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2 years ago

Suppose that Sin(θ)=5/13 and cos(θ)=-12/13. Find the value of tan(θ), cot (θ), CSC (θ), and sec (θ). Find the value of tan(2θ)

Mathematics

1 answer:

horrorfan [7]2 years ago

5 0

Answer:

Because θ lies in quadrant II, 2θ must lie in quadrant IV. This means the tangent of 2θ is negative.

The adjacent side to θ is 7 because √(25²-24²)=7, so tanθ=7/24.

The double angle formula for tangent is tan 2θ = (2 tan θ) / (1 − tan² θ).

Substituting the value for tanθ in and keeping in mind that this is in quadrant IV, we get tan 2θ = -(2(7/24)/(1-(7/24)²)).

Simplified, this becomes tan 2θ = -336/527.

Therefore, the answer is C. -336/527.

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3 years ago

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Craig earns extra money as a lifeguard. He earns $37 for 4 hours and $64.75 for 7 hours. Explain how you know whether the relati

nordsb [41]

Answer:No it is not proportional

Step-by-step explanation:

4 hours=$37

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3 years ago

Can someone please find the volume of this shape and show work for how you did it??

castortr0y [4]

Answer:

$221.4m^{3}$

Step-by-step explanation:

To work out the volume of this shape you will first need to work out the area of the triangle. To work out the area of the triangle you would multiply the base of 12 by the height of 4.1, which gives you 49.2. You would then need to divide 49.2 by 2, which gives you 24.6. This is because the formula for the area of a triangle is the same as that of a square, however a triangle is half of a square with the same length and width. To work out the volume you would multiply the area of the triangle, which is 24.6 by the height of 9, which gives you 221.4 $m^{3}$

1) Multiply 12 by 4.1.

$12*4.1=49.2$

2) Divide 49.2 by 2.

$49.2/2=24.6$

3) Multiply 24.6 by 9.

$24.6*9=221.4m^{3}$

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3 years ago

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Bess [88]

Answer:

Its 9

Step-by-step explanation:

I did this in 5th grade... All you gotta do is count and there's 4 in the bottom aka 2 rows of 4 and there's 1 block on the top its 9

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3 years ago

I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

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3 years ago