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olganol [36]
2 years ago
10

East high schools student council plans to buy some stools and chairs for a new student center they need to buy 25 more chairs a

nd stools the chairs cost $32 each and the stools cost $28 each if the budget is $2620 how many chairs can they buy
Mathematics
1 answer:
vova2212 [387]2 years ago
7 0

Answer:

let

s = the number of stools bought, s is integer

c = the number of chairs bought, c is integer

they need to buy 25 more chairs than stools

c = s + 25 => s = c - 25

the chairs cost $32 each and the stools cast $28 each and the budget is $2,620

32c + 28s <= 2620

if i plug s = c - 25 in the above inequality i get

32c + 28(c - 25) <= 2620

by solving the inequality we find

c <= 55.33

since c is integer we can write c<=55

they can buy 55 chairs the most.

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Sales An apparel shop sells skirts for $45 and blouses for $35. Its entire stock is worth $51,750. But sales are slow and only h
inysia [295]

Answer:

260 skirts and 270 blouses are left in the store after the sale.

Step-by-step explanation:

Let x be the number of skirts and y be the number of blouses.

An apparel shop sells skirts for $45 and blouses for $35

Cost of entire stock = $51,750

Thus, we can write the equation,

45x + 35y = 51750

Half the skirts and two-thirds of the blouses are sold for a total of $30,600.

Thus, we can write the equation,

45(\dfrac{x}{2}) + 35(\dfrac{2y}{3}) = 30600\\\\135x + 140y = 183600

Solving the two equations by elimination method, we have,

135x + 140y-(135x + 105y) = 183600 - 155250\\35y = 28350\\\\y = \dfrac{28350}{35}  =810\\\\x = \dfrac{51750-35(810)}{45} = 520

Thus, there were 520 skirts and 810 blouses in the store.

Skirt left =

520-\dfrac{520}{2} = 260

Blouses left =

810 - \dfrac{2}{3}\times 810 = 270

Thus, 260 skirts and 270 blouses are left in the store after the sale.

5 0
3 years ago
-9+5 What is the answer of this question? It is Integers
qaws [65]

Answer:

The answer is -4

Step-by-step explanation:

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Answer/Step-by-step explanation:

Given, b(x) = (\frac{6}{7})^{x}

The table for the function are:

When x = -2

b(-2) = (\frac{6}{7})^{-2}

b(-2) = \frac{1}{(\frac{6}{7})^{2}}

b(-2) = \frac{1}{(\frac{36}{49})}

b(-2) = 1*\frac{49}{36}

b(-2) = \frac{49}{36}

When x = -1

b(-1) = (\frac{6}{7})^{-1}

b(-1) = \frac{1}{(\frac{6}{7})}

b(-1) = 1*\frac{7}{6}

b(-2) = \frac{7}{6}

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b(0) = (\frac{6}{7})^{0}

b(0) = \frac{6^0}{7^0}

b(0) = \frac{1}{1}

b(0) = 1

When x = 1

b(1) = (\frac{6}{7})^{1}

b(1) = \frac{6}{7}

When x = 2

b(2) = (\frac{6}{7})^{2}

b(2) = \frac{6^2}{7^2}

b(2) = \frac{36}{49}

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3 years ago
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What rotation is shown below?
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Answer:

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Step-by-step explanation:

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