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n200080 [17]
3 years ago
8

Consider the sequence whose first five terms are shown. -59 ,-67 ,-65 ,-53 ,-31 ,... The first term is ____ To calculate the sec

ond term, add _____ To calculate each additional term, increase the amount added by _____ (the _____ are blanks)
Mathematics
1 answer:
GaryK [48]3 years ago
7 0

Step-by-step explanation:

The sequence is as follows :

-59 ,-67 ,-65 ,-53 ,-31 ,...

The first term = -59

The second term = -67

Common difference = second term - first term

=-67-(-59)\\\\=-8

To calculate the second term add (-8).

To calculate each additional term, increase the amount added by -8.

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Answer:

80

Step-by-step explanation:

25+55=80

5 0
3 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]
<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.

3 0
4 years ago
The trapezoid ABCD has angle C=70 degrees and D=50 degrees. Find the size of angle A and angle B
uranmaximum [27]

Angle A = 130° and Angle B = 110°

Solution:

Given ABCD is a trapezoid with ∠C = 70° and ∠D = 50°

If ABCD is a trapezoid, then AB is parallel to CD.

AD is a transversal to AB and CD and

BC is a tranversal to AB ad CD.

Sum of the interior angles on the same side are supplementary.

∠A + ∠D = 180°

⇒ ∠A + 50° = 180°

Subtract 50° on both sides to equal the expression.

⇒ ∠A = 180° – 50°

⇒ ∠A = 130°

Similary, ∠B + ∠C = 180°

⇒ ∠B + 70° = 180°

Subtract 50° on both sides to equal the expression.

⇒ ∠B = 180° – 70°

⇒ ∠B = 110°

Hence, angle A = 130° and angle B = 110°.

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