All categories

3 years ago

If cos omega = 3 5 find tan omega

Mathematics

2 answers:

Citrus2011 [14]3 years ago

4 0

Answer:

mmm 30

Step-by-step explanation:

Andrew [12]3 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

Differentiate with respect to x and simplify your answer. Show all the appropriate steps? 1.e^-2xlog(ln x)^3 2.e^-2x(log(ln x))^

serious [3.7K]

(1) I assume "log" on its own refers to the base-10 logarithm.

$\left(e^{-2x}\log(\ln x)^3\right)'=\left(e^{-2x}\right)'\log(\ln x)^3+e^{-2x}\left(\log(\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{e^{-2x}}{\ln10(\ln x)^3}\left((\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10(\ln x)^3}\left(\ln x\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10\,x(\ln x)^3}$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}}{\ln10\,x\ln x}$

Note that writing $\log(\ln x)^3=3\log(\ln x)$ is one way to avoid using the power rule.

(2)

$\left(e^{-2x}(\log(\ln x))^3\right)'=(e^{-2x})'(\log(\ln x))^3+e^{-2x}\left(\log(\ln x))^3\right)'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2(\log(\ln x))'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2\dfrac{(\ln x)'}{\ln10\,\ln x}$

$=-2e^{-2x}(\log(\ln x))^3+\dfrac{3e^{-2x}(\log(\ln x))^2}{\ln10\,x\ln x}$

(3)

$\left(\sin(xe^x)^3\right)'=\left(\sin(x^3e^{3x})\right)'=\cos(x^3e^{3x}(x^3e^{3x})'$

$=\cos(x^3e^{3x})((x^3)'e^{3x}+x^3(e^{3x})')$

$=\cos(x^3e^{3x})(3x^2e^{3x}+3x^3e^{3x})$

$=3x^2e^{3x}(1+x)\cos(x^3e^{3x})$

(4)

$\left(\sin^3(xe^x)\right)'=3\sin^2(xe^x)\left(\sin(xe^x)\right)'$

$=3\sin^2(xe^x)\cos(xe^x)(xe^x)'$

$=3\sin^2(xe^x)\cos(xe^x)(x'e^x+x(e^x)')$

$=3\sin^2(xe^x)\cos(xe^x)(e^x+xe^x)$

$=3e^x(1+x)\sin^2(xe^x)\cos(xe^x)$

(5) Use implicit differentiation here.

$(\ln(xy))'=(e^{2y})'$

$\dfrac{(xy)'}{xy}=2e^{2y}y'$

$\dfrac{x'y+xy'}{xy}=2e^{2y}y'$

$y+xy'=2xye^{2y}y'$

$y=(2xye^{2y}-x)y'$

$y'=\dfrac y{2xye^{2y}-x}$

8 0

3 years ago

3_3/4 divided by 2 please?

belka [17]

Answer:

Step-by-step explanation:

4/2 is 2. lol

5 0

4 years ago

Read 2 more answers

Please help with 1 question thank you.

frosja888 [35]

Let's replace the first, second, third, and home with letters.
first base = F
second base = S
Third base = T
home base = H
We have two triangles, ΔHTS and ΔHFS
We know HS = HS because it's a shared side.
We know FS = TS
We know that ∠HST and ∠HSF are equal.
Because of SAS, the two triangles are congruent.

According to CPCTC, TH must be congruent to FH.
Thus, the distance from home plate to third base and the distance from home plate to first base must be the same.

Have an awesome day! :)

7 0

4 years ago

How many solutions exist for 0.75(x+40)=0.35(x+20)+0.35(x+20)

ExtremeBDS [4]

Hey there since its first degree polynomial only one

6 0

3 years ago

Read 2 more answers

Find the point that partitions CD in the ratio 1:2 c(-1,7) and d(5, 1)

Amiraneli [1.4K]

Coordinates of the point that partitions the line into the given ratio is; (1, 5)

<h3>How to divide a line segment?</h3>

If the ratio dividing a line segment represents a:b, then the formula to divide the line segment is;

Coordinates = {(bx₁ + ax₂)/(a + b), (by₁ + ay₂)/(a + b)}

From our question, we have the ratio as;

1:2. Thus; a = 1 and b = 2

Our coordinate of CD is; C(-1,7) and D(5, 1)

Thus;

x₁ = -1

x₂ = 5

y₁ = 7

y₂ = 1

Coordinates of the point that partitions the line into the given ratio is;

Coordinate = {(2*-1 + 1*5)/(1 + 2), (2*7 + 1*1)/(1 + 2)}

Coordinate = (3/3, 15/3) = (1, 5)

Read more about Division of Line Segment at; brainly.com/question/17374569

#SPJ1

3 0

2 years ago