According to the secant-tangent theorem, we have the following expression:
Now, we solve for <em>x</em>.
Then, we use the quadratic formula:
![x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Where a = 1, b = 6, and c = -315.
![\begin{gathered} x_{1,2}=\frac{-6\pm\sqrt[]{6^2-4\cdot1\cdot(-315)}}{2\cdot1} \\ x_{1,2}=\frac{-6\pm\sqrt[]{36+1260}}{2}=\frac{-6\pm\sqrt[]{1296}}{2} \\ x_{1,2}=\frac{-6\pm36}{2} \\ x_1=\frac{-6+36}{2}=\frac{30}{2}=15 \\ x_2=\frac{-6-36}{2}=\frac{-42}{2}=-21 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_%7B1%2C2%7D%3D%5Cfrac%7B-6%5Cpm%5Csqrt%5B%5D%7B6%5E2-4%5Ccdot1%5Ccdot%28-315%29%7D%7D%7B2%5Ccdot1%7D%20%5C%5C%20x_%7B1%2C2%7D%3D%5Cfrac%7B-6%5Cpm%5Csqrt%5B%5D%7B36%2B1260%7D%7D%7B2%7D%3D%5Cfrac%7B-6%5Cpm%5Csqrt%5B%5D%7B1296%7D%7D%7B2%7D%20%5C%5C%20x_%7B1%2C2%7D%3D%5Cfrac%7B-6%5Cpm36%7D%7B2%7D%20%5C%5C%20x_1%3D%5Cfrac%7B-6%2B36%7D%7B2%7D%3D%5Cfrac%7B30%7D%7B2%7D%3D15%20%5C%5C%20x_2%3D%5Cfrac%7B-6-36%7D%7B2%7D%3D%5Cfrac%7B-42%7D%7B2%7D%3D-21%20%5Cend%7Bgathered%7D)
<h2>Hence, the answer is 15 because lengths can't be negative.</h2>
You can use the identity
cos(x)² +sin(x)² = 1
to find sin(x) from cos(x) or vice versa.
(1/4)² +sin(x)² = 1
sin(x)² = 1 - 1/16
sin(x) = ±(√15)/4
Then the tangent can be computed as the ratio of sine to cosine.
tan(x) = sin(x)/cos(x) = (±(√15)/4)/(1/4)
tan(x) = ±√15
There are two possible answers.
In the first quadrant:
sin(x) = (√15)/4
tan(x) = √15
In the fourth quadrant:
sin(x) = -(√15)/4
tan(x) = -√15
Answer:
89
Step-by-step explanation:
it is basically counting by 2s but rather by 20s.
20, 40,60,80
all you do is add a 9 at the end.
The height of the rocket is found in terms of the angle as
.. h/(3 mi) = tan(θ)
.. h = (3 mi)*tan(θ)
Then the rate of change of height (vertical velocity) is
.. h' = (3 mi)*sec(θ)^2*θ'
.. h' = (3 mi)*4*(1.5 rad/min)
.. h' = 18 mi/min
The rocket's velocity is 18 miles per minute at that moment.
