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Wewaii [24]
3 years ago
7

Kelly runs a bakery that sells Chocolate Blizzards and Mint Breezes. The bakery must make at least 25 dozen and at most 45 dozen

Chocolate Breezes. The bakery must make at least 2 dozen and at most 15 dozen Mint Breezes. Each dozen of Chocolate Blizzards requires 5 ounces of flour while each dozen of Mint Breezes requires 9 ounces of flour. The bakery only has 315 ounces of flour available. If a dozen Chocolate Blizzards generates $2.92 in income, and a dozen Mint Breezes generates $2.11 in income, how many dozen of each dessert should Kelly make to maximize profit? Show all work. I just need the equations I would need to get this answer.
Mathematics
1 answer:
drek231 [11]3 years ago
3 0
Yeah Thanks I love to see that and the answer A
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Simplify completely quantity 4 x squared minus 7 x plus 3 all over x squared plus 5 x minus 6. quantity 4 x plus 1 over quantity
irinina [24]

Answer:

The answer is (4x - 3)/(x + 6)

Step-by-step explanation:

∵ (4x² - 7x + 3)/(x² + 5x - 6)

Factorize the numerator:

4x² - 7x + 3 = (4x - 3)(x - 1)

Factorize the denominator:

x² + 5x - 6 = (x - 1)(x + 6)

∴ (4x - 3)(x - 1)/(x - 1)(x + 6) ⇒ reduce (x - 1) from in up and down

∴ The answer is (4x - 3)/(x + 6) ⇒ the fourth answer

8 0
3 years ago
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Virty [35]

Answer:

Step-by-step explanation:

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5 0
3 years ago
Find the cube roots of 27(cos 330° + i sin 330°)
Aleksandr-060686 [28]

Answer:

See below for all the cube roots

Step-by-step explanation:

<u>DeMoivre's Theorem</u>

Let z=r(cos\theta+isin\theta) be a complex number in polar form, where n is an integer and n\geq1. If z^n=r^n(cos\theta+isin\theta)^n, then z^n=r^n(cos(n\theta)+isin(n\theta)).

<u>Nth Root of a Complex Number</u>

If n is any positive integer, the nth roots of z=rcis\theta are given by \sqrt[n]{rcis\theta}=(rcis\theta)^{\frac{1}{n}} where the nth roots are found with the formulas:

  • \sqrt[n]{r}\biggr[cis(\frac{\theta+360^\circ k}{n})\biggr] for degrees (the one applicable to this problem)
  • \sqrt[n]{r}\biggr[cis(\frac{\theta+2\pi k}{n})\biggr] for radians

for  k=0,1,2,...\:,n-1

<u>Calculation</u>

<u />z=27(cos330^\circ+isin330^\circ)\\\\\sqrt[3]{z} =\sqrt[3]{27(cos330^\circ+isin330^\circ)}\\\\z^{\frac{1}{3}} =(27(cos330^\circ+isin330^\circ))^{\frac{1}{3}}\\\\z^{\frac{1}{3}} =27^{\frac{1}{3}}(cos(\frac{1}{3}\cdot330^\circ)+isin(\frac{1}{3}\cdot330^\circ))\\\\z^{\frac{1}{3}} =3(cos110^\circ+isin110^\circ)

<u>First cube root where k=2</u>

<u />\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(2)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+720^\circ}{3})\biggr]\\3\biggr[cis(\frac{1050^\circ}{3})\biggr]\\3\biggr[cis(350^\circ)\biggr]\\3\biggr[cos(350^\circ)+isin(350^\circ)\biggr]

<u>Second cube root where k=1</u>

\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(1)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+360^\circ}{3})\biggr]\\3\biggr[cis(\frac{690^\circ}{3})\biggr]\\3\biggr[cis(230^\circ)\biggr]\\3\biggr[cos(230^\circ)+isin(230^\circ)\biggr]

<u>Third cube root where k=0</u>

<u />\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(0)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ}{3})\biggr]\\3\biggr[cis(110^\circ)\biggr]\\3\biggr[cos(110^\circ)+isin(110^\circ)\biggr]

4 0
2 years ago
What is the area of the figure?<br><br><br><br> 128 in2<br> 118 in2<br> 108 in2<br> 48 in2
lys-0071 [83]

Answer:

108 in2

Step-by-step explanation:

5 0
3 years ago
If f and g are continuous functions with f(0) = 2 and the following limit, find g(0). lim_(x-&gt;0)[2 f(x) - g(x)] = -3
ladessa [460]

We have by distributivity of limits over sums, and continuity of f(x) and g(x), that

\displaystyle \lim_{x\to0} \bigg(2f(x) - g(x)\bigg) = 2 \lim_{x\to0} f(x) - \lim_{x\to0} g(x) = 2 f(0) - g(0)

Given f(0)=2, it follows that

2^2 - g(0) = -3 \implies g(0) = \boxed{7}

5 0
2 years ago
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