Well u would multiply 20 x 15 to find out the highest if the range which is 300 . So that knocks it down to B or D. But since they're $15 each, It would be based of the multiples of 15 making your answer B .

The range is all the y values.....or the f(x) values .

So the practical range would be the multiples of 15 between 0 and 300,Inclusive .

<h2>Hope this helps you!! </h2>

5 0

3 years ago

A set of 3 consecutive integers has a sum of 30. Which integers are they?

riadik2000 [5.3K]

You can only do it like '7.5 plus 9.5 plus 13

8 0

4 years ago

Read 2 more answers

Use l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Ove

steposvetlana [31]

Answer:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}$

Step-by-step explanation:

The limit is:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}$

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

$\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}$

by replacing, and applying repeatedly you obtain:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}$

hence, the limit of the function is -1/14

8 0

3 years ago

Find a solution to the following system of equations. –2x + 5y = 8 2x + 2y = 6

aksik [14]

-2x + 5y = 8
2x + 2y = 6
----------------add
7y = 14
y = 14/7
y = 2

2x + 2y = 6
2x + 2(2) = 6
2x + 4 = 6
2x = 6 - 4
2x = 2
x = 2/2
x = 1

solution is (1,2)

4 0

3 years ago

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Px + qx + py +qy factorize​

Px + qx + py +qy factorize