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malfutka [58]
3 years ago
8

Define A = {x : 0 ≤ x ≤ 1}, B = {x : 0 ≤ x ≤ 3}, and C = {x : −1 ≤ x ≤ 2}. Draw diagrams showing each of the following sets of p

oints: (a) AC ∩ B ∩C (b) AC ∪(B ∩C) (c) A ∩ B ∩CC (d) [(A ∪ B)∩CC ] C

Mathematics
1 answer:
Verizon [17]3 years ago
4 0

Answer: (a) {0,1}; (b) {0,1, 2}; (c) {0,1}; (d) {0, 1, 2}

Step-by-step explanation: Defining the sets A,B and C:

A = {0,1};  B = {0, 1, 2, 3}; C = {-1, 0, 1, 2}

(a) A ∩ B ∩ C = {0, 1}, because it represents the intersection of the sets, the items that appears in all the three sets.

(b) A ∪ (B ∩ C) = {0, 1} ∪ {0, 1, 2} = {0, 1, 2}. In this case, it represents the union of the sets A with the intersection of sets B and C.

The (c) and (d) are a combination between union and intersection of the sets.

The Venn Diagrams are the representations of the sets and their interactions and are represented in the attachments

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The first number plus the second number equals 83, the first number
timama [110]

Answer:

I hope this helps

Step-by-step explanation:

So the way to handle a problem like this is to rewrite the three numbers in terms of their relationship to one of the three numbers.  Let's call the numbers x, y, and z, where x is the first number, y is the second, and z is the third.

We know x + y + z = 83.

Since the third number is twice the second, we say that z = 2y.

Since the second number is seven less than the first, we say that y = x - 7.  We can rewrite this as x = y + 7.

Now all three can be written in terms of y.

x + y + z = 83

(y + 7) + y + 2y = 83

4y + 7 = 83

4y = 76

y = 19

The second number is 19.  The first is 19 + 7 or 26, and the third is 2 times 19 or 38.

A check reveals that 26 + 19 + 38 = 83.

5 0
3 years ago
Jaime en un joven que vive en puyo con su padre. A jaime le gusta cosechar chonta con su padre los sabados en la mañana jaime ha
xenn [34]

Answer:

\large \boxed{\text{23 s}}

Step-by-step explanation:

Jaime is at point A and wants to get across the river to Point B.

They must head upstream towards Point C.

1. Calculate the net speed across the river.

∆ABC is a right triangle, so

\begin{array}{rcl}AB^{2} + BC^{2} & = & AC^{2} \\AB^{2} +10^{2} & = & 12^{2} \\AB^{2}+100& = & 144 \\AB^{2}& = & 44\\AB & = & \sqrt{44}\\& = & 2\sqrt{11}\\& \approx &\textbf{6.633 m/s}\\\end{array}

2. Calculate the time to cross the river

\text{Time} = \text{150 m} \times \dfrac{\text{1 s}}{\text{6.633 m}} \approx \textbf{23 s}\\\\\text{It will take $\large \boxed{\textbf{23 s}}$ to cross the river.}

7 0
3 years ago
How many solutions do these equations have y= 3x+4 y+6=3x
Nataly [62]

Answer:

Step-by-step explanation:

hello :

the system is :  y= 3x+4....(*)

                         y+6=3x....(**)

by (**) : y = 3x - 6

so : y= 3x+4

      y= 3x - 6

means :  3x+4 = 3x - 6

                     4= -6.......(false)

conclusion : no reals  solutions

5 0
3 years ago
I will give Brainlyest!!! Key features of graphing the parabola: y=x^2+12x+32 A. AxisofSymmetry: x = − b/2a B. Vertex: C. ​ Zero
slamgirl [31]

Answer:

Axis of Symmetry: -6

Vertex: (-6, -4)

x- int: (-4, 0) (-8, 0)

y-int: (0, 32)

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3 0
3 years ago
The probability of drawing a red candy at random from a bag of 25 candies is 2/5. After 5 candies are removed from tehe bag, wha
yaroslaw [1]

Given:

The probability of drawing a red candy at random from a bag of 25 candies is \dfrac{2}{5}.

To find:

The probability of randomly drawing a red candy from the bag after removing 5 candies from the bag.

Solution:

Let n be the number of red candies in the bag. Then, the probability of getting a red candy is:

P(Red)=\dfrac{\text{Number of red candies}}{\text{Total candies}}

\dfrac{2}{5}=\dfrac{n}{25}

\dfrac{2}{5}\times 25=n

10=n

After removing the 5 candies from the bag, the number of remaining candies is 25-5=20 and the number of remaining red candies is 10-5=5.

Now, the probability of randomly drawing a red candy from the bag is:

P(Red)=\dfrac{5}{20}

P(Red)=\dfrac{1}{4}

Therefore, the required probability is \dfrac{1}{4}.

3 0
3 years ago
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