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3 years ago

PLEASE SOLVE FOR ME!!!! I am stuck on question... 36/5x - 10 < 5

1 answer:

3 0

$*\ \dfrac{36}{5x} - 10 \ \textless \ 5 \qquad x \neq 0 \\ \\ \\ \dfrac{36}{5x}\ \textless \ 5 + 10 \\ \\ \\ \dfrac{36}{5x}\ \textless \ 15 \\ \\ \\ 36\ \textless \ 15 * (5x) \\ \\ 36 \ \textless \ 75x \\ \\ \dfrac{36}{75}\ \textless \ x \qquad \dfrac{12}{25}\ \textless \ x \qquad \boxed{x\in ( \frac{12}{25}; +\infty) }$

$*\ \dfrac{36}{5x-10} \ \textless \ 5 \qquad x \neq 2 \\ \\ \\36 \textless \ 5 (5x-10) \\ \\ 36 \textless \ 25x-50 \\ \\ 36+50\ \textless \25x \\ \\ 86 \ \textless \ 25x \\ \\ \dfrac{86}{25}\ \textless \ x\qquad \to \qquad \boxed{x\in ( \frac{86}{25}; +\infty) }$

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20 pts awarded and brainliest chosen!!!!

g100num [7]

Answer:

D. x <= -3 or x >= 5

Step-by-step explanation:

3 | x-1 | > = 12

3(x - 1) >= 12

3x - 3 >= 12

3x >= 15

x >= 5

-3(x - 1) >= 12

-3x + 3 >= 12

-3x >= 9

x <= -3

Answer

x <= -3 or x >= 5

4 0

2 years ago

You have just opened a new dance club, Swing Haven, but are unsure of how high to set the cover charge (entrance fee). One week

bonufazy [111]

Answer:

a) The demand function is

$q(p) = -4 p + 107$

b) The nightly revenue is

$R(p) = -4 p^2 + 107 p$

c) The profit function is

$P(p) = -4 p^2 + 133.75 p - 939$

d) The entrance fees that allow Swing Haven to break even are between 10.03 and 23.41 dollars per guest.

Step-by-step explanation:

a) Lets find the slope s of the demand:

$s = \frac{79-43}{7-16} = \frac{36}{-9} = -4$

Since the demand takes the value 79 in 7, then

$q(p) = -4 (p-7) + 79 = -4 p + 107$

b) The nightly revenue can be found by multiplying q by p

$R(p) = p*q(p) = p*( -4 p + 107) = -4 p^2 + 107 p$

c) The profit function is obtained from substracting the const function C(p) from the revenue function R(p)

$P(p) = R(p) - C(p) = p*q(p) = -4 p^2 + 107 p - (-26.75p + 939) = \\\\-4 p^2 + 133.75 p - 939$

d) Lets find out the zeros and positive interval of P. Since P is a quadratic function with negative main coefficient, then it should have a maximum at the vertex, and between the roots (if any), the function should be positive. Therefore, we just need to find the zeros of P

$r_1, r_2 = \frac{-133.75 \,^+_-\, \sqrt{133.75^2-4*(-4)*(-939)} }{-8} = \frac{-133.75 \,^+_-\, 53.526}{-8} \\r_1 = 10.03\\r_2 = 23.41$

Therefore, the entrance fees that allow Swing Haven to break even are between 10.03 and 23.41 dollars per guest.

7 0

2 years ago

Factor completely 2x^2 − 2x − 40. 2(x − 5)(x 4) (2x − 10)(x 4) (x − 5)(2x 8) 2(x − 4)(x 5)

Zina [86]

2x^2 - 2x - 40 = 2x^2 -10x + 8x - 40 = 2x(x - 5) + 8(x - 5)

= (x-5)(2x+8) = 2(x-5)(x+4)

3 0

3 years ago

If Quadrilateral Q is a scaled copy of Quadrilateral P created with a scale factor of 3

Stolb23 [73]

I don't see a drawing of the quadrilaterals, so I don't know what the perimeter of quadrilateral P is. But whatever the perimeter of P is, Q will be 1/3 of that. Perimeter is a length, so even though it may pertain to a 2-dimensional object, it is still a 1-dimensional, linear measure. When two objects are similar (same shape, but scaled up or down by a scale factor), all corresponding linear measures have the same scale factor.

If you were asked about area or volume, that would be a different matter. In the case of area, you would square the scale factor, and in the case of volume, you would cube the scale factor.

7 0

2 years ago

What is the input-output table for the function f(x) = 3x^2-x+4

mojhsa [17]

Answer:

Step-by-step explanation:

The input-output table can be made by putting value of x and finding the value of f(x)

f(x) = 3x^2-x+4

f(0) = 3(0)^2-0+4 = 0-0+4 = 4

f(1) = 3(1)^2-1+4 = 3-1+4 = 2+4 =6

f(2) = 3(2)^2-2+4 = 3(4)-2+4 = 12-2+4 = 10+4 = 14

f(3) = 3(3)^2-3+4 = 3(9)-3+4 = 27-3+4 = 24+4 = 28

So put value of x and find f(x) and fill the input-output table.

x f(x)

0 4

1 6

2 14

3 28

3 0

2 years ago