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2 years ago

Which trigonometric functions are negative in the fourth (IV) quadrant?

Mathematics

1 answer:

anzhelika [568]2 years ago

7 0

<h3>Answers are: sine, tangent, cosecant, cotangent</h3>

Explanation:

On the unit circle we have some point (x,y) such that x = cos(theta) and y = sin(theta). The sine corresponds to the y coordinate of the point on the circle. Quadrant IV is below the x axis which explains why sine is negative here, since y < 0 here.

Since sine is negative, so is cosecant as this is the reciprocal of sine

csc = 1/sin

In quadrant IV, cosine is positive as x > 0 here. So the ratio tan = sin/cos is going to be negative. We have a negative over a positive when we divide.

Because tangent is negative, so is cotangent.

The only positive functions in Q4 are cosine and secant, which is because sec = 1/cos.

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No.

Step-by-step explanation:

No, because 23 > 10 + 10.

If any side is equal or greater than the sum of the other 2 sides no triangle can be drawn.

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Step-by-step explanation:not b

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2 years ago

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that

Helga [31]

Answer:

$\sum^\infty_{n=0} -5 (\frac{x+2}{2})^n$

Step-by-step explanation:

Rn(x) →0

f(x) = 10/x

a = -2

Taylor series for the function <em>f </em>at the number a is:

$f(x) = \sum^\infty_{n=0} \frac{f^{(n)}(a)}{n!} (x - a)^n$

$f(x) = f(a) + \frac{f'(a)}{1!}(x-a)+\frac{f"(a)}{2!} (x-a)^2 + ...$ ............ equation (1)

Now we will find the function <em>f </em> and all derivatives of the function <em>f</em> at a = -2

f(x) = 10/x f(-2) = 10/-2

f'(x) = -10/x² f'(-2) = -10/(-2)²

f"(x) = -10.2/x³ f"(-2) = -10.2/(-2)³

f"'(x) = -10.2.3/x⁴ f'"(-2) = -10.2.3/(-2)⁴

f""(x) = -10.2.3.4/x⁵ f""(-2) = -10.2.3.4/(-2)⁵

∴ The Taylor series for the function <em>f</em> at a = -4 means that we substitute the value of each function into equation (1)

So, we get $\sum^\infty_{n=0} - \frac{10(x+2)^n}{2^{n+1}}$ Or $\sum^\infty_{n=0} -5 (\frac{x+2}{2})^n$

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3 years ago

A thermometer reading 7 °C is brought into a room with a constant temperature of 34°C. If the thermometer reads 14 °C after 4 m

never [62]

For this case we are going to define the following variable:
x: time in minutes
We write the linear function that represents the problem:
t (x) = (14/4) x + 7
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t (6) = (14/4) * (6) + 7
t (6) = 28 ° C
For x = 11 we have:
t (11) = (14/4) * (11) + 7
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Answer:
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3 years ago

Eva sees a dolphin 3.23.23, point, 2 meters below sea level and a bird \dfrac{47}{10} 10 47 start fraction, 47, divided by, 10

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Answer:

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Step-by-step explanation:

Given

Dolphin = 3.2

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Required

What is the vertical distance between both

Using a vertical scale;

Represent the bird "above sea level" position with positive and dolphin at "below sea level" position with negative.

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Distance = +[Above The Sea] - [Below The Sea]

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Distance = 47/10 - 3.2

Hence, the vertical distance between both is 47/10 - 3.2

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