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3 years ago

What value or values of x satisfy 2x + 5.31 = 4.2?

Mathematics

1 answer:

Softa [21]3 years ago

3 0

Answer: x can equal -3.2 or -7.4.

Step-by-step explanation:

The correct expression is:

2|x+5.3|=4.2

So, we have an absolute value equation, where it can be positive or negative:

Positive

2 (x+5.3)=4.2

2x+10.6 =4.2

2x=4.2-10.6

2x = -6.4

x= -6.4/2

x = -3.2

Negative

2[-(x+5.3)]=4.2

-2x-10.6 =4.2

-2x= 4.2+10.6

-2x= 14.8

x= 14.8/-2

x= -7.4

x can equal -3.2 or -7.4.

Feel free to ask for more if needed or if you did not understand something.

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3 years ago

Hamburger Hut sells regular hamburgers as well as a larger burger. Either type can include cheese, relish, lettuce, tomato, must

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Answer:

a) 40 different hamburgers can be ordered with exactly three extras

b) 20 different regular hamburgers can be ordered with exactly three extras

c) 7 different regular hamburgers can be ordered with at least five extras

Step-by-step explanation:

The order in which the extras are ordered is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this problem:

2 options of hamburger(regular or larger)

6 options of extras(cheese, relish, lettuce, tomato, mustard, or catsup.).

(a) How many different hamburgers can be ordered with exactly three extras?

1 hamburger type, from a set of 2.

3 extras, from a set of 6. So

$C_{2,1}*C_{6,3} = \frac{2!}{1!(2-1)!}*\frac{6!}{3!(6-3)!} = 2*20 = 40$

40 different hamburgers can be ordered with exactly three extras

(b) How many different regular hamburgers can be ordered with exactly three extras?

3 extras, from a set of 6. So

$C_{6,3} = \frac{6!}{3!(6-3)!} = 20$

20 different regular hamburgers can be ordered with exactly three extras

(c) How many different regular hamburgers can be ordered with at least five extras?

Five extras:

5 extras, from a set of 6. So

$C_{6,5} = \frac{6!}{5!(6-5)!} = 6$

Six extras:

6 extras, from a set of 6. So

$C_{6,6} = \frac{6!}{6!(6-6)!} = 1$

6 + 1 = 7

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