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3 years ago

12

Solve the equation a/a^2-9 +3/a-3=1/a+3

1 answer:

MAXImum [283]3 years ago

7 0

Answer is: a= -4
STEP
1
:

1
Simplify —————
a + 3
Equation at the end of step
1
:

a 3 1
(————————+—————)-——— = 0
((a2)-9) (a-3) a+3
STEP
2
:

3
Simplify —————
a - 3
Equation at the end of step
2
:

a 3 1
(————————+———)-——— = 0
((a2)-9) a-3 a+3
STEP
3
:

a
Simplify ——————
a2 - 9
Equation at the end of step
3
:

a 3 1
(————————————————— + —————) - ————— = 0
(a + 3) • (a - 3) a - 3 a + 3

Equation at the end of step
4
:

(4a + 9) 1
————————————————— - ————— = 0
(a + 3) • (a - 3) a + 3
Pull out like factors :

3a + 12 = 3 • (a + 4)

Equation at the end of step
6
:

3 • (a + 4)
————————————————— = 0
(a + 3) • (a - 3)

3•(a+4)
——————————— • (a+3)•(a-3) = 0 • (a+3)•(a-3)
(a+3)•(a-3)

a+4 = 0

Subtract 4 from both sides of the equation :
a = -4

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What is the sign of f on the interval -2

TEA [102]

Answer:

$f$ is sometimes positive and sometimes negative.

Step-by-step explanation:

$f(x)=(x-3)(x+2)(x+4)(x+4)(x-1)(2x-9)$

Take $x=-1\in(-2,\frac{9}{2})\ as-2$

$f(-1)=(-1-3)(-1+2)(-1+4)(-1-1)(-2-9)\\\\=(-4)(1)(3)(-2)(-11)\\\\=-264\\\\f(-1)$

Take $x=2\in(-2,\frac{9}{2})\ as\ -2$

$f(2)=(2,-3)(2+2)(2+4)(2-1)(2\times2-9)\\\\=(-1)(4)(6)(1)(-5)\\\\=120\\\\f(2)>0$

Hence $f(x)$ for $x=-1$ and $f(x)>0$ for $x=2$

So $f$ is sometimes positive and sometimes negative in the interval.

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3 years ago

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sleet_krkn [62]

Answer:

7 and 14

Step-by-step explanation:

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3 years ago

Read 2 more answers

A skier is trying to decide whether or not to buy a season ski pass. A daily pass costs $68. A season ski pass costs $300. The

lina2011 [118]

first, ignore the rent for the skies. we could add it to both, but since it's the same, it would not change the number of days here.

this information is meant to test your decision making on what's important for the question.

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5 0

2 years ago

Let f(x)=x^3 and g(x)= √x, and h(x)=x/3. Find each of the following: f(g(h(6)))

rewona [7]

Answer:

go from inside out.

h(x)=x/3

$(\sqrt{2})^{3}$ => h(6) = 6/3 = 2

g(x) = $\sqrt{x}$

=> g(h(6)) = g(2) = $\sqrt{2}$

f(x) = $x^{3}$

=> f(g(h(6))) = f( $\sqrt{2}$ ) = $(\sqrt{2} )^{3}$

6 0

3 years ago

algol [13]

Answer: Option A is the correct answer, y = x + 2

y - 3 = x-1

Step-by-step explanation:

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2 years ago