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Evgesh-ka [11]
3 years ago
6

Voce 2x-u - 1x2-3x+3 = 2​

Mathematics
1 answer:
kumpel [21]3 years ago
5 0

Answer:

u=−x2−x+1

Step-by-step explanation:

Let's solve for u.

2x−u−1x^2−3x+3=2

Step 1: Add x^2 to both sides.

−x2−u−x+3+x2=2+x2

−u−x+3=x2+2

Step 2: Add x to both sides.

−u−x+3+x=x2+2+x

−u+3=x2+x+2

Step 3: Add -3 to both sides.

−u+3+−3=x2+x+2+−3

−u=x2+x−1

Step 4: Divide both sides by -1.

−u/−1=x2+x−1

−1/u=−x2−x+1

Answer:

u=−x2−x+1

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catering service offers 8 ​appetizers, 11 main​ courses, and 7 desserts. A banquet committee is to select 7 ​appetizers, 8 main​
guapka [62]

Answer:  The required number of ways is 46200.

Step-by-step explanation:  Given that a catering service offers 8 ​appetizers, 11 main​ courses, and 7 desserts.

A banquet committee is to select 7 ​appetizers, 8 main​ courses, and 4 desserts.

We are to find the number of ways in which this can be done.

We know that

From n different things, we can choose r things at a time in ^nC_r ways.

So,

the number of ways in which 7 appetizers can be chosen from 8 appetizers is

n_1=^8C_7=\dfrac{8!}{7!(8-7)!}=\dfrac{8\times7!}{7!\times1}=8,

the number of ways in which 8 main courses can be chosen from 11 main courses is

n_2=^{11}C_8=\dfrac{11!}{8!(11-8)!}=\dfrac{11\times10\times9\times8!}{8!\times3\times2\times1}=165

and the number of ways in which 4 desserts can be chosen from 7 desserts is

n_3=^7C_4=\dfrac{7!}{4!(7-4)!}=\dfrac{7\times6\times5\times4!}{4!\times3\times2\times1}=35.

Therefore, the number of ways in which the banquet committee is to select 7 ​appetizers, 8 main​ courses, and 4 desserts is given by

n=n_1\times n_2\times n_3=8\times165\times35=46200.

Thus, the required number of ways is 46200.

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