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ale4655 [162]
3 years ago
8

Simplify the given polynomial and use it to complete the statement.

Mathematics
2 answers:
lutik1710 [3]3 years ago
7 0

Answer:

Option A,C,B

The polynomial simplifies to an expression that is a quadratic binomial with the degree 2.

Step-by-step explanation:

Given : Expression (-5x^2-2x+4)+(8x^2-x-1)-(x+2)(x-5)

To find : Simplify the given polynomial and use it to complete the statement?

Solution :

Step 1 - Write the expression

(-5x^2-2x+4)+(8x^2-x-1)-(x+2)(x-5)

Step 2 - Open the last bracket using distributive property (a+b)(c+d)=ac+ad+bc+bd

=(-5x^2-2x+4)+(8x^2-x-1)-(x^2-5x+2x-10)

Step 3 - Open all the parenthesis

=-5x^2-2x+4+8x^2-x-1-x^2+5x-2x+10

Step 4 - Add the like terms and solve

=2x^2+13

So, The simplified expression is

(-5x^2-2x+4)+(8x^2-x-1)-(x+2)(x-5)=2x^2+13

The simplified form contains two elements hence the given polynomial is binomial and the highest power is 2 so, the degree is 2.

The polynomial simplifies to an expression that is a quadratic binomial with the degree 2.

Lorico [155]3 years ago
4 0

Answer:

1)  2x^{2}+13

2) Option C.

3) Option B.

Step-by-step explanation:

1. You must apply the Distributive property as following:

(-5x^2-2x+4)+(8x^2-x-1)-(x^{2}-5x+2x-10)

2. Now, you must distribute the negative sign, then you have:

-5x^2-2x+4+8x^2-x-1-x^{2}+5x-2x+10

3. Finally, you must add the like terms. Then you obtain the polynomial:

2x^2+13

4. By definition, a polynomial that has two terms is classified as a binomial. Therefore, the answer is the option C.

5. The degree of a polynomial is determined by highest exponent of the variable. So, it is a polynomial of degree 2 (option B).

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The cost, C dollars, of hiring a taxi is given by the formula C=2.00+320n, where n is the number of tenths of a mile driven. Wha
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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.
sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then  

\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

\bf G(x,y) = (x, y, 3-x^2-y^2)

with 0≤ x≤1 and  0≤ y≤1

So

\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)

we also have

\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)

and so

\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and  0≤ y≤1

\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

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3 years ago
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