The problem presents 2 variables and 2 conditions to follow to determine the approach in solving the problem. The variables are 52 cards, and 9 cards. The 2 conditions presented would be the teacher giving out one card to each student at a time to each student until all of them are gone. The second variable is more likely made as a clue and the important variable that gives away the approach to be used. The approach to be used is division. This is to ensure that there will be students receiving the 9 cards. Thus, we do it as this: 52 / 9 = ?
The answer would be 5.77778 (wherein 7 after the decimal point is infinite and 8 would just be the rounded of number). This would ensure us that there will be 5 students that can receive 9 cards but there will be 7 cards remaining which goes to the last student, which is supposed to be 8 since she gives one card to each student at a time to each student. So the correct answer would be just 4 students. The fifth student will only receive 8 cards and the last student would have 8, too.
Answer:
<h2>30</h2>
Step-by-step explanation:
![3x+4y^2\\\\\text{Substitute the value of x = -2 and y = 3 to the expression:}\\\\3(-2)+4(3^2)\qquad\text{use PEMDAS}\\\\=-6+4(9)=-6+36=30](https://tex.z-dn.net/?f=3x%2B4y%5E2%5C%5C%5C%5C%5Ctext%7BSubstitute%20the%20value%20of%20x%20%3D%20-2%20and%20y%20%3D%203%20to%20the%20expression%3A%7D%5C%5C%5C%5C3%28-2%29%2B4%283%5E2%29%5Cqquad%5Ctext%7Buse%20PEMDAS%7D%5C%5C%5C%5C%3D-6%2B4%289%29%3D-6%2B36%3D30)
The greatest common factor is 28.
The factors of 56 are: 1, 2, 4, 7, 8, 14, 28, 56
The factors of 84 are: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Then the greatest common factor is 28.
Answer:
![sin\theta_1 = - \frac{2\sqrt{70}}{19}](https://tex.z-dn.net/?f=sin%5Ctheta_1%20%3D%20%20-%20%5Cfrac%7B2%5Csqrt%7B70%7D%7D%7B19%7D)
Step-by-step explanation:
We are given that
is in <em>fourth</em> quadrant.
is always positive in 4th quadrant and
is always negative in 4th quadrant.
Also, we know the following identity about
and
:
![sin^2\theta + cos^2\theta = 1](https://tex.z-dn.net/?f=sin%5E2%5Ctheta%20%2B%20cos%5E2%5Ctheta%20%3D%201)
Using \theta_1 in place of \theta:
![sin^2\theta_1 + cos^2\theta_1 = 1](https://tex.z-dn.net/?f=sin%5E2%5Ctheta_1%20%2B%20cos%5E2%5Ctheta_1%20%3D%201)
We are given that ![cos\theta_1 = \frac{9}{19}](https://tex.z-dn.net/?f=cos%5Ctheta_1%20%3D%20%5Cfrac%7B9%7D%7B19%7D)
![\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}](https://tex.z-dn.net/?f=%5CRightarrow%20sin%5E2%5Ctheta_1%20%2B%20%5Cdfrac%7B9%5E2%7D%7B19%5E2%7D%20%3D%201%5C%5C%5CRightarrow%20sin%5E2%5Ctheta_1%20%3D%201%20-%20%5Cdfrac%7B81%7D%7B361%7D%5C%5C%5CRightarrow%20sin%5E2%5Ctheta_1%20%3D%20%20%5Cdfrac%7B361-81%7D%7B361%7D%5C%5C%5CRightarrow%20sin%5E2%5Ctheta_1%20%3D%20%20%5Cdfrac%7B280%7D%7B361%7D%5C%5C%5CRightarrow%20sin%5Ctheta_1%20%3D%20%20%5Csqrt%7B%5Cdfrac%7B280%7D%7B361%7D%7D%5C%5C%5CRightarrow%20sin%5Ctheta_1%20%3D%20%20%2B%5Cdfrac%7B2%5Csqrt%7B70%7D%7D%7B19%7D%2C%20-%5Cdfrac%7B2%5Csqrt%7B70%7D%7D%7B19%7D)
is in <em>4th quadrant </em>so
is negative.
So, value of ![sin\theta_1 = - \frac{2\sqrt{70}}{19}](https://tex.z-dn.net/?f=sin%5Ctheta_1%20%3D%20%20-%20%5Cfrac%7B2%5Csqrt%7B70%7D%7D%7B19%7D)
By flipping the fraction of -3/7, you will get the fraction of -7/3 or - 2 1/3.
Therefore, the reciprocal of -3/7 is - 2 1/3.
Hope this helps!