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ollegr [7]
3 years ago
5

Cell Phone Plus and Phones R Us offer special prices for new customers. Each cell phone company’s advertisement is shown below.

Mathematics
1 answer:
AVprozaik [17]3 years ago
3 0
2 months, 50+15x =60+10x when x=2
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P=s1+s2+s3 solve for s3
tresset_1 [31]

P=s1+s2+s3

Move +s1 to the other side. Sign changes from +s1 to -s1

P-s1=s1-s1+s2+s3

p-s1=s2+s3

Move s2 to the other side. Sign changes from s2 to -s2

p-s1-s2=s2-s2+s3

p-s1-s2=s3

Answer: p-s1-s2=s3

6 0
3 years ago
What is t if 3/5=2/3t.
Dvinal [7]
<span><span>т = <span><span>9/10

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6 0
3 years ago
PLEASE PLEASE HELP ASAP WILL MARK BRAINLIEST
Lelu [443]

Answer:

The answer is 216.

Step-by-step explanation:

4 0
3 years ago
This figure is 14 of a circle. What is the best approximation for the perimeter of the figure? use 3.14 to aproximate pi.
Lana71 [14]
I'm pretty sure its 3.1 inches
4 0
3 years ago
This is finding exact values of sin theta/2 and tan theta/2. I’m really confused and now don’t have a clue on how to do this, pl
Lostsunrise [7]

First,

tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>)

and given that 90° < <em>θ </em>< 180°, meaning <em>θ</em> lies in the second quadrant, we know that cos(<em>θ</em>) < 0. (We also then know the sign of sin(<em>θ</em>), but that won't be important.)

Dividing each part of the inequality by 2 tells us that 45° < <em>θ</em>/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(<em>θ</em>/2) > 0 and sin(<em>θ</em>/2) > 0.

Now recall the half-angle identities,

cos²(<em>θ</em>/2) = (1 + cos(<em>θ</em>)) / 2

sin²(<em>θ</em>/2) = (1 - cos(<em>θ</em>)) / 2

and taking the positive square roots, we have

cos(<em>θ</em>/2) = √[(1 + cos(<em>θ</em>)) / 2]

sin(<em>θ</em>/2) = √[(1 - cos(<em>θ</em>)) / 2]

Then

tan(<em>θ</em>/2) = sin(<em>θ</em>/2) / cos(<em>θ</em>/2) = √[(1 - cos(<em>θ</em>)) / (1 + cos(<em>θ</em>))]

Notice how we don't need sin(<em>θ</em>) ?

Now, recall the Pythagorean identity:

cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1

Dividing both sides by cos²(<em>θ</em>) gives

1 + tan²(<em>θ</em>) = 1/cos²(<em>θ</em>)

We know cos(<em>θ</em>) is negative, so solve for cos²(<em>θ</em>) and take the negative square root.

cos²(<em>θ</em>) = 1/(1 + tan²(<em>θ</em>))

cos(<em>θ</em>) = - 1/√[1 + tan²(<em>θ</em>)]

Plug in tan(<em>θ</em>) = - 12/5 and solve for cos(<em>θ</em>) :

cos(<em>θ</em>) = - 1/√[1 + (-12/5)²] = - 5/13

Finally, solve for sin(<em>θ</em>/2) and tan(<em>θ</em>/2) :

sin(<em>θ</em>/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)

tan(<em>θ</em>/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2

3 0
3 years ago
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