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3 years ago

How many sides does a polygon have?

2 answers:

6 0

Hello!

A polygon has as many angles as it has sides. For example, a triangle has 3 sides and 3 angles. A pentagon has 5 sides and 5 angles. An octadecagon has 18 sides and 18 angles!

Blizzard [7]3 years ago

4 0

A polygon has at least 3 sides, it has to have 3 or more sides in order for it to be a polygon

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In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= root 2 cm.

Nata [24]

Given CD is an altitude such that AD=BC , AB=3 cm and CD= √2 cm.

Let AD=x, Since given AB=3

AD+DB=3

x+DB = 3

DB = 3-x

Since ΔBCD is rght angle triangle, let's apply Pythagoras theorem

$BC^{2} = DB^{2} +CD^{2}$

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

$BC^{2} =(3-x)^{2} +2$

Since given AD=BC,let us plugin BC=x in above step.

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

$x^{2} =9-6x+x^{2} +2$

6x=11

x= $\frac{11}{6}$

Now we know AD=x= $\frac{11}{6}$ and given CD=√2.

Let us apply Pythagoras theorem for ΔACD

$AC^{2} =AD^{2} +DC^{2}$

$AC^{2} = (\frac{11}{6} )^{2} +(\sqrt{2} )^{2}$

$AC^{2} =\frac{193}{36}$

$AC=\sqrt{\frac{193}{36} }$ = 2.315cm

3 0

3 years ago

How long does a person going 10 mph (16 kph ) to go .6 miles

Olegator [25]

the answer will be 5

8 0

3 years ago

0.361, 0.35, 1/5, 0.36 greatest to least

bixtya [17]

.361 .36 .35 1/5

1/5 = .20

6 0

3 years ago

Read 2 more answers

Based on the diagram shown, find θ to the nearest degree.

sattari [20]

Answer:

θ = 38°

Step-by-step explanation:

The lower right triangle is congruent to the upper left triangle, so we have θ and 20° being the two acute angles in the triangle. The law of sines tells you ...

sin(θ)/9 = sin(20°)/5

sin(θ) = (9/5)sin(20°)

θ = arcsin(9/5·sin(20°)) ≈ 38°

___

Another solution to the triangle is θ = 180° -38° = 142°. The diagram clearly shows θ as an acute angle, so we take this second solution to be extraneous.

8 0

3 years ago

At most, how many unique roots will a third-degree polynomial function have?

beks73 [17]

The fundamental theorem of algebra states that a polynomial with degree n has at most n solutions. The "at most" depends on the fact that the solutions might not all be real number.

In fact, if you use complex number, then a polynomial with degree n has exactly n roots.

So, in particular, a third-degree polynomial can have at most 3 roots.

In fact, in general, if the polynomial $p(x)$ has solutions $x_1,\ x_2,\ldots x_n$ , then you can factor it as

$p(x) = (x-x_1)(x-x_2)\ldots (x-x_n)$

So, a third-degree polynomial can't have 4 (or more) solutions, because otherwise you could write it as

$p(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)$

But this is a fourth-degree polynomial.

7 0

3 years ago