Answer: B. 350.
Step-by-step explanation:
Given: Total number of fluorescent colors = 7
Total number of pastel colors = 5
Since , he want three fluorescent colors and two pastel colors.
Number of ways to select r things out of n things = ![^nC_r=\dfrac{n!}{r!(n-r)!}](https://tex.z-dn.net/?f=%5EnC_r%3D%5Cdfrac%7Bn%21%7D%7Br%21%28n-r%29%21%7D)
Number of ways to choose colors = ![^7C_3\times\ ^5C_2](https://tex.z-dn.net/?f=%5E7C_3%5Ctimes%5C%20%5E5C_2)
![=\dfrac{7!}{3!4!}\times\dfrac{5!}{2!3!}\\\\=\dfrac{7\times6\times5\times4!}{3\times2\times1\times4!}\times\dfrac{5\times4\times3!}{2\times3!}\\\\=350](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B7%21%7D%7B3%214%21%7D%5Ctimes%5Cdfrac%7B5%21%7D%7B2%213%21%7D%5C%5C%5C%5C%3D%5Cdfrac%7B7%5Ctimes6%5Ctimes5%5Ctimes4%21%7D%7B3%5Ctimes2%5Ctimes1%5Ctimes4%21%7D%5Ctimes%5Cdfrac%7B5%5Ctimes4%5Ctimes3%21%7D%7B2%5Ctimes3%21%7D%5C%5C%5C%5C%3D350)
Hence, the correct option is B. 350.
Answer:
? = 13
Step-by-step explanation:
Using Pythagoras' identity in the right triangle
?² = 5² + 12² = 25 + 144 = 169 ( take the square root of both sides )
? =
= 13
Answer:
x = -5, y = -6, z = -3
Step-by-step explanation:
Given the system of three equations:
![\left\{\begin{array}{l}x-2y+3z=-2\\6x+2y+2z=-48\\x+4y+3z=-38\end{array}\right.](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7Dx-2y%2B3z%3D-2%5C%5C6x%2B2y%2B2z%3D-48%5C%5Cx%2B4y%2B3z%3D-38%5Cend%7Barray%7D%5Cright.)
Write the augmented matrix for the system of equations
![\left(\begin{array}{ccccc}1&-2&3&|&-2\\6&2&2&|&-48\\1&4&3&|&-38\end{array}\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Cbegin%7Barray%7D%7Bccccc%7D1%26-2%263%26%7C%26-2%5C%5C6%262%262%26%7C%26-48%5C%5C1%264%263%26%7C%26-38%5Cend%7Barray%7D%5Cright%29)
Find the reduced row-echelon form of the augmented matrix for the system of equations:
![\left(\begin{array}{ccccc}1&-2&3&|&-2\\6&2&2&|&-48\\1&4&3&|&-38\end{array}\right)\sim \left(\begin{array}{ccccc}1&-2&3&|&-2\\0&-14&16&|&36\\0&-6&0&|&36\end{array}\right)\sim \left(\begin{array}{ccccc}1&3&-2&|&-2\\0&16&-14&|&36\\0&0&-6&|&36\end{array}\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Cbegin%7Barray%7D%7Bccccc%7D1%26-2%263%26%7C%26-2%5C%5C6%262%262%26%7C%26-48%5C%5C1%264%263%26%7C%26-38%5Cend%7Barray%7D%5Cright%29%5Csim%20%5Cleft%28%5Cbegin%7Barray%7D%7Bccccc%7D1%26-2%263%26%7C%26-2%5C%5C0%26-14%2616%26%7C%2636%5C%5C0%26-6%260%26%7C%2636%5Cend%7Barray%7D%5Cright%29%5Csim%20%5Cleft%28%5Cbegin%7Barray%7D%7Bccccc%7D1%263%26-2%26%7C%26-2%5C%5C0%2616%26-14%26%7C%2636%5C%5C0%260%26-6%26%7C%2636%5Cend%7Barray%7D%5Cright%29)
Thus, the system of three equations is
![\left\{\begin{array}{r}x+3z-2y=-2\\16z-14y=36\\-6y=36\end{array}\right.](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Br%7Dx%2B3z-2y%3D-2%5C%5C16z-14y%3D36%5C%5C-6y%3D36%5Cend%7Barray%7D%5Cright.)
From the last equation:
![y=-6](https://tex.z-dn.net/?f=y%3D-6)
Substitute it into the second equation:
![16z-14\cdot (-6)=36\\ \\16z=36-84\\ \\16z=-48\\ \\z=-3](https://tex.z-dn.net/?f=16z-14%5Ccdot%20%28-6%29%3D36%5C%5C%20%5C%5C16z%3D36-84%5C%5C%20%5C%5C16z%3D-48%5C%5C%20%5C%5Cz%3D-3)
Substitute y = -6 and z = -3 into the first equation:
![x+3\cdot (-3)-2\cdot (-6)=-2\\ \\x=-2+9-12\\ \\x=-5](https://tex.z-dn.net/?f=x%2B3%5Ccdot%20%28-3%29-2%5Ccdot%20%28-6%29%3D-2%5C%5C%20%5C%5Cx%3D-2%2B9-12%5C%5C%20%5C%5Cx%3D-5)