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2 years ago

Advance Algebra: please help!

Mathematics

2 answers:

Irina18 [472]2 years ago

8 0

Answer:

= -x^2 -x +6

D: all real values

Step-by-step explanation:

f(x) = 4-x^2

g(x) = 2-x

f+g = 4-x^2+ 2-x

Combine like terms

= -x^2 -x +4+2

= -x^2 -x +6

The domain is what values can x take

D: all real values

balandron [24]2 years ago

6 0

Step-by-step explanation:

Ok, so what we need to do is add f(x) to g(x). like so:

$(4-x^2)+(2-x)$

Since $x^2$ is squared and x isn't, we can't add them.

But we can add the 4 and 2 and get this:

$-x^2-x+6$

As far as I know, the domain is all real numbers.

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What number goes I the blanket make this equation true 17-9 = 6+ ____

NeTakaya

Answer:

Step-by-step explanation:

well first do 17-9= (17-7)-2=10-2=8

so 8=6+_

subtract 6 from both sides and you get

2=_

so the blank is 2

7 0

1 year ago

Read 2 more answers

What are the solutions to the equation (x-6)(x+8)=0?

Zepler [3.9K]

Answer:

x = 6, -8

Step-by-step explanation:

If (x - 6)(x + 8) = 0, that would imply that either (x - 6) or (x + 8) would equal zero. Using this, we can find that solving the two equations:

x - 6 = 0

and

x + 8 = 0

would yield the two solutions to the equation.

x - 6 = 0

Add 6 to both sides of the equation.

x = 6

So one of the solutions would be x = 6.

x + 8 = 0

Subtract 8 from both sides of the equation.

x = -8

So the other solution would be x = -8.

The two solutions are x = 6 and x = -8.

I hope you find my answer and explanation to be helpful. Happy studying.

6 0

2 years ago

Find the value of (– 3 /8 ) × (+8 /15). A. 11/23 B. – 1 /5 C. 11/15 D. – 1 /3

PtichkaEL [24]

(-3/8)(+8/15)

Answer will be a negative number. Whenever multiplying a - negative number * a + positive = - negative number.

(-3/8)(+8/15)

Cross out -3 and 15 . Divide by 3. -3/3,=-1 15/3

= -1/5

Cross out 8 and 8. Divide by 8.

8/8= 1 , 8/8= 1

-1/5*1/1= -1/5

Answer : -1/5

4 0

2 years ago

Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of x ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As x goes to infinity, each of the 1/x ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$