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Alex73 [517]
3 years ago
6

In the coordinate plane, ABCD has vertices with coordinates A(1,-1), B(-5,3), C(-3,6),

Mathematics
1 answer:
Nana76 [90]3 years ago
6 0

Answer:

C

Step-by-step explanation:

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Which statement can be used to prove that a given parallelogram is a rectangle?
RSB [31]

The <em>correct answer</em> is:


The diagonals of the parallelogram are congruent.


Explanation:


In every parallelogram, opposite angles are congruent. This would not mean it is a rectangle.


Consecutive sides of a parallelogram are only congruent if the parallelogram is a rhombus or a square; this would not be a rectangle.


The diagonals of every parallelogram bisect each other. This would not mean it is a rectangle.


The diagonals of a rectangle bisect each other. If we know this is true about our parallelogram, this means our parallelogram is a rectangle.

7 0
3 years ago
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PLEASE HURRY PLEASE PLEASE PLEASE
hram777 [196]

Answer:

b

Step-by-step explanation:

make a formula, its the easiest way to solve problems like these also its very efficent

3 0
3 years ago
1/4 as a fraction and a decimal
storchak [24]

Answer: 1/4 fraction .25 decimal

Hope this helps :)

8 0
3 years ago
X + Y = 3,500 X - Y = 2,342
LUCKY_DIMON [66]
X + y = 3500 -------------- (1)
x - y = 2342 --------------- (2)

(1) + (2):
2x = 5842
x = 2921 -------------- Sub into (1)
x+ y = 3500
2921 + y = 3500
y = 3500 - 2921
y = 579

----------------------------------------------
Answer: x = 2921, y = 579
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8 0
3 years ago
Consider the differential equation
Gennadij [26K]

The first solution is quadratic, so its derivative y' on the left side is linear. But the right side would be a polynomial of degree greater than 1, so this is not the correct choice.

The third solution has a similar issue. The derivative of √(x² + 1) will be another expression involving √(x² + 1) on the left side, yet on the right we have y² = x² + 1, so that the entire right side is a polynomial. But polynomials are free of rational powers, so this solution can't work.

This leaves us with the second choice. Recall that

1 + tan²(x) = sec²(x)

and the derivative of tangent,

(tan(x))' = sec²(x)

Also notice that the ODE contains 1 + y². Now, if y = tan(x³/3 + 2), then

y' = sec²(x³/3 + 2) • x²

and substituting y and y' into the ODE gives

sec²(x³/3 + 2) • x² = x² (1 + tan²(x³/3 + 2))

x² sec²(x³/3 + 2) = x² sec²(x³/3 + 2)

which is an identity.

So the solution is y = tan(x³/3 + 2).

4 0
2 years ago
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