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3 years ago

What is the quotient of the following division problem?

Mathematics

2 answers:

LekaFEV [45]3 years ago

6 0

The quotient of x + 1 StartLongDivisionSymbol x squared + 3 x + 2 EndLongDivisionSymbol is x+2

Step-by-step explanation:

We need to find quotient of x^2+3x+2 divided by x+1

The division is shown in figure attached.

Quotient: x+2

Remainder: 0

The quotient of x + 1 StartLongDivisionSymbol x squared + 3 x + 2 EndLongDivisionSymbol is x+2

Mekhanik [1.2K]3 years ago

5 0

Answer:

x+2

Step-by-step explanation:

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3(x-4) = x + 14<br> How do you solve this equation

Dmitry_Shevchenko [17]

Answer:

x=13

Step-by-step explanation:

you have to isolate the variable

3(x-4)=x+14

distribute

3x-12=x+14

subtract 14 from each side

3x-26=x

subtract 3x from each side

-26=-2x

divide each side by -2

13=x

flip sides

x=13

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3 years ago

What value of x is in the solution set of 9(2x + 1) < 9x - 18?

Alex17521 [72]

Answer:

b. -3

Step-by-step explanation:

9(2x+1) < 9x-18

(distribute the 9)

18x+9 <9x-18

(subtract 9 from both sides)

18x<9x-27

(subtract 9x from both sides)

9x<-27

(divide both sides by 9)

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3 years ago

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Who was responsible for inventing the first working four-stroke engine?<br>

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Step-by-step explanation:

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4 years ago

plain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = x2 − 2x x2 − 4 if x ≠ 2 1 if x =

ElenaW [278]

Answer with Step-by-step explanation:

We are given that

$f(x)=\left\{\begin{matrix}\dfrac{x^2-2x}{x^2-4}&,if\ \ x\neq 2 \\ 1&,if\ \ x=2\end{matrix}\right.$

We have to explain that why the function is discontinuous at x=2

We know that if function is continuous at x=a then LHL=RHL=f(a).

$f(x)=\frac{x(x-2)}{(x+2)(x-2)}=\frac{x}{x+2}$

LHL=Left hand limit when x <2

Substitute x=2-h

where h is small positive value >0

$\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2-h}{2-h+2}$

$\lim_{h\rightarrow 0}\frac{2-h}{4-h}=\frac{2}{4}=\frac{1}{2}$

Right hand limit =RHL when x> 2

Substitute

x=2+h

$\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2+h}{2+h+2}=\lim_{h\rightarrow 0}\frac{2+h}{4+h}$

$=\frac{2}{4}=\frac{1}{2}$

LHL=RHL= $\frac{1}{2}$

f(2)=1

$LHL=RHL\neq f(2)$

Hence, function is discontinuous at x=2

4 0

4 years ago