Hi!
Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 30: 1, 2, 3, 5, 6, 10, 30
Explanation: <u><em>This question is super super easy. This question should be the least common multiples. First find the prime factorization of 12. It gave us 2*2*3=12. Then the prime factorization of 30. It gave us 2*3*5=30. You can also multiply each factors the greater number of times and it occurs to in steps 1 or step 2 and it gave us LCM=2*2*3*5=60. The least common multiple factors is 60 is the right answer. Hope this helps! And thank you for posting your question at here on Brainly. And have a great day. -Charlie</em></u>
Answer:
99 points
Step-by-step explanation:
Average of a data a is the sum of the data divided by the number of data.
Average = sum of data ÷ number of data
A = T/n .....1
For the average of 5 test to be 90.
The total sum of score must be;
Number of tests n = 5
Average A = 90
Substituting into equation 1
90 = T/5
T = 90×5 = 450 points
The total score must be equal to 450 point to have an average of 90 points
The sum of the first four test is;
88+84+88+91 = 351
Let x represent the first score,
351 + x = 450
x = 450 - 351
x = 99 points
The minimum score she can get in the fifth test is 99 points
Collection or percolation
We are given Elena’s bedroom door's width = 0.8 m.
Also the scale drawing is in the ratio of 1 to 50 that is 1/50.
<em>In order to find the width of scale drawing, we need to multiply original width of the door by 1/50.</em>
If we multiply 0.8 by 1/50, we get
0.8 × 1/50 = 0.8/50 = 0.016 meter.
So, we can say 0.016 meter wide should the door be on the scale drawing, if the ratio is 1 to 50.
![\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
