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vova2212 [387]
4 years ago
6

A polynomial that has a degree of 0 is called ___.

Mathematics
2 answers:
Ganezh [65]4 years ago
5 0

Answer:

Constant

Step-by-step explanation:

Degree 0 means the highest power of the variable is 0, which also means there's only one term

(Because the powers have to non-negative integers, and 0 is the smallest)

Ax⁰ is the form of a degree-0 polynomial

x⁰ = 1

Ax⁰ = A

Which is a constant

Natasha2012 [34]4 years ago
3 0

Answer:

constant or constant polynomial

Some call is the zero polynomial

Step-by-step explanation:

Let x be raised to the 0 power

x^0 = 1

This is constant

3x^0 = 3*1 = 3

This is still constant

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ruslelena [56]

Answer:

Step-by-step explanation:

48 all u have to do is multiple

3 0
4 years ago
Covert 7 pi over 5 to degrees
hjlf

Answer:

Step-by-step

To convert radians to degrees, multiply by 180π , since a full circle is 360° or 2π radians.

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4 years ago
Find the zero(s) of the rational expression below. fraction numerator 5 x squared plus 26 x plus 24 over denominator 3 x squared
SVETLANKA909090 [29]
For this case we have the following rational expression:
 \frac{5x^2 + 26x + 24}{3x^2 + 10x - 8}
 What we must do for this case, is to factor the numerator and the denominator.
 We have then:
 \frac{(5x+6)(x+4)}{(3x-2)(x+4)}
 Then, canceling similar terms we have:
 \frac{(5x+6)}{(3x-2)}
 Then, the function is not defined for the values that make the denominator zero.
 We have then:
 3x-2 = 0&#10;&#10;3x = 2&#10;&#10;
 x =  \frac{2}{3}
 And on the other hand, a zero of the function is:
 5x+6 = 0&#10;&#10;5x = -6
 x =  \frac{-6}{5}
 Answer:
 
a zero of the function is:
 
x = \frac{-6}{5}
 the function is not defined for the value:
 
x = \frac{2}{3}
4 0
4 years ago
Someone pls help, this is really important
Bumek [7]

Answer:

they are addcicent 35 cdfrs

4 0
3 years ago
How to integrate this
Radda [10]

To integrate, let's first clean out the inside, and simplify:

16\int {(sin x)^4} \ dx

We can now safely substitute the value of u:

u = sin x

du = cos (x) dx

Now, we can write that:

16\int{u^4} \ du

This is much easier to integrate:

16\int{u^4} \ du = 16[\frac{1}{5}u^5] + C

Simplify:

\frac{16}{5}u^5 + C

Replace u (final answer):

\frac{16}{5}sin^5x +C

Hope I could help with your integration.

4 0
3 years ago
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