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VARVARA [1.3K]
3 years ago
9

NEED HELP ASAP!!!! please !!!!

Mathematics
1 answer:
tekilochka [14]3 years ago
8 0
I would help you but there isn’t any questions or picture of the problem you need help with ????
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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]
<span>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 </span>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 <span>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) </span>cos \theta=\frac{\overline{OD}}{\overline{OB}}=\frac{\overline{OD}}{r}=\frac{\overline{OD}}{1}=\overline{OD}
<span>
(3) </span>tan \theta=\frac{\overline{FD}}{\overline{OC}}=\frac{\overline{FC}}{r}=\frac{\overline{FC}}{1}=\overline{FC}
<span>
Triangles </span>ΔOFC and ΔOBD are similar, so it is true that:

\frac{\overline{FC}}{\overline{OF}}= \frac{\overline{BD}}{\overline{OB}}<span>

</span>∴ \overline{OF}= \frac{\overline{FC}}{\overline{BD}}= \frac{tan \theta}{sin \theta}= \frac{1}{cos \theta} \rightarrow \boxed{sec \theta= \frac{1}{cos \theta}}<span>

b.1 </span>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 <span>Use the representation of tangent as a clue for what to show for cotangent. 
</span>
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.

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4 years ago
The time spent studying, in hours, on a daily basis of 9 different students was used to create the box plot shown above. Which o
Alexeev081 [22]

Answer:

Step-by-step explanation:

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3 years ago
Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock resistance. The results from 100 disks are sum
mrs_skeptik [129]

Answer:

Part a

The probability that the event disk has high shock resistance is 0.86

Part b

The probability that a disk has high scratch resistance given that the disk has high shock resistance is 0.8140

Step-by-step explanation:

(a).

From the given information,

Let A denote the event that a disk has high shock resistance,

and let B denote the event that a disk has high scratch resistance.

                          SHOCK HIGH(A)     SHOCK LOW(A)          TOTAL

SCRATCH  

HIGH(B)               70                               9                                79

RESISTANCE       16                               5                                  21

LOW(B)

TOTAL                86                                 14                                100

Compute P(A).

Therefore, the probability value of the event A is 0.86.

Part a

The probability that the event disk has high shock resistance is 0.86

Explanation | Hint for next step

Based on the given information, the probability that the event disk has high shock resistance is 0.86. That means it is approximately equal to 86%.

Step 2 of 2

(b)

From the given information,

Let A denote the event that a disk has high shock resistance,

and let B denote the event that a disk has high scratch resistance.

 

                      SHOCK HIGH(A)     SHOCK LOW(A)            TOTAL

SCRATCH

HIGH(B)             70                               9                               79

SCRATCH

LOW(B)             16                                 5                               21

TOTAL             86                                15                               100

ComputeP(B/A) = P(A∩B) /P(A)  ;P(A) > 0

P(A∩B)  =70/100

​  =0.70

​  

From the part [a], the probability value of the event A is P(A) =0.86 .

Therefore,

P( {B/A}  = P(A∩B)  /P(A)

​=  0.70 /0.86

​  

=0.8140

​  

Part b

The probability that a disk has high scratch resistance given that the disk has high shock resistance is 0.8140

Explanation | Common mistakes

The disk with high scratch resistance is found to be 81.40% with the condition that the disk with high shock resistance is maintained.

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3 years ago
Find the volume of the cone
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Answer:

132::

Step-by-step explanation:

The power of g00gle <3

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Math question! Please help!!
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3 years ago
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