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Zinaida [17]
2 years ago
7

What is a triangle with three congruent sides?

Mathematics
1 answer:
Marta_Voda [28]2 years ago
5 0
A triangle with three congruent sides is an equilateral triangle. An equilateral triangle's all sides are always equal to each other, and the angles are also congruent.
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Which statement is true about this question -4(2p+5)+8p=-11
aleksandrvk [35]
The answer to this problem is “C “
6 0
3 years ago
Why are (1,2) and(2,1) not equal ordered pair? ​
vladimir1956 [14]

Answer:

The pairs ( 1 , 2 ) and ( 2 , 1 ) are not equal because their respective elements are not equal.

Step-by-step explanation:

The pair of elements which are in specific order is called an ordered pair. The pair ( 1 , 2 ) is not same as the pair ( 2 , 1 ). In the pair ( 1 , 2 ) 1 is in the first position and 2 is in the second position. In the pair ( 2 , 1 ), 2 is in the first position and 1 is in the second position.

Two ordered pairs ( a , b ) and (c , d ) are said to be equal if a = c and b = d. We write ( a , b ) = ( c , d ).

Hope I helped!

Best regards!!

8 0
3 years ago
How to convert a non-terminating number into a fraction
AleksandrR [38]
<span>divide the numerator by the denominator to find its decimal equivalent.</span>
6 0
3 years ago
where does the graph of the function y=tan(x) have asymptotes? at the values of x where cos(x)=0 at the values of x where sin(x)
Gnom [1K]
Answer: at the values where cos(x) = 0

Justification:

1) tan(x) = sin(x) / cos(x).

2) functions have vertical asymptotes at x = a if Limit of the function x approaches a is + or - infinity.

3) the limit of tan(x) approaches +/- infinity where cos(x) approaches 0.

Therefore, the grpah of y = tan(x) has asymptotes where cos(x) = 0.

You can see the asympotes at x = +/- π/2 on the attached graph. Remember that cos(x) approaches 0 when x approaches +/- (n+1) π/2, for any n ∈ N, so there are infinite asymptotes.



7 0
3 years ago
Read 2 more answers
What is the solution to the equation -3d/a^2-2d-8 + 3/d-4 = -2/ d+2
Alex73 [517]

Answer:

d=1

Step-by-step explanation:

\frac{-3d}{d^2-2d-8} +\frac{3}{d-4} =\frac{-2}{d+2}

Lets factor the denominator d^2 -2d-8

d^2 - 2d - 8 = (d-4)(d+2)

\frac{-3d}{(d-4)(d+2)} +\frac{3}{d-4} =\frac{-2}{d+2}

Now make the denominators same

LCD: (d-4)(d+2)

\frac{-3d}{(d-4)(d+2)} +\frac{3(d+2)}{(d-4)(d+2)} =\frac{-2(d-4)}{(d+2)(d-4)}

Denominators are same on both sides

So equate the numerators

-3d +3(d+2) = -2(d-4)

-3d +3d +6 = -2d +8

6 = -2d + 8

subtract 8 on both sides

-2 = -2d

So d=1




3 0
2 years ago
Read 2 more answers
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