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3 years ago

15

Stephen has a bag of candy that he is dividing up between 4 friends. If Stephen and his friends each got at least 30 pieces of c

andy. Write and solve an inequality to determine the minimum amount of candy in the bag.

1 answer:

ikadub [295]3 years ago

8 0

4 x 30 is 120 so Amount they get or x > 30
=

X > 30
=

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Step-by-step explanation:

The number before x is always the slope and the other number, the eight in this case is the y-intercept

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On a coordinate plane, the segment with endpoints (10, 40) and (70, 120) is

OlgaM077 [116]

Answer:

The length of the resulting segment is 500.

Step-by-step explanation:

Vectorially speaking, the dilation is defined by following operation:

$P'(x,y) = O(x,y) + k\cdot [P(x,y)-O(x,y)]$ (1)

Where:

$O(x,y)$ - Center of dilation.

$P(x,y)$ - Original point.

$k$ - Scale factor.

$P'(x,y)$ - Dilated point.

First, we proceed to determine the coordinates of the dilated segment:

( $P(x,y) = (10, 40)$ , $Q(x,y) = (70, 120)$ , $O(x,y) = (0,0)$ , $k = 5$ )

$P'(x,y) = O(x,y) + k\cdot [P(x,y)-O(x,y)]$

$P(x,y) = (0,0) +5\cdot [(10,40)-(0,0)]$

$P'(x,y) = (50,200)$

$Q'(x,y) = O(x,y) + k\cdot [Q(x,y)-O(x,y)]$

$Q' (x,y) = (0,0) +5\cdot [(70,120)-(0,0)]$

$Q'(x,y) = (350, 600)$

Then, the length of the resulting segment is determined by following Pythagorean identity:

$l_{P'Q'} = \sqrt{(350-50)^{2}+(600-200)^{2}}$

$l_{P'Q'} = 500$

The length of the resulting segment is 500.

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3 years ago