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aleksandrvk [35]
1 year ago
13

Solve x + 5 = 7.

Mathematics
1 answer:
Rudiy271 year ago
5 0
Answer: C

|x+5| = 7 (Spilt into 2 problems)

X+5=7
X+5=-7

Solve for x on both

x=2 and x=-12
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Shakespear's Pizza sells 1 comma 100 large Vegi Pizzas per week for ​$16 a pizza. When the owner offers a​ $5 discount, the week
zimovet [89]

Answer:

(a) A(x) = 100x + 1100

(b) x = 2.5 USD

(c) A(x) = 100x + 1100 and x = -1.5 USD

Step-by-step explanation:

(a) Since there's a linear relation between weekly sales A(x) and the discount, we can model it as the following expression

A(x) = mx + b

where x is the discount, m is the slope and b is the intersect of the linear equation. We can solve for slope first

m = \frac{s_2 - s_1}{x_2 - x_1} = \frac{1600 - 1100}{5 - 0} = \frac{500}{5} = 100

To solve for b, we can just plug in m = 100, x = 0 and A(0) = 1100

1100 = 100*0 + b

b = 1100

Therefore A(x) = 100x + 1100

With discount x, the actual price would then be p - x. Where p is the original undiscounted pricing ($16 in this case)

And with sale A(x), the revenue would then be

R = (p - x)A(x) = (p - x)*(100x + 1100)

R = 100xp + 1100p - 100x^2 - 1100x

R = -100x^2 + (100p - 1100)x + 1100p

To find the maximum value of this, we can take the 1st derivative and set it to 0

R' = -200x + 100p - 1100 = 0

200x = 100p - 1100

x = \frac{100p - 1100}{200} = 0.5p - 5.5

when p = $16 then x = 0.5*16 - 5.5 = 8 - 5.5 = 2.5 USD

(c) when the price p is $8 then A(x) would sill be 100x + 1100 because it doesn't depend on p, but the discount. But to maximize the revenue, x will need to be

0.5p - 5.5 = 0.5*8 - 5.5 = 4 - 5.5 = -1.5 USD

So the Pizza owner would need to raise the price by 1.5 USD because originally the pizza is already inexpensive.

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Answer:

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3. (1)x + y =20

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Step-by-step explanation:

Just trust me and its too much work explaining and im lazy so just ask your teacher or a friend if you can.

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Suppose that many large samples are taken from a population and that the sample proportions are normally distributed, with a mea
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Answer:

[0.16316 , 0.27684]

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