If a necklace was on discount 10% and the original price was $20 what was the amount of discount?

Mathematics

2 answers:

netineya [11]2 years ago

8 0

Answer:

its 10$

Step-by-step explanation:

Andreas93 [3]2 years ago

7 0

Answer:

10$

Step-by-step explanation:

dollars hope it helps

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Alyaa gets 15 signatures each day for 12 day. One the thirteenth day Alyaa get 55 signatures. Alyaa is trying to get 500 signatu

amid [387]

Okay dso if she got 15 points for 12 days then you would multiply them (or just add 15+15+15+15 and so on but multiplying is easier) then you are going to add 55 because of the thirteenth day.
15(12)=180
180+55=235 is how many you have.
now you are going to subtract 500 from that number to see how many more you need. :)

5 0

3 years ago

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

Please help will mark Brainly

Brilliant_brown [7]

Answer:

Step-by-step explanation:

This is not a linear relationship, it doesn't have a set increase or uniform slope and it seems more like an exponential or quadratic curve to me when I graph it. As you can see you cant draw a straight line through the data points so it cannot be a linear relationship.