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sasho [114]
3 years ago
13

Six times a number plus another number tripled

Mathematics
2 answers:
crimeas [40]3 years ago
8 0

Step-by-step explanation:

Hello there!

This question is ambiguous.

I'm assuming you want to write an equation for the given expression.

These are just a few of the ways that we can write it:

3(6n+x)

6n+3x

:)

charle [14.2K]3 years ago
4 0

Answer:

(6x2)+(3x3)=

Step-by-step explanation:

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Find four rational number between 1/4 and 2/3.
olya-2409 [2.1K]

Answer:

4/12, 5/12, 6/12, 7/12

Step-by-step explanation:

1/4 x 3/3 = 3/12

2/3 x 4/4 = 8/12

between 3/12 and 8/12

4/12, 5/12, 6/12, 7/12

you can simplify these if you wish

Hope that helped!!! k

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Step-by-step explanation:

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Car A: 3x + 6

Car B: 4x+8

Step-by-step explanation:

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2 years ago
In 2000 the population of a country was 4,580,000 by 2015 the population had increased by 18%
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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
2 years ago
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