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The oblique asymptote of f(x) is y = x/2 - 3, not y = x/2 - 2.

Answer is A) 125 Inches Squared

Put Adi in group 1, and Noor in group 2

Choose in 10C2 ways the other members for group 1

.

Choose in 8C2 ways the other members for group 2.

Choose 6C3 ways the members for group 3

Finally we can fill group 4 with the remaining members in 3C3 ways.

Groups 1 and 2 are always unique, groups 3 and 4 are duplicated, so we should be divide by 2!:

Final result: (10C2*8C2*6C3*3C3) /2! = 12,600 ways.

Sorry, but the answer is wrong. Thank you for trying though! This is the explanation for the correct answer:

The vertical asymptote of \(f(x)\) is \(x=2.\) Hence, \(d=2.\)

By long division,\(f(x) = \frac{1}{2} x - 2 - \frac{2}{2x - 4}.\)
Thus, the oblique asymptote of \(f(x)\) is \(y = \frac{1}{2} x - 2,\) which passes through \((0,-2).\) Therefore, the oblique asymptote of \(g(x)\) is \(y = -2x - 2.\)
Therefore, \(g(x) = -2x - 2 + \frac{k}{x - 2}\) for some constant \(k.\)
Finally, \(f(-2) = \frac{(-2)^2 - 6(-2) + 6}{2(-6) - 4} = -\frac{11}{4},\) so \(g(-2) = -2(-2) - 2 + \frac{k}{-2 - 2} = -\frac{11}{4}.\)
Solving, we find \(k = 19.\) Hence, \(g(x) = -2x - 2 + \frac{19}{x - 2} = \frac{-2x^2 + 2x + 23}{x - 2}.\)
We want to solve \(\frac{x^2 - 6x + 6}{2x - 4} = \frac{-2x^2 + 2x + 23}{x - 2}.\)
Then \(x^2 - 6x + 6 = -4x^2 + 4x + 46,\) or \(5x^2 - 10x - 40 = 0.\) This factors as \(5(x + 2)(x - 4) = 0,\) so the other point of intersection occurs at \(x = 4.\) Since \(f(4) = \frac{4^2 - 6 \cdot 4 + 6}{2(4) - 4} = -\frac{1}{2},\) the other point of intersection is \(\boxed{\left( 4, -\frac{1}{2} \right)}.\)

Still, thank you for trying, Guest!

thank you

Let's start by adding x and 5 to both sides of the inequality. This gives us:

x \le 17

The solution is all real numbers less than or equal to 17. In interval notation, this is written as (−∞,17]​.

I'm gonna round some numbers drastically, so  

the products are gonna be appriximations, but  

this is just to give you an idea of my approach.  

use tan(30^o) to find BC = 6.9  

area ABC = (1/2)(12)(6.9) = 41.6  

use tan(30^o) to find CX = 4  

area BXC = (1/2)(6.9)(4) = 13.7  

area BXA = area ABC – area BXC  

area BXA = 27.9  

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that wasn't correct....

Since angle A is 30 degrees and angle B is 60 degrees, then angle C is 90 degrees. Therefore, triangle ABC is a 30-60-90 triangle.

The ratio of the sides of a 30-60-90 triangle is 1:sqrt(3):2. Since AC is 12, then AB is 12/2 = 6 and BC is 12sqrt(3)/2 = 6sqrt(3).

Since BX bisects angle ABC, then BX is also the altitude of triangle ABC. The area of a triangle is equal to 1/2baseheight. Since the base of triangle BXA is BX and the height of triangle BXA is BX, then the area of triangle BXA is 1/2BX^2 = 1/2(6sqrt(3))^2 = 18*3 = 54.

Therefore, the area of triangle BXA is 54.

I drove to the beach at a rate of 40 miles per hour.  If I had driven at a rate of 50 miles per hour instead, then I would have arrived 45 minutes later.  How many miles did I drive?  

You mean 45 minutes earlier.  Obviously, if you drive faster, you get there faster.  

This problem makes use of

the following relationship:                   Distance = Velocity x Time 

                                                           D  =  V • T  

case 1                                                 D  =  (40) • (T)  

case 2                                                 D  =  (50) • (T – 45)  

Since the Distance, D, is the  

same for both cases, let's set       

the "V•T"s equal to each other.             (50)(T – 45)  =  (40)(T)  

                                                               50T – 2250  =  40T  

Subtract 40T from both sides                  10T – 2250  =  0  

Add 2250 to both sides                                       10T  =  2250  

Divide both sides by 10                                           T  =  225   (this is time in minutes)  

Divide minutes by 60 to get hours          225 minutes  =  3.75 hours  

Plug this T back into original equation                     D  =  (40 mi/hr) • (3.75 hr)  =  150 miles  

_.

3a+4b=5a-6b+24-3a

combine like terms

3a+4b=2a-6b+24

-2a on both sides

a+4b=-6b+24

-4b on both sides to isolate a

a=-10b+24

Answer: a = - 10 b + 24

We expand (4x2−7y3)n using the Binomial Theorem. The term of the form cx10y6 will come from the term in the expansion that has 10 factors of x and 6 factors of y. The coefficient of this term is C(n,10)*4^10*-7^6

For this term to be non-zero, we must have n≥10. Also, for this term to be the largest power of x in the expansion, we must have 10 of the n factors of x come from the 4x^2 term, and the other 2 factors of x must come from the other 2x^2 terms in the expansion. This means that we must have n≥10 and n≡2(mod3).

The smallest positive integer that satisfies both of these conditions is n=11​.

Don't you have to know the exact term such as: x^4 y^6, because x and y appear together in every term except in the first term and in the last term and c would be the coefficient.

I don't think the question has an answer, as 0 is not more than 0. 

There are 1440 ways to split the 12 members of the Underwater Basket-Weaving club into four groups of 3 for a friendly speed basket-weaving competition, if Adi and Noor must be on different teams.

To solve this, we can use the following steps:

First, we can count the number of ways to split the 12 members into four groups of 3, without considering the restriction that Adi and Noor must be on different teams. This can be done using the following formula:

n! / (4! * (n - 12)!)

where n is the total number of members, which is 12.

Plugging this in, we get:

12! / (4! * 8!) = 495

Now, we need to subtract the number of ways to split the members into four groups of 3, with Adi and Noor on the same team. This can be done by counting the number of ways to choose two members from the 10 remaining members, and then dividing by 2, since Adi and Noor can be in either order on the same team.

10! / (2! * 8!) = 45

Finally, we need to add back the number of ways to split the members into four groups of 3, with Adi and Noor on different teams, but in the same group. This can be done by counting the number of ways to choose one member from the 10 remaining members, and then dividing by 2, since Adi and Noor can be in either order in the same group.

10! / (2! * 8!) = 20

Adding these three numbers together, we get the total number of ways to split the members into four groups of 3, with Adi and Noor on different teams:

495 - 45 + 20 = 1440

There are 120 rectangles that consists of an even number of unit squares in a 6 x 9 grid.

To solve this, we can use the following steps:

Find the total number of rectangles that can be formed in a 6 x 9 grid. This can be done by multiplying the number of horizontal lines (6) by the number of vertical lines (9), which gives us 54 rectangles.

Subtract the number of rectangles that contain an odd number of unit squares. This can be done by finding the number of rectangles that can be formed by choosing two horizontal lines and two vertical lines such that the difference between the horizontal lines is odd and the difference between the vertical lines is odd. There are 18 ways to choose two horizontal lines such that the difference is odd, and there are 36 ways to choose two vertical lines such that the difference is odd. This gives us a total of 64 rectangles.

Subtract the number of rectangles that contain both an odd number of horizontal units and an odd number of vertical units. There is only one way to choose two horizontal lines and two vertical lines such that both the difference between the horizontal lines and the difference between the vertical lines is odd. This gives us one rectangle.

The total number of rectangles that consists of an even number of unit squares is 54 - 64 + 1 = 120.

I hope this helps!

The actual answer is -50.

There are 10 paths from A to B in the grid. Each path is a sequence of 3 steps, where each step is either a move to the right or a move up.

There are 17 line segments in the grid. Each line segment is a sequence of 2 steps, where each step is either a move to the right or a move up.

If we choose a line segment at random, then the probability that it is on one of the paths is equal to the number of paths that pass through it divided by the total number of line segments.

The number of paths that pass through a line segment is equal to the number of ways to choose 2 steps out of 3 steps. There are 3C2 = 3 ways to do this.

The total number of line segments is 6.

Therefore, the probability that a randomly chosen line segment is on one of the paths is 3/6 = 1/2.

Therefore, the expected number of paths that pass through a randomly chosen line segment is 1/2 * 4 = 2.

(2000 (1/0.025 - 1/(0.025 1.025^6))) + (100000/1.025^6)

==$97,245.94 - current market price of the bond.

One ordered pair (a,b) satisfies the two equations ab^4 = 48 and ab = 72. What is the value of b in this ordered pair?  

To find b, consider                                                  ab⁴  =  48  

We will divide both sides by ab.  

Since ab=72, we will divide the left side  

by "ab" and the right side by its equal 72.  

                                                                              ab⁴         48  

                                                                             ——   =   ——  

                                                                              ab           72  

Note that ab⁴ = (ab) * (b³)  

Cancel ab out of the left side.  

Reduce 48/72 on the right side.  

                                                                               b³           2  

                                                                             ——   =   ——  

                                                                                1            3  

                                                                                 b   =   cube root of (2 / 3)  

_.

Each member paints at the same rate. 

6000 ft^2 is triple the amount of work.

Thus, we would need triple the amount of time.

40 minutes*3=120 minutes