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ExtremeBDS [4]
2 years ago
5

THIS IS DUE IN 5 MINUTES

Mathematics
1 answer:
PilotLPTM [1.2K]2 years ago
7 0

Answer:

The answer is 6!!!

Step-by-step explanation:

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A brand of clay brick tiles has the dimensions 3 5 8 by 7 5 8 by 1 2 inch. These thin bricks are commonly used for floors. Estim
raketka [301]

Answer: The cost to tile the floor is $318.24.

Step-by-step explanation:

Since we have given that

Dimensions of floor is 12 by 17 foot.

So, Area of floor would be

12\times 17\\\\=204\ sq.\ foot

As we know that 1 foot = 12 inch

So, Area of floor would be

204\times (12)^2\\\\=204\times 144\\\\=29376\ sq.\ foot

Dimension of brick tiles is given by

3\dfrac{5}{8}\ by\ 7\dfrac{5}{8}\ by\ \dfrac{1}{2}

So, Area of brick tiles is given by

2(lb+bh+lh)\\\\=2(\dfrac{29}{8}\times \dfrac{61}{8}+\dfrac{61}{8}\times \dfrac{1}{2}+\dfrac{29}{8}\times \dfrac{1}{2})\\\\=2(\dfrac{1769}{64}+\dfrac{61}{16}+\dfrac{29}{16})\\\\=2\times \dfrac{1769+244+116}{64}\\\\=\dfrac{2129}{32}

So, Number of clay brick would be

29376\div \dfrac{2129}{32}\\\\=29376\times \dfrac{32}{2129}\\\\=\dfrac{29376\times 32}{2129}\\\\=441.53\\\\\approx 442

cost of each clay brick = $0.72

So, Total cost of flooring would be

442\times 0.72\\\\=\$318.24

Hence, The cost to tile the floor is $318.24.

7 0
2 years ago
Joel is making sauces at his restaurant.He has 8 gallons of chicken broth. He puts it into 5 pots so each has the same amount. H
GarryVolchara [31]

Answer:1 r3

Step-by-step explanation:8 Divided by five

4 0
3 years ago
1/3(4 + 18)-2²<br>PLIS I need help :(​
liraira [26]

Answer:

10/3

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

Step-by-step explanation:

<u>Step 1: Define</u>

1/3(4 + 18) - 2²

<u>Step 2: Evaluate</u>

  1. (Parenthesis) Add:                                                                                          1/3(22) - 2²
  2. Exponents:                                                                                                      1/3(22) - 4
  3. Multiply:                                                                                                           22/3 - 4
  4. Subtract:                                                                                                          10/3
5 0
3 years ago
Every radius of a circle is a chord
Ugo [173]
False

A chord touches the  circle at 2 points.
5 0
3 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:

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

  • If x has an even number of 1's, then the last digit of y has to be 0, and f(x) = x c {0} = y
  • 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
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