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3 years ago

a nonlinear system of inequalities consists of two or more inequalities. ____ of these is/are nonlinear

Mathematics

2 answers:

GrogVix [38]3 years ago

3 0

Answer:

A nonlinear system of inequalities consists of two or more inequalities. At least one of these is/are nonlinear

Step-by-step explanation:

System of non-linear inequalities--

A system of non-linear inequalities is a system of inequalities consisting of two or more variables such that it contain at least one inequality which is non-linear.

The difference between the system of linear inequalities and a system of non-linear inequalities is it may cover more region of the solution as compared to system of linear inequalities.

katrin2010 [14]3 years ago

3 0

The answer for apex is, one.

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The rate to rent a small moving van is $30 per day, plus $0.25 per mile. The Sparkling Appliance Company rented a van because it

Zigmanuir [339]

Total rental cost = $390 = 2($30/day) + ($0.25/mile)x (x=number of miles)
330
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0.25

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3 years ago

Find the value of a:<br> a(x+3)<5x+15-x<br> a=

Vinil7 [7]

Answer:

a= 4x+15/x+3

Step-by-step explanation:

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3 years ago

Read 2 more answers

Mrs. Owens was making Thanksgiving dinner for 10

saul85 [17]

Well if she is making dinner for 10 people and each person want 2 pieces then you need to multiple 10 x 2 = 20. So she is going to need to make 20 pieces.

If each of her pumpkin pies only has 8 pieces then she is going to have to make 3 pies.

3 pies x 8 pieces = 24 pieces of pumpkin pie. So she has to make 3 pies because she needs 20 pieces since everyone wants 2.

In conclusion Mrs. Owens has to make 3 pies and she is going to have 4 pieces left over.

7 0

3 years ago

Expand (2x+2)^6<br> How would you find the answer using the binomial theorem?

Yanka [14]

Answer:

Step-by-step explanation:

$\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.$

$\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\$

$(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+ \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.$

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1 year ago

Solve each equation 11.3 - 7.2f = -3.82

Tema [17]

11.3-7.2f = -3.82
-11.3 -11.30
_____________
-7.2f = -15.12
____ _____
-7.2 -7.2
f = 2.1

6 0

3 years ago