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2 years ago

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) " .
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Explanation:
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Factor: " x² − 5x − 14" .
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What are factors of: "-14" ; that add up to "-5" ?

-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.

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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.

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3 years ago

Read 2 more answers

What is the solution of the equation 14^x=5? 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
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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 3 $M_3 = \{3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69\}$

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