Which number can be classified as a terminating decimal number?

Help please !~ I’ll give brainliest! Any help is appreciated :)

jeka94

In most approximate way it can never be the area of the circle (Accurate area)

This is why because after creating base lines of each triangle it forms an inscribed polygon .

Some little spaces left over each side or b_n

So

h is already approaching to r

and b_1+b_2..b_n is approaching to the perimeter i.e 2πr

Option B

5 0

2 years ago

Please help me how to solve and what it asks me to do

Anna007 [38]

Higher Order Thinking problems require you to think about the facts that are given to you in any problem and do something with those facts. This involves understanding them, infer from them, connect them to other facts and concepts, categorize them, manipulate them, put them together and apply them to find new solutions.

For example the first problem which asks to evaluate the expression 0.1(3.7). This can also mean that 1/10*3.7 = 0.37.

I hope it helps, Regards.

6 0

4 years ago

The diagram shows how cos θ, sin θ, and tan θ relate to the unit circle. Copy the diagram and show how sec θ, csc θ, and cot θ r

DIA [1.3K]

Copy the diagram and show how sec θ, csc θ, and cot θ relate to the unit circle.

The representation of the diagram is shown if Figure 1. There's a relationship between sec θ, csc θ, and cot θ related the unit circle. Lines green, blue and pink show the relationship.

a.1 First, find in the diagram a segment whose length is sec θ.

The segment whose length is sec θ is shown in Figure 2, this length is the segment $\overline{OF}$ , that is, the line in green.

a.2 Explain why its length is sec θ.

We know these relationships:

(1) $sin \theta=\frac{\overline{BD}}{\overline{OB}}=\frac{\overline{BD}}{r}=\frac{\overline{BD}}{1}=\overline{BD}$

(2) $cos \theta=\frac{\overline{OD}}{\overline{OB}}=\frac{\overline{OD}}{r}=\frac{\overline{OD}}{1}=\overline{OD}$

(3) $tan \theta=\frac{\overline{FD}}{\overline{OC}}=\frac{\overline{FC}}{r}=\frac{\overline{FC}}{1}=\overline{FC}$

Triangles ΔOFC and ΔOBD are similar, so it is true that:

$\frac{\overline{FC}}{\overline{OF}}= \frac{\overline{BD}}{\overline{OB}}$ 

∴ $\overline{OF}= \frac{\overline{FC}}{\overline{BD}}= \frac{tan \theta}{sin \theta}= \frac{1}{cos \theta} \rightarrow \boxed{sec \theta= \frac{1}{cos \theta}}$ 

b.1 Next, find cot θ

The segment whose length is cot θ is shown in Figure 3, this length is the segment $\overline{AR}$ , that is, the line in pink.

b.2 Use the representation of tangent as a clue for what to show for cotangent.

It's true that:

$\frac{\overline{OS}}{\overline{OC}}= \frac{\overline{SR}}{\overline{FC}}$

But:

$\overline{SR}=\overline{OA}$
$\overline{OS}=\overline{AR}$

Then:

$\overline{AR}= \frac{1}{\overline{FC}}= \frac{1}{tan\theta} \rightarrow \boxed{cot \theta= \frac{1}{tan \theta}}$

b.3 Justify your claim for cot θ.

As shown in Figure 3, θ is an internal angle and ∠A = 90°, therefore ΔOAR is a right angle, so it is true that:

$cot \theta= \frac{\overline{AR}}{\overline{OA}}=\frac{\overline{AR}}{r}=\frac{\overline{AR}}{1} \rightarrow \boxed{cot \theta=\overline{AR}}$

c. find csc θ in your diagram.

The segment whose length is csc θ is shown in Figure 4, this length is the segment $\overline{OR}$ , that is, the line in green.