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Aleonysh [2.5K]
3 years ago
15

The area in square feet of a rectangle field is X squared -100 X +2100. The width, and feet, is X -30. What is the length in fee

t
Mathematics
1 answer:
nignag [31]3 years ago
4 0
Answer:  The length is:  "(x − 70)"  ft. 
__________________________________________________

Area (A) = Length (L) * width (w) ;  that is:

   A = L * w ;

We want to solve for the "Length" (L)" ; 

So, we can rearrange the equation:
 
   A / w  = (L* w) / w ;

to get:  A / w = L ;

↔  L = A / w ;  

A = (x² <span>− 100x + 2100) ;

w = (x </span><span>− 30)
_____________________________
  L = (</span>x² − 100x + 2100) / (x − 30) ;
________________________________

So;  (x − 30) * _?__ = x² − 100x + 2100 ?? ,

________________________________________

-30 * _?_ = 2100 ?  ;  →  2100 ÷ (-30) = -70 ;

-30 − 70 = ? -100?? Yes!

So, the answer is:  "(x − 70)" ; 

That is:  " (x − 30) (x − 70)  =  x² − 100x + 2100 .

The Length is:  "(x − 70)"  ft.
______________________________________
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Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x. (
True [87]
So it tells us that g(3) = -5 and g'(x) = x^2 + 7.

So g(3) = -5 is the point (3, -5)
Using linear approximation
g(2.99) is the point (2.99, g(3) + g'(3)*(2.99-3))

now we just need to simplify that
(2.99, -5 + (16)*(-.01)) which is (2.99, -5 + -.16) which is (2.99, -5.16)
So g(2.99) = -5.16 

Doing the same thing for the other g(3.01)
(3.01, g(3) + g'(3)*(3.01-3))
(3.01, -5 + 16*.01) which is (3.01, -4.84)
So g(3.01) = -4.84

So we have our linear approximation for the two. 

If you wanted to, you could check your answer by finding g(x).  Since you know g'(x), take the antiderivative and we will get 
g(x) = 1/3x^3 + 7x + C
Since we know g(3) = -5, we can use that to solve for C
1/3(3)^3 + 7(3) + C = -5 and we find that C = -35
so that means g(x) = (x^3)/3 + 7x - 35

So just to check our linear approximations use that to find g(2.99) and g(3.01)

g(2.99) = -5.1597
g(3.01) = -4.8397

So as you can see, using the linear approximation we got our answers as
g(2.99) = -5.16
g(3.01) = -4.84
which are both really close to the actual answer.  Not a bad method if you ever need to use it. 
5 0
3 years ago
The nth term of a sequence is given by 3n² + 11 Calculate the difference between the 6th term and the 9th term of the sequence.​
katovenus [111]

The difference between the 6th term and the 9th term of the sequence is 135

<h3>How to determine the difference</h3>

Given that the nth term is;

3n² + 11

For the 6th term, the value of n is 6

Let's solve for the 6th term

= 3( 6)^2 + 11

= 3 × 36 + 11

= 108 + 11

= 119

For the 9th term, n = 9

= 3 (9)^2 + 11

= 3( 81) + 11

= 243 + 11

= 254

The difference between the 6th and 9th term

= 254 - 119

= 135

Thus, the difference between the 6th term and the 9th term of the sequence is 135

Learn more about algebraic expressions here:

brainly.com/question/4344214

#SPJ1

4 0
2 years ago
Thomas bikes at a rate of 24 mi in 3 h. Thomas’s friend De’Monte bikes 9.5 mi each hour. De’Monte and Thomas are 105 mi apart an
Murrr4er [49]
6 hours
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8 0
3 years ago
Help and explain ///////////////////////////
Ilia_Sergeevich [38]

Hello,

First, you must understand that

(f-g)(x) means f(x)-g(x) it is the difference of two functions :

f(x)=2x+4 and g(x)=3x-7

f(5)=2*5+4=10+4=14

g(5)=3*5-7=15-7=8

So, (f-g)(5)= f(5)-g(5)=14-8<u>=6</u>

Answer A: none of the choices are correct.

An other way to do it:

(f-g)(x)=f(x)-g(x)=2x+4-(3x-7)= 2x-3x+4+7=-x+11

if x= 5 then (f-g)(5)=-5+11<u>=6</u>

4 0
3 years ago
FOR ALL YOU STAR WARS FANS OUT THERE MY TEACHER GAVE US A STAR WARS MATH QUESTION. CHECK MY ANSWER MYBE? He asked how fast a bla
kvv77 [185]
(Links in comments below)

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At around frame 15 - about 0.233s into the clip - one can barely make out a green projectile being launched from one of the Devastator's batteries. It takes the projectile between 19 and 20 frames (I'll round up to 20) - now at frame 35, about 0.333s later - for it to reach the left edge of the camera's field of view.

Let's suppose the Tantive IV was situated in roughly the same physical position in frame 1 as the projectile was in frame 35.

Now, the Tantive IV exits the screen at around frame 182 - about 3.017s into the clip. According to link [2], the standard CR90 "Corellian" Corvette can reach speeds of up to 950km/h. The Rebels are fleeing the Empire, so it's reasonable to assume that they are moving as fast as they can. Let's also assume they're traveling at a constant rate, and that the tractor beam aboard the Devastator hasn't been activated yet.

Note: Link [2] refers to this maximum speed as "atmosphere speed". I can't find any references as to what this means, so I assume it means something along the lines of "same conditions as Earth's atmosphere". The two ships are flying above Tatooine, which is inhabited in part by humans, so it's probably a safe bet that the atmospheres of Tatooine and Earth aren't that dissimilar.

So if it takes the Tantive IV about 3.017s to move (approximately) the same distance as the Devastator's projectile, and if we assume the Tantive IV is moving at maximum speed, then it must have traveled a distance d of (approximately)

d=\dfrac{950\text{ km}}{1\text{ h}}\times(3.017\text{ s})\implies d\approx0.796\text{ km}

However, it's quite clear from the previous shot that the assumption above is not all that accurate. So let's consider "your" assumption, that the distance between the Tantive IV and the Devastator is about 10 times its own length, i.e. d\approx2.296\text{ km}. Now, if the projectile takes about 0.333s to travel the same distance, then it must have moving at a speed r of (approximately)

2.296\text{ km}=r\times(0.333\text{ s})\implies r\approx24,821.6\dfrac{\text{km}}{\text{h}}
3 0
3 years ago
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