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Taya2010 [7]
4 years ago
8

How do you construct a square?

Mathematics
2 answers:
BARSIC [14]4 years ago
8 0

Answer:

You have to measure the sides and make sure they are the same and you shape them to look like a square

Step-by-step explanation:

amid [387]4 years ago
7 0

Answer:

Step-by-step explanation: First draw a line to any direction then another then another and then another and then connect them to together.

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Which of the following numbers is a solutikn for inequality below k >3
Leya [2.2K]

Answer:

Step-by-step explanation:

It is any number that is 3 or less

Hope that helps! :)

8 0
3 years ago
Read 2 more answers
What is cos(37pi)?<br> A. 1<br> B. 0<br> C. -1<br> D. -2
Svetllana [295]
37pi = 116.238
Cos(116.238) = -1
7 0
3 years ago
How do you do this problem?
icang [17]

Step One

Put the right side over a common denominator.


\dfrac{a(x+ 1)}{*(x + 1)} - \dfrac{bx}{x(x + 1)} =  \dfrac{ax+a - bx}{x(x + 1)} = \dfrac{ax - bx + a}{x(x + 1)}


Step Two

Remove the brackets on the numerator in the last equality.


Step three.

This will seem kind of odd but take out the common factor (x) in the first two terms of the numerator.

\dfrac{x(a - b) + a}{x(x + 1)}


Step Four

What you do next comes up very frequently in your second or third calculus course.


You have to notice that there are no xs in the numerators on the left side of the original expression


Therefore you have to make a - b = 0 so that the xs disappear on the right.

a - b =0

a = b


Step Five

Find a

a is all by itself on the numerator of the final result.

The original question has a 1 in the numerator. Therefore a = 1


Step Six

Solve for b

a = b

1 = b Answer



7 0
3 years ago
Andy's Cab Service charges a $6 fee plus $0.50 per mile. His twin brother Randy starts a rival business
Mazyrski [523]
What are you asking for? Better deal? Equation? What?
7 0
3 years ago
What happens when you have a square root on top of another square root?
worty [1.4K]
Just put the numbers under one square root symbol and try to simplify.
3 0
3 years ago
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