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tigry1 [53]
3 years ago
7

What is the degree of the polynomial?

Mathematics
1 answer:
DIA [1.3K]3 years ago
8 0

Answer:

2

Step-by-step explanation:

The degree of the polynomial is the highest exponent of an expression. When more than one variable is present, its is the sum of exponents on one term in the expression.

The polynomial has terms xy, 3x^2, -7 and x. The term with the highest exponent sum is xy or 3x^2. Both have degree 2. The degree of the polynomial is 2.

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What is the sum of the rational expressions below? x+3/x-2 + 3x-8/x-2
Mamont248 [21]

Answer:      \bold{\dfrac{4x-5}{x-2}}

<u>Step-by-step explanation:</u>

Since the expressions have the same denominator, we simply need to add the numerators.

\dfrac{x+3}{x-2}+\dfrac{3x-8}{x-2}\quad =\quad \dfrac{x+3+3x-8}{x-2}\quad = \quad \large\boxed{\dfrac{4x-5}{x-2}}

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The moon is about 240,000 miles from Earth. What is this distance written as a whole number multiplied by a power of ten?
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Consider the functions z = 4 e^x ln y, x = ln (u cos v), and y = u sin v.
Katen [24]

Answer:

remember the chain rule:

h(x) = f(g(x))

h'(x) = f'(g(x))*g'(x)

or:

dh/dx = (df/dg)*(dg/dx)

we know that:

z = 4*e^x*ln(y)

where:

y = u*sin(v)

x = ln(u*cos(v))

We want to find:

dz/du

because y and x are functions of u, we can write this as:

dz/du = (dz/dx)*(dx/du) + (dz/dy)*(dy/du)

where:

(dz/dx)  = 4*e^x*ln(y)

(dz/dy) = 4*e^x*(1/y)

(dx/du) = 1/(u*cos(v))*cos(v) = 1/u

(dy/du) = sin(v)

Replacing all of these we get:

dz/du = (4*e^x*ln(y))*( 1/u) + 4*e^x*(1/y)*sin(v)

          = 4*e^x*( ln(y)/u + sin(v)/y)

replacing x and y we get:

dz/du = 4*e^(ln (u cos v))*( ln(u sin v)/u + sin(v)/(u*sin(v))

dz/du = 4*(u*cos(v))*(ln(u*sin(v))/u + 1/u)

Now let's do the same for dz/dv

dz/dv = (dz/dx)*(dx/dv) + (dz/dy)*(dy/dv)

where:

(dz/dx)  = 4*e^x*ln(y)

(dz/dy) = 4*e^x*(1/y)

(dx/dv) = 1/(cos(v))*-sin(v) = -tan(v)

(dy/dv) = u*cos(v)

then:

dz/dv = 4*e^x*[ -ln(y)*tan(v) + u*cos(v)/y]

replacing the values of x and y we get:

dz/dv = 4*e^(ln(u*cos(v)))*[ -ln(u*sin(v))*tan(v) + u*cos(v)/(u*sin(v))]

dz/dv = 4*(u*cos(v))*[ -ln(u*sin(v))*tan(v) + 1/tan(v)]

5 0
2 years ago
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