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just olya [345]
2 years ago
5

What is 2/3 + 1/2 In fraction form please .

Mathematics
2 answers:
sergejj [24]2 years ago
8 0

Answer:

<h2>\frac{7}{6} or 1\frac{1}{6}</h2>

Step-by-step explanation:

<h2 /><h2>give me brainliest please</h2>

first get common denominators:

2/3 + 1/2

4/6 + 3/6

then add

4/6 + 3/6

7/6

simplify

7/6   or   1 and 1/6

vazorg [7]2 years ago
7 0

Answer: 7/6 or 1 and 1/6

Step-by-step explanation:

2     1

  +

3     2

you have to make them have the same denominators(the number on the bottom of a fraction)

2/3--> 4/6

1/2---> 3/6

4/6+3/6=7/6 or 1 and 1/6

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Answer:

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Step-by-step explanation:

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Read 2 more answers
the eagle trucking company must deliver 7/8 ton of cement blocks and 5/8 ton of bricks how much will this load weigh?
Mademuasel [1]
This load will weigh 1.5 TONS ALTOGETHER.

So, the truck is delivering a \frac{7}{8} ton of cement blocks, and a \frac{5}{8} of bricks. To find how much this load weighs, we'd have to add the two fractions together, which proves to be quite easy, because the denominators are already the same, so we wouldn't have to change anything.

\frac{7}{8} + \frac{5}{8} = \frac{12}{8}

And now, we would want to convert that fraction into a whole number, or at least a decimal, to find how many tons the truck is carrying. So we would divide. Once the work is done, here is what would happen:

12 ÷ 8 = 1.5

And there's your answer. The truck is carrying 1.5 TONS.
4 0
2 years ago
Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are
Svetlanka [38]

Answer:

(a) Number of sets B given that

  • A⊆B⊆C: 2¹⁰.  (That is: A is a subset of B, B is a subset of C. B might be equal to C)
  • A⊂B⊂C: 2¹⁰ - 2.  (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

<h3>(a)</h3>

Let x_1, x_2, \cdots, x_{20} denote the 20 elements of set X.

Let x_1, x_2, \cdots, x_{10} denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain x_1, x_2, \cdots, x_{10}.

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from x_1, x_2, \cdots, x_{20}.

\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}&  \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}.

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus 2^{10} = 1024 possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves 2^{10} -2 = 1024 - 2 = 1022 possibilities.

<h3>(b)</h3>

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}&  \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}.

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus 2^{10} = 1024 possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's 2^{20} = 1048576 possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only 2^{20} - 2^{10} possibilities left.

5 0
2 years ago
Han can run 100 meters in 20 seconds how long will it take him to run 3000 metres
Juliette [100K]

Answer:

600 seconds

Step-by-step explanation:

In this case we have to calculate the speed;

Speed = Distance/Time

Speed = 100/20 = 5 mph

Next to get our answer we have to calculate the amount of time

Time = Distance/Speed

Time = 3000/5 = 600 seconds

4 0
3 years ago
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