Answer:
yes it is I'm sure of it ok.
Answer:
7.7
Step-by-step explanation:
To solve this question we must use Pythagoras (a²+b²=c²) Let those two dots be A and B respectively.Imagine a line is drawn that pases through both dots,
Imagine that the line given is the hypotenuse (the longest side of a right-angled triangle) of a right-angled triangle and the other sides as a line from A going to the right, and a line from B going downwards, making the final vertexes for the triangle at coordinates: 0,-5 4,1 and 4,-5 (the vertex we made). now we find the lengths for the sides: 4 and 6 so we use these to plug them into pythagoras's equation:
4²+6²=c²
24+36=c²
√60=c
c=7.74....
But all we need is until the tenths so the answer is 7.7 (as the hundredths digit, 4, rounds down)
D because it is the most specific
![\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta) \\\\\\ tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\ -----------------------------\\\\ 2cos(A)=3tan(A)\implies 2cos(A)=3\cfrac{sin(A)}{cos(A)} \\\\\\ 2cos^2(A)=3sin(A)\implies 2[1-sin^2(A)]=3sin(A) \\\\\\ 2-2sin^2(A)=3sin(A)\implies 2sin^2(A)+3sin(A)-2](https://tex.z-dn.net/?f=%5Cbf%20sin%5E2%28%5Ctheta%29%2Bcos%5E2%28%5Ctheta%29%3D1%5Cimplies%20cos%5E2%28%5Ctheta%29%3D1-sin%5E2%28%5Ctheta%29%0A%5C%5C%5C%5C%5C%5C%0Atan%28%5Ctheta%29%3D%5Ccfrac%7Bsin%28%5Ctheta%29%7D%7Bcos%28%5Ctheta%29%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%0A2cos%28A%29%3D3tan%28A%29%5Cimplies%202cos%28A%29%3D3%5Ccfrac%7Bsin%28A%29%7D%7Bcos%28A%29%7D%0A%5C%5C%5C%5C%5C%5C%0A2cos%5E2%28A%29%3D3sin%28A%29%5Cimplies%202%5B1-sin%5E2%28A%29%5D%3D3sin%28A%29%0A%5C%5C%5C%5C%5C%5C%0A2-2sin%5E2%28A%29%3D3sin%28A%29%5Cimplies%202sin%5E2%28A%29%2B3sin%28A%29-2)
![\bf \\\\\\ 0=[2sin(A)-1][sin(A)+2]\implies \begin{cases} 0=2sin(A)-1\\ 1=2sin(A)\\ \frac{1}{2}=sin(A)\\\\ sin^{-1}\left( \frac{1}{2} \right)=\measuredangle A\\\\ \frac{\pi }{6},\frac{5\pi }{6}\\ ----------\\ 0=sin(A)+2\\ -2=sin(A) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5C%5C%5C%5C%5C%5C%0A0%3D%5B2sin%28A%29-1%5D%5Bsin%28A%29%2B2%5D%5Cimplies%20%0A%5Cbegin%7Bcases%7D%0A0%3D2sin%28A%29-1%5C%5C%0A1%3D2sin%28A%29%5C%5C%0A%5Cfrac%7B1%7D%7B2%7D%3Dsin%28A%29%5C%5C%5C%5C%0Asin%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%3D%5Cmeasuredangle%20A%5C%5C%5C%5C%0A%5Cfrac%7B%5Cpi%20%7D%7B6%7D%2C%5Cfrac%7B5%5Cpi%20%7D%7B6%7D%5C%5C%0A----------%5C%5C%0A0%3Dsin%28A%29%2B2%5C%5C%0A-2%3Dsin%28A%29%0A%5Cend%7Bcases%7D)
now, as far as the second case....well, sine of anything is within the range of -1 or 1, so -1 < sin(A) < 1
now, we have -2 = sin(A), which simply is out of range for a valid sine, so there's no angle with such sine
so, only the first case are the valid angles for A
