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DedPeter [7]
3 years ago
8

What is larger? -1 or -4/3

Mathematics
2 answers:
ryzh [129]3 years ago
6 0

Answer: -1 is ;larger

IgorC [24]3 years ago
3 0

Answer:

-4/3

Step-by-step explanation:

-4/3 is equal to -1 1/3 which is greater than just -1

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The boundary lines for the system of inequalities is given in the graph.
diamong [38]

Answer:

C)

region C

Step-by-step explanation:

We have to use what is called the zero-interval test [test point] in order to figure out which portion of the graph these inequalities share:

\displaystyle y ≤ -x + 2

0 ≤ 2 ☑ [We shade the portion of the graph that CONTAIN THE ORIGIN, which is the bottom portion.]

\displaystyle y ≥ 2x - 3

0 ≥ −3 ☑ [We shade the portion of the graph that CONTAINS THE ORIGIN, which is the left side.]

So, now that we got that all cleared up, we can tell that both graphs share a region where the ORIGIN IS VISIBLE. Therefore region C matches the above inequalities.

I am joyous to assist you anytime.

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−1/5 ÷ 35 ?<br><br> I know this is easy. :)
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Answer:

-1/175

Step-by-step explanation:


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6x4x8x5=<br> that is the
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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
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<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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