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NemiM [27]
3 years ago
11

A number that has exactly two factors,1 and itself.is a____________________________.

Mathematics
2 answers:
chubhunter [2.5K]3 years ago
6 0
The answer to this question is A prime number. A prime number has only two factors one an itself. Opposite to composite which has more than two. Example!
Prime:2,3,5,7
Composite:4,6,8,9
den301095 [7]3 years ago
6 0

A multiple that has two or more factors is a composite number. A multiple that has two factors itselfs and 1 is a prime number. Hope this helps!
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Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify
irina [24]

Answer:

x-coordinates of relative extrema = \frac{-1}{2}

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}

Differentiate with respect to x

f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}

f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}

Differentiate f'(x) with respect to x

f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1

At x = \frac{-1}{2},

f''\left ( \frac{-1}{2} \right )=\frac{4\left ( -1+4\left ( \frac{-1}{2} \right ) \right )}{3\left ( \frac{-1}{2} \right )^{\frac{5}{3}}}>0

We know that if f''(a)>0 then x = a is a point of minima.

So, x=\frac{-1}{2} is a point of minima.

For inflexion points:

Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.

So, x-coordinates of the inflexion points are 0, 1

7 0
3 years ago
I need help on this one
umka2103 [35]
You seem to have gotten m∠2. Remember that ∠1 and ∠2 are alternate interior angles, meaning they're both equal. Since they gave you m∠1 as being 26°, you now know the measure of ∠2.

As for m∠3 and m∠4, If you look at ∠3 you'll see that it is complementary to ∠1 (They both add up to 90°), so if you subtract m∠1 from 90° you'll have found m∠3. You find m∠4 the same way.

Hope this helped.
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3 years ago
PLEASE HELP LOTS OF POINTS
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