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3 years ago

Write and graph a system of inequalities that models the situation.

Mathematics

1 answer:

liubo4ka [24]3 years ago

7 0

Answer:

The feasible region in the attached figure

Step-by-step explanation:

Let

x ----> the number of shirts

y ----> the number of pants

we know that

The cost of the shirts (number of shirts multiplied by the cost of one shirt) plus the cost of the pants (number of pants multiplied by the cost of one pant) must be less than or equal to $80

The inequality that represent this problem is

$5x+8y\leq 80$

using a graphing tool

The feasible region is the triangular shaded area

see the attached figure

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No Solutions<br> 5x - 2x + 7 - x = _x + _

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Step-by-step explanation:

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2 years ago

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Given that f(x) = x2 − 3x + 3 and g(x) = the quantity of x minus one, over four , solve for f(g(x)) when x = 5. (1 point)

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Rewrite g(x) as x-1

------

and then substitute this result for x in f(x) = x^2 - 3x + 3:

f(g(x)) = (x-1)^2 / 4^2 - 3(x-1)/4 + 3.

At this point we can substitute the value 5 for x:

f(g(5)) = (5-1)^2 / 4^2 - 3(5-1)/4 + 3

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3 years ago

Please solve with explanation I’ve been asking all day (this is not a multiple choice question)

Anastaziya [24]

Answer:

a) $SA = 522.9~cm^2$

b) $V_{cone} = 670.2~cm^3$

c) $V_{empty} = 1340.4~cm^3$

Step-by-step explanation:

For a cone,

$SA = \pi r (L + r)$

where L = slant height

$L = \sqrt{r^2 + h^2}$

We have r = 8 cm; h = 10 cm

$L = \sqrt{(8~cm)^2 + (10~cm)^2}$

$L = \sqrt{164~cm^2}$

$SA = (\pi)(8~cm)(\sqrt{164~cm^2} + 8~cm)$

$SA = 522.9~cm^2$

$V_{cone} = \dfrac{1}{3}\pi r^2 h$

$V_{cone} = \dfrac{1}{3}(\pi)(8~cm)^2(10~cm)$

$V_{cone} = 670.2~cm^3$

$V_{cylinder} = \pi r^2 h$

empty space = volume of cylinder - volume of cone

$V_{empty} = V_{cylinder} - V_{cone}$

$V_{empty} = \pi r^2 h - \dfrac{1}{3}\pi r^2 h$

$V_{empty} = (\pi)(8~cm)^2(10~cm) - \dfrac{1}{3}(\pi)(8~cm)^2(10~cm)$

$V_{empty} = 1340.4~cm^3$

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3 years ago