried a few until I hit it.  

_.

First, we try to rewrite 675 as a power of 5. We can see that 675=3⋅3⋅3⋅3⋅5=54⋅33. Since we don't care about the factor of 3, this shows that 675 can also be written as 54.

We now have the equation 53x+2=54. Since for any real numbers a and b with a=0, am=an implies m=n, we must have 3x+2=4. Solving for x, we find x = 2/3.

Alternatively, since we want to isolate 5x, taking the logarithm of both sides (with an appropriate base) would be another approach. Note that this would ultimately lead to the same solution.

In a row of five squares, each square is to be colored either red, yellow, or blue, so that no two consecutive squares have the same color. How many ways are there to color the five squares, if there must be at least three yellow squares?  

In order not to have two yellows next to each other,  

the three yellows must occupy squares 1, 3, and 5. 

That leaves squares 2 and 4 to fill, and two colors  

to do it with. There are only four ways. 

Blue Blue     Red Red     Blue Red     Red Blue  

So, the answer is four ways.  

_.

1. The equation is simplified to to find the enclosed area.
 

2. The equation is further rewritten in standard form as (x - 8)^2 + (y + 11)^2 = 271, representing a circle centered at (8, -11) with a radius of the square root of 271 units.

 3. The area enclosed by the circle is calculated using the formula A = 271π square units.

4. Therefore, the area of the region enclosed by the given graph is 271π square units.

An arithmetic sequence has a common difference between each term

Let the common difference be n

The first term = 2

The second term = 2+n

The third term = 2+2n

The fourth term = 2+3n

The fifth term = 2+4n

The sixth term = 2+5n

And so on...

The sum of the third and sixth terms is 25 so:

2+2n+2+5n = 25

7n+4 = 25

7n = 21

n = 3

Thus, the fourth term is 2+3n = 2+9 = 11

Here is my attempt 

1. The leading term of f(x) * g(x) is -x^5.

2. The leading coefficient of f(x) * g(x) is -1.

3. The degree of f(x) * g(x) is 5.

4. The constant term of f(x) * g(x) is 0.

5. The coefficient of x^2 in f(x) * g(x) is 0.