2 years ago

Do all squares have two pairs of parallel sides?

Mathematics

2 answers:

vovangra [49]2 years ago

7 0

Yes. I need at least 20 characters for this to be an answer so I'm typing this to answer your question

Keith_Richards [23]2 years ago

4 0

Not all it depends what type of square

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62,591 + 7.1 X 10-2 in standard notations

quester [9]

Answer:

0.000071

Step-by-step explanation:

The given number that is given in scientific notation is, ............(1)

We need to write this number in standard notation. A number can be written in scientific notation as :

..........(2)

In standard notation,

So, is equivalent to . Hence, the correct option is (d)

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When working in trigonometry, you typically measure angles in

Neporo4naja [7]

Answer:

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Step-by-step explanation:

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2 years ago

Convert the fraction to higher terms. Enter the value of the missing numerator. 3/11= /77?

lianna [129]

the answer is 21/77 because you multiplied 11 by 7 to get 77, you multiply 7 by which gives you 21/77

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Ram is seven years old

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2 years ago

Find the value of x² - x+1 when x=√3 -1, in the form a+b√3 where a and b are integers

lidiya [134]

Answer:

$6-3\sqrt{3}$

If you compare this to $a+b\sqrt{3}$ , then $a=6$ and $b=-3$ .

Step-by-step explanation:

We are given $x=\sqrt{3}-1$ and we are asked to find the value of $x^2-x+1$ .

So we will need to find $x^2$ .

$x=\sqrt{3}-1$

Square both sides:

$x^2=(\sqrt{3}-1)^2$

Expand using identity: $(x-a)^2=x^2-2ax+a^2$ .

$x^2=(\sqrt{3})^2-2(\sqrt{3})(1)+1^2$

$x^2=3-2\sqrt{3}+1$

$x^2=4-2\sqrt{3}$

Let's go back to the full $x^2-x+1$ .

$x^2-x+1$

$(4-2\sqrt{3})-(\sqrt{3}-1)+1$

$4-2\sqrt{3}-\sqrt{3}+1+1$

$6-3\sqrt{3}$

If you compare this to $a+b\sqrt{3}$ , then $a=6$ and $b=-3$ .

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3 years ago