Answer:
19.1 feet
Step-by-step explanation:
This forms a right triangle
use the pythagorean theorem
a^2 + b^2 = c^2
a^2 + 6^2 = 20^2
a^2 + 36 = 400
a^2 = 364
a = 19.0787840283
Rounded
19.1 feet
Hello !
Here are some rules for logarithms : log ₐ b = n ⇔ aⁿ = b
log ₐ aⁿ = n
log ₐ (b·c) = logₐ b + log ₐ c
log ₐ bⁿ = n· log ₐ b ____________________________________________________
25=5²
625=5⁴
100 = 5² · 2²
____________________________________________________
(log₅ 25 + log₅ 625) / log₅100 =
(log₅ 5²+ log₅ 5⁴) / log₅ (5² · 2²) =
(log₅ 5²+ log₅ 5⁴) /( log₅ 5² + log₅ 2²) =
(2 + 4)/(2+log₅ 2² ) =
6/ (2+2·log₅2)=
6/2 + 6/ (2·log₅2) =
3+ 3/log₅2
log₅2 = 0.5
So 3+ 3/log₅2 = 3 + 3/0.5 =3 + 6 = 9
Answer : 9
The angle for shooting laser gun should be 89.98°.
The vertical height where a tiny but horrible alien is standing is given as 443443443 meters. The horizontal distance of a men in black agent from building is given in the question as 181818 meters. So, let us assume the angle formed to be theta (θ).
Writing the formula for tan theta, that relates the given height and distance
Now, tan θ = vertical height ÷ horizontal distance
Keep the values in above mentioned equation of theta to find the angle.
tan θ = 
Performing division to find the tan theta and subsequently calculate the angle
tan θ = 2438.94
θ = tan⁻¹ 2438.94
Calculating tan inverse to find the value of theta
θ = 89.98°
Therefore, the angle for shooting laser gun by a men in black agent should be 89.98°.
Learn more about calculation of angle -
brainly.com/question/21865829
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Triangularizing matrix gives the matrix that has only zeroes above or below the main diagonal. To find which option is correct we need to calculate all of them.
In all these options we calculate result and write it into row that is first mentioned:
A)R1-R3
![\left[\begin{array}{ccc}-1&0&0|0\\0&1&1|6\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%260%260%7C0%5C%5C0%261%261%7C6%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
B)2R2-R3
![\left[\begin{array}{ccc}1&0&1|1\\-2&2&1|4\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%7C1%5C%5C-2%262%261%7C4%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
C)-2R1+R3
![\left[\begin{array}{ccc}0&0&-1|-1\\0&1&1|6\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%26-1%7C-1%5C%5C0%261%261%7C6%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
D)2R1+R3
![\left[\begin{array}{ccc}4&0&3|3\\0&1&1|6\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%260%263%7C3%5C%5C0%261%261%7C6%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
E)3R1+R3
![\left[\begin{array}{ccc}5&0&4|4\\0&1&1|6\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%260%264%7C4%5C%5C0%261%261%7C6%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
None of the options will triangularize this matrix. The only way to <span>triangularize this matrix is
R3-2R1
</span>
![\left[\begin{array}{ccc}1&0&1|1\\0&1&1|6\\0&0&-1|-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%7C1%5C%5C0%261%261%7C6%5C%5C0%260%26-1%7C-1%5Cend%7Barray%7D%5Cright%5D%20%20)
<span>
This equation is similar to C) but in reverse order. Order in which rows are written is important.</span>