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Anna35 [415]
3 years ago
6

the wilsons back yard is rectangular plot that has a length of 100 feet and a width of 80 feet .the family planted a garden with

a length and width of 60 feet. the family planted the lawn in the remaining area of the back yard . what is the area in square feet
Mathematics
1 answer:
Irina-Kira [14]3 years ago
8 0

the area in square feet is 4400. (because to find the area of the rectangle you multiply the 2 side lengths, 100 and 80 which gives you 8000, then you find the area of the square, multiply 60 by 60 which is 3600, then you subtract the total area of the backyard by the area taken up by the garden, 8000-3600, which gives you your total of 4400.)

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working together, shawn and hector can install a tile floor in 6 hours. it would take shawn 9 hours to do the job alone. what is
Vikki [24]
First, let x be the number of hours it will take for Hector to finish the same job alone. This means that every hour, Hector can do 1/x of the job. The amount of work done by Shawn and Hector in 6 hours should be equal to 1 complete work. This can be expressed as,
                      (1/9 + 1/x)(6) = 1
The value of x is 18. Thus, Hector can to 1/18 of the job in one hour. 
8 0
4 years ago
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Pls help i dont understand :( will give Brainliest + 25 Points
Aleks04 [339]

Answer:

The answer is B

Step-by-step explanation:

The best way to solve is to plug in the x points into each equation to see if there is only one that works for the first point and if more than one works do the same for the second point

5 0
3 years ago
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How do you do these problems?
Angelina_Jolie [31]

Answer:

18e⁶

⁵/₁₂₈

Step-by-step explanation:

Rₙ(x) = f⁽ⁿ⁺¹⁾(c) / (n+1)! (x − a)ⁿ⁺¹, and a < c < x.

f(x) = eˣ, a = 0, and n = 1.  Thus R₁ is:

R₁(x) = f"(z)/2! x²

R₁(x) = eᶻ/2 x²

|R₁| is a maximum when |f"(z)| is a maximum.  On the domain 0 < z < 6, that maximum is e⁶.  At x = 6, the upper bound of |R₁| is:

|R₁| = 18e⁶

This time, f(x) = 1 / √(1 + x) = (1 + x)^-½.  a = 0, and n = 2.

R₂(x) = f⁽³⁾(z)/3! x³

Find f⁽³⁾(x):

f'(x) = -½ (1 + x)^-³/₂

f"(x) = ¾ (1 + x)^-⁵/₂

f⁽³⁾(x) = -¹⁵/₈ (1 + x)^-⁷/₂

On the domain -½ < z < 0, |f⁽³⁾(z)| is a maximum at z = 0.

|f⁽³⁾(z)| = ¹⁵/₈

Therefore, at x = -½, the upper bound of R₂ is:

|R₂| = (¹⁵/₈)/6 |(-½)³|

|R₂| = ⁵/₁₂₈

8 0
3 years ago
PLEASE HELP ME!!!!
12345 [234]

This is what I got, but you might want to double check me because I'm not that good at math (this is what I got for the discontinuity);

1. f(x)= x^2+6+8/x+4

*x+4=0                                      

-4 -4                                      

 x= -4

This means that x cannot equal -4

2. f(x)= (x+2)(x-4)/(x+2)(x+2)

*Note: When I did my calculations, I cancelled out all (x+2) binomials except for one in the denominator*

f(x)= x-4/x+2

*x+2=0                                      

   -2 -2                                      

     x= -2

This means x cannot equal -2

*You may want to check my work, but I believe that your answer is going to be either C or D. Personally, I assume it's C, but that's just me. Anyway, it's probably going to be C or D, or at least that's what I think.*

3 0
3 years ago
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