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10 + 8 - 5x + 6 - 14, or 10 - 5x.

2y = 17, so y = 8.5. Double checking, 55.5 = 55.5. Also, this is not college algebra. 

35% of 100 is 35. Also, this is not a website where you post random math problems. They have to be complicated, such as calculus, algebra, geometry, etc. 

we'll start with something very easy then.

factor the following equation:

\(8x^5y^6-2x^3y^9-24x^5y^4+6x^3y^7\)

Just list all the numbers with a 9 in them and then count them:

9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 109, 119, 129, 139, 149> Total  = 25 "nines"

Hello, 
I have solve the ans, you can read this

Let's analyze the problem step by step:

The hypotenuse of a right triangle is equal to one of its legs: $AB = AC$.
The area of square $BCDE$ is given as $144$. Since the square has sides equal to the legs of the right triangle, its area is also equal to the product of the legs: $AB \cdot AC = 144$.
We are asked to find integer values of $AB$ and $AC$ that are each less than $20$.
Now, we can find the possible integer pairs of $(AB, AC)$ that satisfy the conditions.

Since $AB = AC$, we have $AB^2 = 144$, which implies $AB = \sqrt{144} = 12$.

Therefore, there is only one possibility for the lengths of $AB$ and $AC$, which is $AB = AC = 12$.

So, there is only one possibility for the lengths of $AB$ and $AC$ that satisfies the given conditions.
 

I hope my ans. is good. 

To find the number of times the digit 9 appears in the list of all integers from 1 to 150, we can consider the following: 1. Count the number of times 9 appears as the units digit: It appears 15 times (9, 19, 29, ..., 149). 2. Count the number of times 9 appears as the tens digit: It appears 15 times (90, 91, 92, ..., 99, 109, 119, ..., 149). Therefore, the digit 9 appears a total of 15 + 15 = 30 times in the list of all integers from 1 to 150.

Find  \(f(f(f(f(f(4)))))\) 

f(4) = 4 -1 = 3

f (f(4)) = f(3) =  3

f (f(f(4))) = f(3)  = 3

f(f(f(f(4)))) =  f(3) = 3

f(f(f(f(f(4))))) = f(3) = 3

Let  A's rate per minute  =  3/R =  the amt of the job that A can do in one minute

Let B's rate per minute = 1/R = the amt of the  job that B can  do in one  minute

And rate * time = part of job done       (a whole job done = 1) 

So  with A alone let the time = T    and we  have

(3/R) * (T ) =  1

3T  = R

T = R/3

With both hoses operating  let the time = T - 4  and we  have

(3/R) * (T - 4)  + (1/R) *  ( T - 4)  =   1

3 ( R/3 - 4) / R  +  (R/3 - 4) /R  =  1               multiply through by R

R - 12 + R/3 - 4  =  R    rearrange as

R + R/3 - R =  16

R/3 =  16

R = 48

B's  rate per minute =  1/R = 1/ 48 of the job done.....so it takes  B 48 minutes to fill the tank alone

Rigorously, continuity is defined below:
 

For every \(k\) in the domain of \(f\);

• \(f(k)\) exists
•  \(\lim_{i\to k}f(i)\) exists (i.e. \(\lim_{i\to k^-}f(i) = \lim_{i \to k+}f(i)\))
• and $\lim_{i\to k}f(i) = f(k)$

Put 3 into  both functions for  x and set them equal

3^2 =  3a (3)

9 = 9a

a  = 1

Put 2 into  each of the  first two functions  for x and set them  equal

a(2) + 3 = 2 - 5

2a = 2 - 5 - 3

2a = -6

a = -3

Put - 2  into  each of the last two  functions for x and  set them equal

-2 - 5 = 2(-2) - b

-7 = -4 - b

b = -4 + 7

b = 3

a + b =  3 + -3  =   0

2x + 1 = 0                                               8 - 4x = 0

2x = -1                                                    8 = 4x

x = -1/2    OK because x   ≤ 3                x = 8/4  = 2    not OK because x must  be > 3

Sum =  -1/2