Answer:
<h2>0 - no solution</h2>
Step-by-step explanation:
![\text{Let}\\\\\left\{\begin{array}{ccc}ax+by=c\\dx+ey=f\end{array}\right\\\\\text{in its simplest form, i.e. that a and b, d and e are relatively first}](https://tex.z-dn.net/?f=%5Ctext%7BLet%7D%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dax%2Bby%3Dc%5C%5Cdx%2Bey%3Df%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5Ctext%7Bin%20its%20simplest%20form%2C%20i.e.%20that%20a%20and%20b%2C%20d%20and%20e%20are%20relatively%20first%7D)
![\text{if}\ a=d,\ b=e,\ c=f,\ \text{then a system of equations has infinitely many solutions}\\\\\text{Example:}\\\\\left\{\begin{array}{ccc}x+3y=6\\-2x-6y=-12&\text{divide both sides by (-2)}\end{array}\right\\\\\left\{\begin{array}{ccc}x+3y=6\\x+3y=6\end{array}\right](https://tex.z-dn.net/?f=%5Ctext%7Bif%7D%5C%20a%3Dd%2C%5C%20b%3De%2C%5C%20c%3Df%2C%5C%20%5Ctext%7Bthen%20a%20system%20of%20equations%20has%20infinitely%20many%20solutions%7D%5C%5C%5C%5C%5Ctext%7BExample%3A%7D%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B3y%3D6%5C%5C-2x-6y%3D-12%26%5Ctext%7Bdivide%20both%20sides%20by%20%28-2%29%7D%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B3y%3D6%5C%5Cx%2B3y%3D6%5Cend%7Barray%7D%5Cright)
![\text{if}\ a=d,\ b=e,\ c\neqf,\ \text{then the system of equations has no solution}\\\\\text{Example:}\\\\\left\{\begin{array}{ccc}3x+9y=12&\text{divide both sides by 3}\\x+3y=1\end{array}\right\\\\\left\{\begin{array}{ccc}x+3y=4\\x+3y=1\end{array}\right](https://tex.z-dn.net/?f=%5Ctext%7Bif%7D%5C%20a%3Dd%2C%5C%20b%3De%2C%5C%20c%5Cneqf%2C%5C%20%5Ctext%7Bthen%20the%20system%20of%20equations%20has%20no%20solution%7D%5C%5C%5C%5C%5Ctext%7BExample%3A%7D%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7D3x%2B9y%3D12%26%5Ctext%7Bdivide%20both%20sides%20by%203%7D%5C%5Cx%2B3y%3D1%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B3y%3D4%5C%5Cx%2B3y%3D1%5Cend%7Barray%7D%5Cright)
![\text{In other cases it has one solution.}](https://tex.z-dn.net/?f=%5Ctext%7BIn%20other%20cases%20it%20has%20one%20solution.%7D)
![\text{We have:}\\\\\left\{\begin{array}{ccc}x+2y=2\\2x+4y=-8&\text{divide both sides by 2}\end{array}\right\\\\\left\{\begin{array}{ccc}x+2y=2\\x+2y=-4\end{array}\right\\\\\text{Conclusion:}\\\\\text{The system of equations has no solution.}](https://tex.z-dn.net/?f=%5Ctext%7BWe%20have%3A%7D%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B2y%3D2%5C%5C2x%2B4y%3D-8%26%5Ctext%7Bdivide%20both%20sides%20by%202%7D%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B2y%3D2%5C%5Cx%2B2y%3D-4%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%5Ctext%7BConclusion%3A%7D%5C%5C%5C%5C%5Ctext%7BThe%20system%20of%20equations%20has%20no%20solution.%7D)
![\text{Now I will show it}](https://tex.z-dn.net/?f=%5Ctext%7BNow%20I%20will%20show%20it%7D)
![\left\{\begin{array}{ccc}x+2y=2\\x+2y=-8&\text{change the sings}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}x+2y=2\\-x-2y=8\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad0=10\qquad\bold{FALSE}](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B2y%3D2%5C%5Cx%2B2y%3D-8%26%5Ctext%7Bchange%20the%20sings%7D%5Cend%7Barray%7D%5Cright%5C%5C%5Cunderline%7B%2B%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B2y%3D2%5C%5C-x-2y%3D8%5Cend%7Barray%7D%5Cright%7D%5Cqquad%5Ctext%7Badd%20both%20sides%20of%20the%20equations%7D%5C%5C.%5Cqquad0%3D10%5Cqquad%5Cbold%7BFALSE%7D)
Answer:
Domain = {x : x ≠ 4 , -4} or (-∞ , -4) ∪ (-4 , 4) ∪ (4 , ∞)
Step-by-step explanation:
<u>TO FIND :-</u>
- Domain of
![C(x) = \frac{x + 9}{x^2 - 16}](https://tex.z-dn.net/?f=C%28x%29%20%3D%20%5Cfrac%7Bx%20%2B%209%7D%7Bx%5E2%20-%2016%7D)
<u>SOLUTION :-</u>
Domain of a function is a value for which the function is valid.
The function
is valid until the denominator is 0.
So make sure that the denominator must not be 0.
![=> x^2 - 16 > 0](https://tex.z-dn.net/?f=%3D%3E%20x%5E2%20-%2016%20%3E%200)
Find the values of x for which the denominator becomes 0. To find it , you'll have to solve the above inequality.
![=>x^2 - 16 + 16 > 0 + 16](https://tex.z-dn.net/?f=%3D%3Ex%5E2%20-%2016%20%2B%2016%20%3E%200%20%2B%2016)
![=> x^2 > 16](https://tex.z-dn.net/?f=%3D%3E%20x%5E2%20%3E%2016)
![=> x > \sqrt{16}](https://tex.z-dn.net/?f=%3D%3E%20x%20%3E%20%5Csqrt%7B16%7D)
![=> \boxed{x > 4} \: or \:\boxed{x > -4}](https://tex.z-dn.net/?f=%3D%3E%20%5Cboxed%7Bx%20%3E%204%7D%20%5C%3A%20or%20%5C%3A%5Cboxed%7Bx%20%3E%20-4%7D)
We can say that <u>4 & -4 can't be domains</u> because these values will make the function undefined.
Now try putting values of x such that -4 < x < 4. You'll observe that the function will be valid for all those values of x between -4 & 4.
<u>CONCLUSION :-</u>
The function will be valid for any value of 'x' except 4 & -4. So in :-
Interval notation , it can be written as → (-∞ , -4) ∪ (-4 , 4) ∪ (4 , ∞)
Set builder notation , it can be written as → {x : x ≠ 4 , -4}
Answer:
![I=\$156.08](https://tex.z-dn.net/?f=I%3D%5C%24156.08)
Step-by-step explanation:
we know that
The compound interest formula is equal to
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
substitute in the formula above
![A=\$1,406.08](https://tex.z-dn.net/?f=A%3D%5C%241%2C406.08)
Find out the interest
I=A-P
substitute
![I=\$1,406.08-\$1,250.00\\I=\$156.08](https://tex.z-dn.net/?f=I%3D%5C%241%2C406.08-%5C%241%2C250.00%5C%5CI%3D%5C%24156.08)
5x7 and 5x 0.3 seperately Then add together.
=36.5
Answer:
A piece of duct measures 48 inches.
In each connection, we lost 1 1/2 inches (i guess that it is what you wanted to write)
Then:
If we have 7 pieces of duct connected we have 5 connections (the ones in the extremes are not connected between them, so the number of connections is equal to the number of ducts minus two.)
So here we have 7 times 48 inches for the pieces, minus 5 times 1 1/2 in (i will write 1 1/2 = 1 + 0.5 = 1.5 in, so the math is easier) for the connections, the length is:
L = 7*48in - 5*1.5in = 328.5 inches.
for the 12 pieces duct, we have 12 pieces and 10 connections, so the length is:
L = 12*48in - 10*1.5in = 561 in
Now, if we want to make only one duct with those two, then we must add their lengths, but if we connect them, we also need to subtract the 1.5in of the new connection:
L = 561in + 328.5in - 1.5in = 888in