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2 years ago

A. Construct a parallelogram PQRS with PQ =20cm, PS = 8cm, and

Mathematics

1 answer:

Romashka-Z-Leto [24]2 years ago

5 0

Answer:

18cm

Step-by-step explanation:

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A circle has a diameter with endpoints (1,6) and (-3, -6). Find its area.

mamaluj [8]

Answer:

(1,6) i think it is $\lim_{n \to \infty} a_n \geq x^{2} x^{2} x^{2} \neq \neq$

Step-by-step explanation:

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3 years ago

An angle measures 79.6° less than the measure of its complementary angle. What is the measure of each angle?

wel

Answer:

y = 259.6/2 = 129.8 degrees. so x = 129.8 - 79.6 = 50.2 degrees.

Step-by-step explanation:

I got this answer from somewhere else. but it is correct. i hope this helped.

8 0

3 years ago

<, =, or >? -9/2 ? -4 1/2

astra-53 [7]

Less than right? Because it’s a negative?

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3 years ago

Find the 61st term of -12, -28, -44,

Citrus2011 [14]

Answer:

$a_{61}=-972$

Step-by-step explanation:

<u>Arithmetic Sequences</u>

The arithmetic sequences are identified because any term n is obtained by adding or subtracting a fixed number to the previous term. That number is called the common difference.

The equation to calculate the nth term of an arithmetic sequence is:

$a_n=a_1+(n-1)r$

Where

an = nth term

a1 = first term

r = common difference

n = number of the term

We are given the first terms of a sequence:

-12, -28, -44,...

Find the common difference by subtracting consecutive terms:

r = -28 - (-12) = -16

r = -44 - (-28) = -16

The first term is a1 = -12. Now we calculate the term n=61:

$a_{61}=-12+(61-1)(-16)$

$a_{61}=-12-60*16=-12-960$

$\mathbf{a_{61}=-972}$

6 0

2 years ago

ZE is the angle bisector of measure YEX and the perpendicular bisector of GF, GX is the angle bisector of measure YGZ and the pe

den301095 [7]

Answer:

Step-by-step explanation:

The center of inscribed circle into triangle is point of intersection of all interior angles of triangle.

The center of circumscribed circle over triabgle is point of intersection of perpendicular bisectors to the sides.

Circumscribed circle always passes through the vertices of the triangle.

Inscribed circle is always tangent to the triangle's sides.

In your case angles' bisectors and perpendicular bisectors intesect at one point, so point A is the center of inscribed circle and the center of corcumsribed circle. Thus, these circles pass through the points X, Y, Z and G, E, F, respectively.

7 0

3 years ago