Cross multiply and solve
1(y^2-9)=6(y-3)
y^2-9= 6y-18
y^2-6y-9+18=0
y^2-6y+9=0
(y-3)(y-3)=0
(y-3)=0
y=3
When you plug y in however, it makes the denominator 0, making both fractions undefined.
Final answer: No real solution
Answer:
Esmeralda has enough music left in her playlist for a 30 minute workout
Step-by-step explanation:
we know that
If Esmeralda accidentally deleted 25% of the music, then the percentage of music left in her playlist equals 100%-25%=75%.

Find out how many minutes represent 75% of 40 minutes
Multiply 40 by 0.75

therefore
Esmeralda has enough music left in her playlist for a 30 minute workout
Answer:
b=2
c=-24.01
Step-by-step explanation:
If I understood your question right you need to find the constants b and c in the polynomial:

such that:

In order to get a system of two equations you can find limes when x goes to 4 from the right side or from the left side.
I solved this using a number that is close to 4. When it goes from right I used 4.1 and from left 3.9. This way you get two equations:

When you solve the system you get the answers above.
Answer:
![\left[\begin{array}{ccc}3&0&0\\0&3&0\\0&0&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%260%260%5C%5C0%263%260%5C%5C0%260%263%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
In order to find out the resulting matrix, we will have to multiply the identity matric and the scalar 3:
The 3x3 identity matrix is:
![\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
Multiplying with scalar 3:
![3\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=3%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
The scalar will be multiplied by each element of the matrix.
Multiplying zeros with scalar 3 will give us zero. So the resulting matrix will be:
![\left[\begin{array}{ccc}3*1&0&0\\0&3*1&0\\0&0&3*1\end{array}\right] = \left[\begin{array}{ccc}3&0&0\\0&3&0\\0&0&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%2A1%260%260%5C%5C0%263%2A1%260%5C%5C0%260%263%2A1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%260%260%5C%5C0%263%260%5C%5C0%260%263%5Cend%7Barray%7D%5Cright%5D)
So the resultant matrix will be a scalar matrix with 3 at diagonal positions..
Answer:
no, it is not the answer is(2,-2)