Tgα = tan(α) = sqrt(2)/1 = opposite/ adjacent
opposite = sqrt(2)
adjacent = 1
hypotenuse = sqrt(opposite^2 + adjacent^2)
= sqrt(sqrt(2)^2 + 1^2) = sqrt(3)
sin(α) = opposite/hypotenuse
= sqrt(2)/sqrt(3)
cos(α) = adjacent/hypotenuse
= 1/sqrt(3)
now we can calculate the value of
5sin²α+3cos²α / 3sin²α-5cos²α
... do the math
Answer:
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Step-by-step explanation:
zuzuch bzhzhs bu sizhshsuhsnd uh s gsv sjizhs
Answer:
you are correct the answer is c
Answer:
The answer is "9 and 7".
Step-by-step explanation:
Given:
![A=\left[\begin{array}{cc}7&9\\0&9\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%269%5C%5C0%269%5Cend%7Barray%7D%5Cright%5D)
Using formula:
![\to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]-\lambda \left[\begin{array}{cc}1&0\\0&1\end{array}\right] |=0\\\\\\ \to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]- \left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right] |=0\\\\\\\to|\left[\begin{array}{cc}7-\lambda &9\\0&9-\lambda\end{array}\right]|=0\\\\\\\to|(7-\lambda)(9-\lambda)|=0\\\\\to (7-\lambda)(9-\lambda)=0\\\\\to 7-\lambda=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9-\lambda=0\\\\](https://tex.z-dn.net/?f=%5Cto%20%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%269%5C%5C0%269%5Cend%7Barray%7D%5Cright%5D-%5Clambda%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%7C%3D0%5C%5C%5C%5C%5C%5C%20%5Cto%20%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%269%5C%5C0%269%5Cend%7Barray%7D%5Cright%5D-%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Clambda%260%5C%5C0%26%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%7C%3D0%5C%5C%5C%5C%5C%5C%5Cto%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7-%5Clambda%20%269%5C%5C0%269-%5Clambda%5Cend%7Barray%7D%5Cright%5D%7C%3D0%5C%5C%5C%5C%5C%5C%5Cto%7C%287-%5Clambda%29%289-%5Clambda%29%7C%3D0%5C%5C%5C%5C%5Cto%20%287-%5Clambda%29%289-%5Clambda%29%3D0%5C%5C%5C%5C%5Cto%207-%5Clambda%3D0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%209-%5Clambda%3D0%5C%5C%5C%5C)
Answer:
a) 96m
b) 2 m/s^2
Step-by-step explanation:
The distance travelled in a velocity time graph as shown in the image is the area of the shape. As such, considering the first 14 seconds, the distance covered will be computed by finding the area of the shape.
Note that the shape to be considered ends t the 14 seconds mark. This shape is a trapezium. The area of a trapezium is given as
A = 1/2(a + b) h
where a and b are the parallel sides of the trapezium and h is the height As such, the distance covered in the first 14 seconds
= 1/2 (14 + 10) * 8
= 4 * 24
= 96m
The acceleration after 4 seconds
Acceleration is the rate of change of velocity with time, at time 0 second, the velocity is also 0 but at time 4 seconds, the velocity is 8 m/s hence, the acceleration after 4 seconds
= 8/4
= 2 m/s^2
