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2 years ago

Give the degree of the polynomial. 10yx^5w^4-w^3v^5-5x^11-6

Mathematics

1 answer:

torisob [31]2 years ago

4 0

10yx⁵w⁴ - w³v⁵ - 5x¹¹ - 6

Degree = biggest exponent = 11

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Which is the solution of set?

Anna [14]

Answer:

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3 years ago

A homogeneous rectangular lamina has constant area density ρ. Find the moment of inertia of the lamina about one corner

frozen [14]

Answer:

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Step-by-step explanation:

By applying the concept of calculus;

the moment of inertia of the lamina about one corner $I_{corner}$ is:

$I_{corner} = \int\limits \int\limits_R (x^2+y^2) \rho d A \\ \\ I_{corner} = \int\limits^a_0\int\limits^b_0 \rho(x^2+y^2) dy dx$

where :

(a and b are the length and the breath of the rectangle respectively )

$I_{corner} = \rho \int\limits^a_0 {x^2y}+ \frac{y^3}{3} |^ {^ b}_{_0} \, dx$

$I_{corner} = \rho \int\limits^a_0 (bx^2 + \frac{b^3}{3})dx$

$I_{corner} = \rho [\frac{bx^3}{3}+ \frac{b^3x}{3}]^ {^ a} _{_0}$

$I_{corner} = \rho [\frac{a^3b}{3}+ \frac{ab^3}{3}]$

$I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

Thus; the moment of inertia of the lamina about one corner is $I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)$

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3 years ago

Answer for all 3 boxes please :)

Cerrena [4.2K]

Answer:

$m=-\frac{5}{3}$

Step-by-step explanation:

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3 years ago

A businessperson is charged a $4.96 monthly finance charge on a bill of $283.15.

gladu [14]

Answer:

1.75%

Step-by-step explanation:

The monthly interest rate is the interest amount divided by the base on which it is computed, expressed as a percentage.

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2 years ago

What are the zeros of f(x)=x^2+x-20?

zzz [600]

F(x) = x² + x - 20 = x² + 5x - 4x - 20 = x(x + 5) - 4(x + 5) = (x + 5)(x - 4)

f(x) = 0 ⇔ (x + 5)(x - 4) = 0 ⇔ x + 5 = 0 or x - 4 = 0 ⇒ x = -5 or x = 4

Answer: C. x = -5 and x = 4.

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3 years ago

Give the degree of the polynomial. 10yx^5w^4-w^3v^5-5x^11-6​

Give the degree of the polynomial. 10yx^5w^4-w^3v^5-5x^11-6