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3 years ago

How did the greeks defend themselves from dorian rules?

Mathematics

1 answer:

muminat3 years ago

3 0

Answer: the Dorians were far less advanced then the Mycenaean Greeks; economy collasped and trade stopped

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Is this correct way to solve the problem?

Shkiper50 [21]

Answer:

10x=40

x=40/10

x=4

4 is the answers, not -4

Step-by-step explanation:

pls mark as brainliest

8 0

3 years ago

Let U= U= Universal set ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9}={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} , A={1, 3, 4, 5, 7, 9} A={1, 3, 4, 5, 7,

blagie [28]

Answer:

a) $A' = [0,2,6,8]$

b) $(AUB)' = [0,2,6]$

c) $(AUB')' = [9]$

d) A∩B′ $= [3,5,9]$

Step-by-step explanation:

Assuming this problem: "Let U= U= Universal set ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} , A={1, 3, 4, 5, 7, 9} , and B={1, 4, 7, 8} . List the elemetns of the following sets in the increasing order: a) A′= b) (A∪B)′={ , , }} c) (A∪B′)′={ }} d) A∩B′={ , , }}"

Part a

For this case we just need to find the elements in the universal set that are not in A. And we see that:

$A' = [0,2,6,8]$

And that represent the complement for A

Part b

For this case we need to find first the Union AUB who are the elements on A or B without repetition and we got:

$AUB = [1,3,4,5,7,8,9]$

And now the complement for (AUB)' are the elements that are not in AUB but are on the universal set and we got:

$(AUB)' = [0,2,6]$

Part c

For this case we need to find B' who are the elements on the universal set that are not in B

$B' = [0,2,3,5,6,9]$

Then we can find the union between AUB' and we got:

$AUB' = [0,1,2,3,4,5,6,7,9]$

And then the complment is just:

$(AUB')' = [9]$

Part d

For this case we just need to see the elements in common between A and B' and we got:

A∩B′ $= [3,5,9]$

6 0

4 years ago

What’s the value of x

Mekhanik [1.2K]

Answer:

Step-by-step explanation:

20-7x=6x-6

20+6=6x+7x

26=13x

x=2

5 0

3 years ago

Find the derivative of the algebraic function

Rama09 [41]

$f(x)=x\left(3-\dfrac{4}{x+7}\right)=3x-\dfrac{4x}{x+7}=\dfrac{3x(x+7)}{x+7}-\dfrac{4x}{x+7}=\\=\dfrac{3x^2+21x-4x}{x+7}=
\dfrac{3x^2+17x}{x+7}\\
f'(x)=\dfrac{(6x+17)(x+7)-(3x^2+17x)}{(x+7)^2}\\
f'(x)=\dfrac{6x^2+17x+42x+119-3x^2-17x}{(x+7)^2}\\
f'(x)=\dfrac{119+42x+3x^2}{(x+7)^2}\Rightarrow \text{A}$

7 0

4 years ago

Need to find the number for a

Shalnov [3]

Answer to the first triangle: 15

To solve this we will use the Pythagorean theorem

Pythagorean theorem formula: $a^{2} + b^{2} =c^{2}$

Lets do the first triangle.

17 is the hypotenuse, so its going to replace c in the formula.

8 is the height and it is going to replace a in the formula.

a is what we are trying to find, so our b in the equation is going to be the unknown.

$8^2 + b^2 = 17^2$

$64 + b^2 = 289

$

$b^2 = 289 - 64$

$b^2 = 225$

$\sqrt{b^2} = \sqrt{225}$

$b = 15$

now we know that the missing length in the first triangle (a) is 15

Can you solve the Other one?

Just comment if you need help on the other one.

4 0

4 years ago