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kompoz [17]
3 years ago
13

What is the product of -4 and 8.1? a) -32.4 b) -12.1 c)12.1 b) 32.4

Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
8 0

Option A

<u>Answer: </u>

The product of -4 and 8.1 is -32.4

<u>Solution: </u>

When we multiply a negative number with a positive number we get the resultant product as a negative number.

Here we have to multiply - 4 which is a negative number and 8.1 which is a positive number.

To simplify the multiplication calculation we can write 8.1 as 8+0.1 and then multiply 4 with both 8 and 0.1

-4 \times 8.1 = -4 \times (8+0.1)

⇒  -4 \times 8.1 = -4 \times 8 – 4 \times 0.1   [Multiplying 4 with 8 and 0.1]

⇒  -4 \times 8.1= -32 - 0.4

⇒  -4 \times 8.1= - 32.4

Thus the product of -4 and 8.1 is -32.4

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Step-by-step explanation:

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2 years ago
A group of friends were working on a student film. They spent all their budget on props and costumes. They spent $693 on props,
sergeinik [125]

Answer:

The total budget is $1260

Step-by-step explanation:

Given

Props = \$693

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Required

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\%Props + \%Costumes = 100\%

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5 0
2 years ago
Using the bijection rule to count binary strings with even parity.
AleksandrR [38]

Answer:

Lets denote c the concatenation of strings. For a binary string <em>a</em> in B9, we define the element f(a) in E10 this way:

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  • f(a) = a c {0} if a has an even number of 1's

Step-by-step explanation:

To show that the function f defined above is a bijective function, we need to prove that f is well defined, injective and surjective.

f   is well defined:

To see this, we need to show that f sends elements fromo b9 to elements of E10. first note that f(a) has 1 more binary integer than a, thus, it has 10. if a has an even number of 1's, then f(a) also has an even number because a 0 was added. On the other hand, if a has an odd number of 1's, then f(a) has one more 1, as a consecuence it will have an even number of 1's. This shows that, independently of the case, f(a) is an element of E10. Thus, f is well defined.

f is injective (or one on one):

If a and b are 2 different binary strings, then f(a) and f(b) will also be different because the first 9 elements of f(a) form a and the first elements of f(b) form b, thus f(a) is different from f(b). This proves that f in injective.

f is surjective:

Let y be an element of E10, Let x be the first 9 elements of y, then f(x) = y:

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  • If x has an odd number of 1's, then the last digit of y has to be a 1, otherwise it wont be an element of E10, and f(x) = x c {1} = y

This shows that f is well defined from B9 to E10, injective, and surjective, thus it is a bijection.

3 0
3 years ago
Help me with this math money problem?? (Number 7)
ZanzabumX [31]
A quarter is .25 cents
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If you were to add them together(.77+.25) it would be $1.02
4 0
3 years ago
Read 2 more answers
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