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77julia77 [94]
3 years ago
8

...........................

Mathematics
1 answer:
Brrunno [24]3 years ago
8 0

Answer:

.. .. .. .. .. .... . ..

Step-by-step explanation:

... .... .. .. ..

You might be interested in
A pair of equations is shown below:
lesya692 [45]

Answer:

The solution to the pair of equations is x=1,y=2

Step-by-step explanation:

The given equations are:

y=6x-4

y=5x-3

To solve the pair of equations graphically, we need to graph the two equations. Their point of intersection is the solution to the pair of equations.

The functions are in the form;

y=mx+b

where m=6 is the slope and b=-4 is the y-intercept of y=6x-4.

and where m=5 is the slope and b=-3 is the y-intercept of y=5x-3.

The two equations have been graphed in the attachment.

They intersected at (1,2).

The solution to the pair of equations is x=1,y=2

3 0
3 years ago
What is the decimal equivalent of the fraction 2?<br> 0 2.125<br> 0 2.4<br> O 2.1<br> O 2.18
grandymaker [24]

Answer:

2 .1 because its closes to two

3 0
3 years ago
What is the slope-intercept form of the linear equation 2x–8y=32 ?
MrMuchimi
The answer is (0,-4) (16,0)
4 0
3 years ago
Read 2 more answers
Hope u can help me plz
svet-max [94.6K]

Step-by-step explanation:

Starting with the first one, we know that 3²=9 and 4²=16, so √10 is between 3 and 4. Similarly, √27 is a little over 5. √3 is close to 2, √2 is close to 1, and √9=3, so for simplicity, we can write this as \frac{3.something*5.something}{1.something*1.something*3}

The 3.something and the 3 kind of cross out, and we're left with 5.something over 2 1.somethings, which can be closer to 2 or 1, which is somewhat unclear -- therefore, our answer can be D or E, and we'll wait on this one

For the second one, we can cross out the √6s to get \frac{8\pi -6\pi }{2} =\pi -- π is a little greater than 3, so this is E, making the first one D

For the third one, we can cross out the √3 to get (3-2)/2 = 1/2, or A as it is between 0 and 1

For the last one, note the multiplication -- cross out the √3 to get (3*2√2)/6, and we get 6√2/6=√2, or C

3 0
4 years ago
Let A = {a, b, c}, B = {b, c, d}, and C = {b, c, e}. (a) Find A ∪ (B ∩ C), (A ∪ B) ∩ C, and (A ∪ B) ∩ (A ∪ C). (Enter your answe
wariber [46]

Answer:

(a)

A\ u\ (B\ n\ C) = \{a,b,c\}

(A\ u\ B)\ n\ C = \{b,c\}

(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}

(A\ u\ B)\ n\ C = (A\ u\ B)\ n\ (A\ u\ C)

(b)

A\ n\ (B\ u\ C) = \{b,c\}

(A\ n\ B)\ u\ C = \{b,c,e\}

(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}

A\ n\ (B\ u\ C) = (A\ n\ B)\ u\ (A\ n\ C)

(c)

(A - B) - C = \{a\}

A - (B - C) = \{a,b,c\}

<em>They are not equal</em>

<em></em>

Step-by-step explanation:

Given

A= \{a,b,c\}

B =\{b,c,d\}

C = \{b,c,e\}

Solving (a):

A\ u\ (B\ n\ C)

(A\ u\ B)\ n\ C

(A\ u\ B)\ n\ (A\ u\ C)

A\ u\ (B\ n\ C)

B n C means common elements between B and C;

So:

B\ n\ C = \{b,c,d\}\ n\ \{b,c,e\}

B\ n\ C = \{b,c\}

So:

A\ u\ (B\ n\ C) = \{a,b,c\}\ u\ \{b,c\}

u means union (without repetition)

So:

A\ u\ (B\ n\ C) = \{a,b,c\}

Using the illustrations of u and n, we have:

(A\ u\ B)\ n\ C

(A\ u\ B)\ n\ C = (\{a,b,c\}\ u\ \{b,c,d\})\ n\ C

Solve the bracket

(A\ u\ B)\ n\ C = (\{a,b,c,d\})\ n\ C

Substitute the value of set C

(A\ u\ B)\ n\ C = \{a,b,c,d\}\ n\ \{b,c,e\}

Apply intersection rule

(A\ u\ B)\ n\ C = \{b,c\}

(A\ u\ B)\ n\ (A\ u\ C)

In above:

A\ u\ B = \{a,b,c,d\}

Solving A u C, we have:

A\ u\ C = \{a,b,c\}\ u\ \{b,c,e\}

Apply union rule

A\ u\ C = \{b,c\}

So:

(A\ u\ B)\ n\ (A\ u\ C) = \{a,b,c,d\}\ n\ \{b,c\}

(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}

<u>The equal sets</u>

We have:

A\ u\ (B\ n\ C) = \{a,b,c\}

(A\ u\ B)\ n\ C = \{b,c\}

(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}

So, the equal sets are:

(A\ u\ B)\ n\ C and (A\ u\ B)\ n\ (A\ u\ C)

They both equal to \{b,c\}

So:

(A\ u\ B)\ n\ C = (A\ u\ B)\ n\ (A\ u\ C)

Solving (b):

A\ n\ (B\ u\ C)

(A\ n\ B)\ u\ C

(A\ n\ B)\ u\ (A\ n\ C)

So, we have:

A\ n\ (B\ u\ C) = \{a,b,c\}\ n\ (\{b,c,d\}\ u\ \{b,c,e\})

Solve the bracket

A\ n\ (B\ u\ C) = \{a,b,c\}\ n\ (\{b,c,d,e\})

Apply intersection rule

A\ n\ (B\ u\ C) = \{b,c\}

(A\ n\ B)\ u\ C = (\{a,b,c\}\ n\ \{b,c,d\})\ u\ \{b,c,e\}

Solve the bracket

(A\ n\ B)\ u\ C = \{b,c\}\ u\ \{b,c,e\}

Apply union rule

(A\ n\ B)\ u\ C = \{b,c,e\}

(A\ n\ B)\ u\ (A\ n\ C) = (\{a,b,c\}\ n\ \{b,c,d\})\ u\ (\{a,b,c\}\ n\ \{b,c,e\})

Solve each bracket

(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}\ u\ \{b,c\}

Apply union rule

(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}

<u>The equal set</u>

We have:

A\ n\ (B\ u\ C) = \{b,c\}

(A\ n\ B)\ u\ C = \{b,c,e\}

(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}

So, the equal sets are:

A\ n\ (B\ u\ C) and (A\ n\ B)\ u\ (A\ n\ C)

They both equal to \{b,c\}

So:

A\ n\ (B\ u\ C) = (A\ n\ B)\ u\ (A\ n\ C)

Solving (c):

(A - B) - C

A - (B - C)

This illustrates difference.

A - B returns the elements in A and not B

Using that illustration, we have:

(A - B) - C = (\{a,b,c\} - \{b,c,d\}) - \{b,c,e\}

Solve the bracket

(A - B) - C = \{a\} - \{b,c,e\}

(A - B) - C = \{a\}

Similarly:

A - (B - C) = \{a,b,c\} - (\{b,c,d\} - \{b,c,e\})

A - (B - C) = \{a,b,c\} - \{d\}

A - (B - C) = \{a,b,c\}

<em>They are not equal</em>

4 0
3 years ago
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