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Sunny_sXe [5.5K]
2 years ago
6

What is the complete factorization of x2 -5x-14

Mathematics
2 answers:
Shkiper50 [21]2 years ago
5 0
X² - 5x - 14 = (x - 7)(x + 2)
zalisa [80]2 years ago
3 0
The answer is:  is:  " (x − 7) (x + 2) " .
____________________________________
Explanation:
____________________________________
Factor:  " x² − 5x <span>− 14" . 
</span>____________________________________

What are factors of:  "-14" ; that add up to "-5<span>" ?
</span>
-14, 1 ;  →  -14 + 1 = -13; No.

14, -1 ;  → 14 + (-1) = 14 - 1 = 13; No. 

-2, 7 ;  →  -2 + 7 = 7 + (-2) = 7 - 2 = 5 ; No. 

 2, -7 ; →  2 + (-7) = -7 + 2 = -5 .  YES! .

So, the complete "factorization" of:  

" x² − 5x − 14" ;   is:  " (x − 7) (x + 2) " .
_____________________________________________
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4x+9xy-3xy=? I really need help please
Lesechka [4]

Answer:

6xy+4x

Step-by-step explanation: You start off by gathering the like terms (which are 9xy & -3xy) and then you just keep 4x there. You then subtract 9xy-3xy=6xy and bring down the 4 since it has no like terms. So it’s 6xy+4x.

8 0
2 years ago
if you needed only 1c of milk, what is your best choice at the grocery store- a quart container, a pint container, or a 1/2 gal
Artist 52 [7]
A pint container because there are 2 cups in a pint, whereas there are 6 cups in 1/2 gallon and 4 cups in a quart.
3 0
3 years ago
Read 2 more answers
What is the solution of the equation 14^x=5?<br> Round your answer to the nearest ten-thousandth.
deff fn [24]
Take ln to both sides
ln 14^x = ln 5
then you'll get
x ln 14 = ln 5
divide both sides by ln 14
x = 0.6098533345
after rounding
x = 0.6099
4 0
3 years ago
Read 2 more answers
Seventy cards are numbered 1 through 70 , one number per card. One card is randomly selected from the deck. What is the probabil
Vanyuwa [196]

Answer:

Probability = \frac{2}{35}

Step-by-step explanation:

Given

Total = 70

First, we need to list the multiples of 5

M_5 = \{5,10,15,20,25,30,35,40,45,50,55,60,65,70\}

Then, multiples of 3M_3 = \{3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69\}

Next, is to list out the common elements in both

M_3\ n\ M_5 = \{15,30,45,60\}

n(M_3\ n\ M_5) = 4

The required probability is then calculated as thus:

Probability = \frac{n(M_3\ n\ M_5)}{Total}

Probability = \frac{4}{70}

Probability = \frac{2}{35}

7 0
3 years ago
Which of the following functions are homomorphisms?
Vikentia [17]
Part A:

Given f:Z \rightarrow Z, defined by f(x)=-x

f(x+y)=-(x+y)=-x-y \\  \\ f(x)+f(y)=-x+(-y)=-x-y

but

f(xy)=-xy \\  \\ f(x)\cdot f(y)=-x\cdot-y=xy

Since, f(xy) ≠ f(x)f(y)

Therefore, the function is not a homomorphism.



Part B:

Given f:Z_2 \rightarrow Z_2, defined by f(x)=-x

Note that in Z_2, -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular f(x)=x

f(x+y)=x+y \\  \\ f(x)+f(y)=x+y

and

f(xy)=xy \\  \\ f(x)\cdot f(y)=xy

Therefore, the function is a homomorphism.



Part C:

Given g:Q\rightarrow Q, defined by g(x)= \frac{1}{x^2+1}

g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1}  \\  \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}

Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.



Part D:

Given h:R\rightarrow M(R), defined by h(a)=  \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)

h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\  \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)

but

h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\  \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.



Part E:

Given f:Z_{12}\rightarrow Z_4, defined by \left([x_{12}]\right)=[x_4], where [u_n] denotes the lass of the integer u in Z_n.

Then, for any [a_{12}],[b_{12}]\in Z_{12}, we have

f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\  \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)

and

f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)

Therefore, the function is a homomorphism.
7 0
3 years ago
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