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Eva8 [605]
3 years ago
10

The Suarez family spends 30% of their monthly income on the house payment. If the house payment is $1200 what is the family's in

come? Please Show Work!

Mathematics
2 answers:
Softa [21]3 years ago
4 0
They make $4000.00 a month

4000.00 x 30% = $1200.00
hram777 [196]3 years ago
3 0
This is the answer to your problem

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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 <em>x</em> 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 <em>x</em> goes to infinity, each of the 1/<em>x</em> 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}}}

8 0
3 years ago
3/2^0<br><br> Someone help ASAP please. Thank you!
Klio2033 [76]
If you mean 3 divided by 2 power 0 then the answer is 3
4 0
3 years ago
Find the area and perimeter
ki77a [65]

Answer:

p=26

a-24

Step-by-step explanation:

I added this answer so you could mark the first person brainliest

5 0
2 years ago
I need help!!!!!!!!!!!!!!!!
SVETLANKA909090 [29]
Y=-3 is your answer
6 0
2 years ago
A skateboarder rides by my office every 8 minutes, and a cyclist rides by every 10 minutes. How long will it take for the skateb
kiruha [24]

Answer:

40 minutes

Step-by-step explanation:

10 times 4 = 40

8 times 5 = 40

so they will cross paths after 40 minutes of riding

5 0
2 years ago
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