Step-by-step explanation:
The shape of quadrilateral ABCD changes as the moving platform swings around, but its side lengths do not change. Both pairs of opposite sides are congruent, so ABCD is a parallelogram by the Parallelogram Opposite Sides Converse. By the definition of a parallelogram, — AB — DC .
Answer:
-3.872983346
Step-by-step explanation:
In this case we have: cos θ = 1/4 Whose possible solutions are: θ = 2 * pi * n - Acos (1/4) θ = 2 * pi * n + Acos (1/4) Where, n belongs to the natural numbers. As sin θ <0 then the solution is: θ = 2 * pi * n - Acos (1/4) For n = 1 we get: θ = 4.965069236 radians Thus: tan θ = tan (4.965069236) = - 3.872983346 Answer: tan θ = -3.872983346
Answer:
By Pythagoras,
\displaystyle{r}=\sqrt{{{x}^{2}+{y}^{2}}}r=x2+y2 \displaystyle=\sqrt{{{\left(-{2}\right)}^{2}+{3}^{2}}}=(−2)2+32 \displaystyle=\sqrt{{{4}+{9}}}=\sqrt{{13}}=4+9=13
For this example, we define the trigonometric ratios for θ in the following way:
\displaystyle \sin{\theta}=\frac{y}{{r}}=\frac{3}{\sqrt{{13}}}={0.83205}sinθ=ry=133=0.83205
\displaystyle \cos{\theta}=\frac{x}{{r}}=\frac{{-{2}}}{\sqrt{{13}}}=-{0.55470}cosθ=rx=13−2=−0.55470
\displaystyle \tan{\theta}=\frac{y}{{x}}=\frac{3}{ -{{2}}}=-{1.5}tanθ=xy=−23=−1.5
\displaystyle \csc{\theta}=\frac{r}{{y}}=\frac{\sqrt{{13}}}{{3}}={1.2019}cscθ=yr=313=1.2019
\displaystyle \sec{\theta}=\frac{r}{{x}}=\frac{\sqrt{{13}}}{ -{{2}}}=-{1.80278}secθ=xr=−213=−1.80278
\displaystyle \cot{\theta}=\frac{x}{{y}}=\frac{{-{2}}}{{3}}=-{0.6667}cotθ=yx=3−2=−0.6667
<span>More please help me solve the problem
The stars have different brightness. The brightest stars are Grade 1 and the least degree luimnoase stars 6. The brightness of the stars shrinks 2.5 times with the passing from one grade to another. Whenever the stars are brighter Grade 1, Grade 6 than the stars?</span>
Answer:
she has a 1/4 of the property remaining
Step-by-step explanation: