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2 years ago

15

lily wants to buy a couch and loveseat. Store A has the couch for $400 and the loveseat for $200. If both items will give a 105

discount. Store B does not sell the couch but sells the loveseat for $170. and lily has a coupon for $20 off. If Lily plans to buy both the couch and the loveseat, where should she purchase the loveseat to get the better deal, and why?

1 answer:

valentina_108 [34]2 years ago

8 0

Store A because with option A, the couch is $400 and loveseat is $200 so to buy both it is $600 but she has the discount of $105 off so it would be $495 with option A. With option B, if she buy the couch for $400 and the loveseat for $170 plus the $20 off, the love seat will be $150. So add $150 to $400 and the price is $550. Option a is a cheaper option by $55

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Please help me with those question please help with out number 1

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Answer:

1) For each value of x, a value of y is increased by 5.

x = 0, y = 5

x = 1, y = 10

x = 2, y = 15

x = 3, y = 20

------------------------------------------------------------------------------------------

2) For each two values of x, a value of y is increased by 10.

x = 0, y = -2

x = 2, y = 8

x = 4, y = 18

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3)

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x = 10, y = 5

-------------------------------------------------------------------------------------------------

4)

x = 0, y = 2

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Step-by-step explanation:

This is as easy as replacing x for the actual value show on the table.

1)

$y=5x+5$

When x = 0, y = ?

$y=5(0)+5\\y=0+5\\y=5$

When x = 1, y = ?

$y=5(1)+5\\y=5+5\\y=10$

When x = 2, y = ?

$y=5(2)+5\\y=10+5\\y=15$

When x = 3, y = ?

$y=5(3)+5\\y=15+5\\y=20$

-------------------------------------------------------------------------------------------------------

2)

$y=5x-2$

When x = 0, y = ?

$y=5(0)-2\\y=0-2\\y=-2$

When x = 2, y = ?

$y=5(2)-2\\y=10-2\\y=8$

When x = 4, y = ?

$y=5(4)-2\\y=20-2\\y=18$

When x = 6, y = ?

$y=5(6)-2\\y=30-2\\y=28$

----------------------------------------------------------------------------------------------------

3)

$y=\frac{2}{5}x +1$

When x = 0, y = ?

$y=\frac{2}{5}(0) +1$

$y=0 +1\\y=1$

When x = 1, y = ?

$y=\frac{2}{5}(1) +1\\y=\frac{2}{5} +1$

$y=\frac{2}{5}+1(\frac{5}{5})\\ y=\frac{2}{5}+\frac{5}{5} \\y=\frac{7}{5}$

When x = 5, y = ?

$y=\frac{2}{5}(5) +1\\y=2+1\\y=3$

When x = 10, y = ?

$y=\frac{2}{5}(10) +1\\y=2(2) +1\\y=4+1\\y=5$

-------------------------------------------------------------------------------------------------

4)

$y=15x+2$

When x = 0, y = ?

$y=15(0)+2\\y=0+2\\y=2$

When x = 1, y = ?

$y=15(1)+2\\y=15+2\\y=17$

When x = 2, y = ?

$y=15(2)+2\\y=30+2\\y=32$

When x = 5, y = ?

$y=15(5)+2\\y=75+2\\y=77$

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spayn [35]

True.
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To solve using the Pythagorean theorem, plug in sides AC and CB into a and b, then solve for c.

23^2 + 31^2 = c^2
529 + 961 = c^2
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I hope this helps!

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3 years ago