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ser-zykov [4K]
3 years ago
11

Find the length of the third side to the nearest tenth.

Mathematics
1 answer:
Dima020 [189]3 years ago
3 0
The answer would be 9.2 . if you use pythagorean’s theorem.
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Which expression is equivalent to 2(a + 2b) - a - 2b? A. 3a + 2b B. 3a - 2b C. a - 2b D. a + 2b
alex41 [277]

Answer:

a +2b

Step-by-step explanation:

2(a + 2b) - a - 2b

Distribute

2a +4b -a -2b

Combine like terms

2a -a + 4b-2b

a +2b

6 0
4 years ago
Read 2 more answers
Here is a pattern made from sticks:<br> 88<br> a) How many sticks would be in pattern number 6?
s2008m [1.1K]

Answer:

32 sticks would be in pattern number 6

8 0
2 years ago
Hamburger Hut sells regular hamburgers as well as a larger burger. Either type can include cheese, relish, lettuce, tomato, must
Studentka2010 [4]

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

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

8 0
3 years ago
A company bought New Year's greeting cards to send to its valued customers. Forty-five of the cards they sent cost $2.40 each. T
ASHA 777 [7]

Answer: $262.80

Step-by-step explanation: 45x2.40 + 36x2.9 + 28x1.8 = 262.8

8 0
3 years ago
I could probably figure this out but I have way too much homework so It'd be nice to finish this pathway.
pentagon [3]

Yeah that's right always be a hardworking person

8 0
3 years ago
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