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777dan777 [17]
3 years ago
7

The stores below each have the same snowboard for an original price of $189.59. At which store can you get the snowboard for the

lowest price? Store A: Sale of 30% off and a successive discount of 10% off. Store B: Sale of 20% off and a successive discount of 20% off. Store C: Sale of 40% off. a. All of the stores have the same discounted price. b. You can get the lowest price at Store A. c. You can get the lowest price at StoreB. d. You can get the lowest price at Store C.
Mathematics
1 answer:
Rama09 [41]3 years ago
7 0

Answer: D.

You can get the lowest price at Store C

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I'm rather stuck on this question
White raven [17]

Answer:

m<a = 30°

m<b = 90°

m<c = 60°

Step-by-step explanation:

Angles a, b, and c are all angles on a straight line.

Angles on a straight line add up to 180°.

Therefore, a + b + c = 180°

Given that a:b:c = 1:3:2, the measure of each can be calculated as follows:

Measure of each angle = the ratio proportion of that angle ÷ total ratio proportion * 180°

m<a = \frac{1}{6} * 180 = 30

m<b = \frac{3}{6} * 180 = 90

m<c = \frac{2}{6} * 180 = 60

8 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
I would like to know how to simplify these fractions with a logical reasoning
Andrei [34K]
The answer is 6 square root of 3.
4 times 12 square rooted minus 2 times square root of 3 = 10.39230485 =
6 square root of 3.

4 0
3 years ago
Find the complementary angle of the following <br>1)69°. 2)39°. 3)17°. 4)40°​
Luden [163]

Answer:

1.)21°

2.)51°

3.)73°

4.)50°

Step-by-step explanation:

1.) 90° - 69°

= 21°

2.) 90° - 39°

= 51°

3.) 90° - 17°

= 73°

4.) 90° - 40°

= 50°

3 0
3 years ago
Q plus 12 minus 2=????
Rashid [163]
Answer is: <span>Q+12-2 =<span> q+10</span></span>
4 0
4 years ago
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