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3 years ago

Can anyone help me? a^2 -2a+2a-4-a^2+2a+3a+6=3a+9

Mathematics

2 answers:

slavikrds [6]3 years ago

8 0

Hi, hope this helps :)

Alinara [238K]3 years ago

6 0

Answer:

a=7/2

Step-by-step explanation:

a^2-2a+2a-4-a^2+2a+3a+6=3a+9

You want to solve for "a"

1. Mover all terms on the right side to the left side setting this equation to zero.

2. Combine Like Terms

3. Divide to isolate your "a"

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SLope = (2 + 2)/(2 + 2) = 4/4 = 1

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3 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

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3 years ago

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The total cost of 3 non member tickets and 5 member tickets

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Huhhhh? Is this question complete?

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3 years ago

A motorboat travels 165 kilometers in 3 hours going upstream and 510 kilometers in 6 hours going downstream. What is the rate of

nikitadnepr [17]

<h3>Rate of the boat in still water is 70 km/hr and rate of the current is 15 km/hr</h3><h3><u>Solution:</u></h3>

Given that,

A motorboat travels 165 kilometers in 3 hours going upstream and 510 kilometers in 6 hours going downstream

Therefore,

Upstream distance = 165 km

Upstream time = 3 hours

<h3><u>Find upstream speed:</u></h3>

$speed = \frac{distance}{time}\\\\speed = \frac{165}{3}\\\\speed = 55$

Thus upstream speed is 55 km per hour

Downstream distance = 510 km

Downstream time = 6 hours

<h3><u>Find downstream speed:</u></h3>

$speed = \frac{510}{6}\\\\speed = 85$

Thus downstream speed is 85 km per hour

<em><u>If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then</u></em>

Speed downstream = u + v km/hr

Speed upstream = u - v km/hr

Therefore,

u + v = 85 ----- eqn 1

u - v = 55 ----- eqn 2

Solve both

Add them

u + v + u - v = 85 + 55

2u = 140

u = 70

<em><u>Substitute u = 70 in eqn 1</u></em>

70 + v = 85

v = 85 - 70

v = 15

Thus rate of the boat in still water is 70 km/hr and rate of the current is 15 km/hr

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3 years ago

I need help with this Maths Homework Question.

Monica [59]

Answer:

1. 41/45.

2. x/(x^2+3x+6)

Step-by-step explanation:

So first we fill the ven diagram.

There are 240 in band, so we fill that. 60 students are in both, we put that in the middle, and there are 110 people in choir.

now, since we want the probability that a student is chosen that is in band, and choir, and both. We add all this up

240 + 60 + 110 = 410.

The total possible outcome is 410, and the total outcome is 450, so the answer is

410/450 = 41/45

First, to get the total outcomes, we have to add all the expressions together.

x(x-2) + x + 2x+8 = x^2 - 2 + x + 2x + 8 = x^2 - 2 + 3x + 8 = x^2 + 3x + 6.

Since that is the total outcome, we have to find the possible outcomes.

The problem wants BOTH from the 20th century and British, so it is x.

x/(x^2+3x+6). We cannot simplify any further, thus x/(x^2+3x+6) is our answer

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2 years ago