Tables from equations with 2 variables

Which of the following statements is false? (N = natural numbers) Ø N {0} N {1} N

Iteru [2.4K]

Assuming each statement is claiming the first set is a subset of N, the second statement is false. 0 is not part of the natural numbers.

3 0

3 years ago

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Round the number 33 and 89

trapecia [35]

33 rounded to the tens place would be 30.
89 rounded to the tens place would be 90.

5 0

3 years ago

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41. Assuming that a man can complete the work alone in x days, his work in four days would be: a) b) X X C d) 4x x 42. If a man

Schach [20]

Percentage and ratio word problems require understanding of the relationship between variables from which the question is formed

The options that give the correct values of the duration of the work are;

$41. \ c) \ \dfrac{4}{x}$

$42. \ d) \ \dfrac{4}{x} + \dfrac{6}{y} = \dfrac{1}{5}$

43. a) 35 days

44. c) 21·a + 28·b = 1

45. c) (42, 56)

Reasons:

41. Number of days it takes a man to complete the work alone = x days

Therefore;

$The \ work \ done \ by \ the \ man \ in \ one \ day = \dfrac{1}{x}$

$The \ work \ done \ in \ four \ days \ by\ the \ man = 4 \times \dfrac{1}{x} = \dfrac{4}{x}$

The correct option is $c) \ \dfrac{4}{x}$

42. Number of days it takes a man to complete the work alone = x days

$Work \ done \ by \ a\ man \ in \ one \ day = \dfrac{1}{x}$

$Work \ done \ by \ four \ men \ in \ one \ day = \dfrac{4}{x}$

Number of days it takes a boy to complete the work alone = y days

$Work \ done \ by \ a \ boy \ in \ one \ day = \dfrac{1}{x}$

$Work \ done \ by \ six \ boys \ in \ one \ day = \dfrac{6}{y}$

4 men and 6 boys work for 5 days to complete the work

Therefore, work done by 4 men and 6 boys in 1 day is therefore;

$\dfrac{4}{x} + \dfrac{6}{y} = \dfrac{1}{5}$

The correct option is therefore;

$d) \ \dfrac{4}{x} + \dfrac{6}{y} = \dfrac{1}{5}$

43. As per the case study, we have;

Case 1

$\dfrac{4}{x} + \dfrac{6}{y} = \dfrac{1}{5}$

Which gives;

$\dfrac{6\cdot x + 4\cdot y}{y \cdot x} = \dfrac{1}{5}$

30·x + 20·y = y·x

Case 2

$\dfrac{3}{x} + \dfrac{4}{y} = \dfrac{1}{7}$

Which gives;

$\dfrac{4\cdot x + 3\cdot y}{y \cdot x} = \dfrac{1}{7}$

28·x + 21·y = y·x

Therefore;

30·x + 20·y = 28·x + 21·y

∴ 2·x = y

Plugging in the value of y = 2·x, in Case 1 gives;

$\dfrac{4}{x} + \dfrac{6}{2 \cdot x} = \dfrac{1}{5}$

$\dfrac{2 \times 4 + 6}{2 \times x} = \dfrac{14}{2 \times x} =\dfrac{7}{x} = \dfrac{1}{5}$

7 × 5 = x

x = 7 × 5 = 35

The number of days, x, it takes a man to complete the work alone, is given by option; a) 35 days

44. For the equation $\dfrac{3}{x} + \dfrac{4}{y} = \dfrac{1}{7}$ , if $a = \dfrac{1}{x}$ , and $b = \dfrac{1}{y}$ , we have;

$3 \cdot a+ 4\cdot y = \dfrac{1}{7}$

21·a + 28·y = 1

The correct option is option C. 21·a + 28·b = 1

45. A solution to the equation $\dfrac{3}{x} + \dfrac{4}{y} = \dfrac{1}{7}$ , is given by the values of x, and y, that gives;

$\dfrac{1}{14} + \dfrac{1}{14} = \dfrac{1}{7}$

We have;

3 × 14 = 42

4 × 14 = 56

Therefore, a solution to the equation is (42, 56)

The correct option is $c) \ \underline{ (42, \ 56)}$

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3 0

2 years ago

At birth, one puppy weighs 15/16 pounds. A second puppy weighs 1 and 1/8 pounds. How much more did the second puppy weigh?

zvonat [6]

Answer:

3/16 pounds more.

Step-by-step explanation:

It is 1 1/8 - 15/16

= 9/8 - 15/16

= 18/16 - 15/16

= 3/16.

5 0

2 years ago

Whats the answer to 6/7 times 21

grandymaker [24]

Answer:

slove 6/7 then times it by 21

Step-by-step explanation:

3 0

3 years ago