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AVprozaik [17]
3 years ago
5

Pleaseeeeee Simplify t12/16

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
3 0
6/8 then 3/4 so your answer is 3/4 cause you can simply by 2
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Can I get some help pls!!!
kramer

Answer:

Step-by-step explanation:

7 0
3 years ago
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NO LINKS!!! Part 1: y = x^3<br><br>Shape:<br>Domain:<br>Range: <br>Locater point:​
LiRa [457]

Answer:

Shape: <em>Not sure what to put</em>

Domain: ( -oo, oo )

Range: ( -oo, oo )

Locater point: ( 0, 0 )

Step-by-step explanation:

Shape - You have the shape on the graph

Domain and Range - Both go on forever since there is no hole/interruption

Locater Point: I would say origin because it is the closest thing to a vertex, which is an example of a locater point.

Hopefully this helps!

Brainliest please?

5 0
3 years ago
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Root 3 cosec140° - sec140°=4<br>prove that<br><br>​
Lorico [155]

Answer:

Step-by-step explanation:

We are to show that \sqrt{3} cosec140^{0} - sec140^{0} = 4\\

<u>Proof:</u>

From trigonometry identity;

cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}

\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\

From trigonometry, 2sinAcosA = Sin2A

= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\=  \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\=  \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2  \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

\frac{4(sin(420-140)) }{sin280}

= \frac{4sin280}{sin280}\\ = 4 \ Proved!

3 0
3 years ago
Solve and how to solve
Gennadij [26K]
You have to use cosine law...

5 0
3 years ago
The table below shows how much it costs per gallon to fill a gas tank with premium
Leni [432]

47.50/10 = 4.75

57/12 = 4.75

76/16 = 4.75

The unit rate is C. $4.75

6 0
3 years ago
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