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daser333 [38]
3 years ago
12

Why was it important for Athens to have a good navy?

Mathematics
2 answers:
vredina [299]3 years ago
8 0

Answer:

this would be answer d

owu when minecraft talks to me its like mr blue horse please tell me why you you lead all the creepers to my Mario world 1. he go brrr on a stick is a dog that wants to be a bottle at night with the sun up and the bird went for a walk in his cage while sleeping so i opened a book and it played a video

Step-by-step explanation:

zhuklara [117]3 years ago
4 0

Answer:

the other person is correct

Step-by-step explanation:

D

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In the diagram below, O is circumscribed about quadrilateral ABCD. what is the value of x?​
cestrela7 [59]

Answer:

104

Step-by-step explanation:

cuz it iz

7 0
3 years ago
Read 2 more answers
Calculate the following limit:
aleksklad [387]
\lim_{x\to\infty}\dfrac{\sqrt x}{\sqrt{x+\sqrt{x+\sqrt x}}}=\\
\lim_{x\to\infty}\dfrac{\dfrac{\sqrt x}{\sqrt x}}{\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt x}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{\dfrac{x+\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{\sqrt{x^2}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{x+\sqrt x}{x^2}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{\sqrt x}{\sqrt{x^4}}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{x}{x^4}}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{1}{x^3}}}}}=\\
=\dfrac{1}{\sqrt{1+\sqrt{0+\sqrt{0}}}}=\\
=\dfrac{1}{\sqrt{1+0}}=\\
=\dfrac{1}{\sqrt{1}}=\\
=\dfrac{1}{1}=\\
1


8 0
3 years ago
John is making himself a lunch. He has 3 different soups to choose from, 4 kinds of pop, and 3 different kinds of fruit. How man
notsponge [240]
36 combinations im pretty sure 3*4*3=36
5 0
3 years ago
Find the absolute maximum and absolute minimum values, if any, of the function. (If an answer does not exist, enter DNE.)f(x) =
lyudmila [28]

Answer:

The absolute minimum value is "-\frac{21}{4}" and the absolute maximum value is "15".

Step-by-step explanation:

Given:

f(x)=x^2-x-5

on,

[0,5]

By differentiating it, we get

⇒ f'(x)=2x-1

Set f'(x)=0

then,

⇒ 2x-1=0

          2x=1

            x=\frac{1}{2} (Critical point)

When x=0,

⇒ f(x)=-5

When x=\frac{1}{2},

⇒ f(x)=-\frac{21}{4} (Absolute minimum)

When x=5

⇒ f(x)=15 (Absolute maximum)

7 0
3 years ago
The figure shows a parallelogram inside a rectangle outline:
Tamiku [17]
B.1/6 

i think that's the answer <span />
4 0
3 years ago
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