Answer:
680
Step-by-step explanation:
Based on the given conditions, formulate: 17 × ( 23 + 17 )
Calculate the sum or difference: 17 × 40
Calculate the product or quotient: 680
get the result: 680
Answer: 680
(Hope this helps can I pls have brainlist (crown)☺️)
Answer: x=13
Step 1: Define isosceles triangle
In an isosceles triangle, two sides and angles are equal. The equal sides will be the legs.
Step 2: Write an equation
To write an equation we must first consider the information discussed in step 1. Since we know that the legs will be equal, we will add that to the equation twice. The base and legs will be on one side of the equation, and the perimeter will be on the other. Now we can write the equation!
x+3x-7+3x-7=77
Step 3: Combine like terms
Like terms are terms that share the same variable, or lack of. Let’s add/subtract these now.
x+3x-7+3x-7=77
7x-14=77
Step 4: Solve for x
Let’s do the last part of the equation and solve for x, our final answer.
*Rewrite equation*
7x-14=77
*Add 14 to both sides*
7x=91
*Divide 7 on both sides*
x=13
This is your answer. Hope this helps! Comment below for more questions.
the frequency of the sinusoidal graph is 2 in 2 π interval
Step-by-step explanation:
The frequency of the graphs refers to the number of the cycles, the graph completes in a given fixed interval.
We already know the formula that
P= (1/ F)
Thus, F= (1/ P)
Where F= frequency and P= Period
Period is the horizontal length (x- axis component) of one complete cycle.
Thus, Observing the above graph
We find that the graph completes 1 cycle in π interval and 2 cycles in 2π interval
Thus, the frequency of the sinusoidal graph is 2 in 2 π interval
Answer:
10000 meters
Step-by-step explanation:
10 km is 10000 meters
Answer:
Remember, a basis for the row space of a matrix A is the set of rows different of zero of the echelon form of A.
We need to find the echelon form of the matrix augmented matrix of the system A2x=b2
![B=\left[\begin{array}{cccc}1&2&3&1\\4&5&6&1\\7&8&9&1\\3&2&4&1\\6&5&4&1\\9&8&7&1\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%263%261%5C%5C4%265%266%261%5C%5C7%268%269%261%5C%5C3%262%264%261%5C%5C6%265%264%261%5C%5C9%268%267%261%5Cend%7Barray%7D%5Cright%5D)
We apply row operations:
1.
- To row 2 we subtract row 1, 4 times.
- To row 3 we subtract row 1, 7 times.
- To row 4 we subtract row 1, 3 times.
- To row 5 we subtract row 1, 6 times.
- To row 6 we subtract row 1, 9 times.
We obtain the matrix
![\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&-6&-12&-6\\0&-4&-5&-2\\0&-7&-14&-5\\0&-10&-20&-8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%263%261%5C%5C0%26-3%26-6%26-3%5C%5C0%26-6%26-12%26-6%5C%5C0%26-4%26-5%26-2%5C%5C0%26-7%26-14%26-5%5C%5C0%26-10%26-20%26-8%5Cend%7Barray%7D%5Cright%5D)
2.
- We subtract row two twice to row three of the previous matrix.
- we subtract 4/3 from row two to row 4.
- we subtract 7/3 from row two to row 5.
- we subtract 10/3 from row two to row 6.
We obtain the matrix
![\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&0&0\\0&0&3&2\\0&0&0&2\\0&0&0&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%263%261%5C%5C0%26-3%26-6%26-3%5C%5C0%260%260%260%5C%5C0%260%263%262%5C%5C0%260%260%262%5C%5C0%260%260%262%5Cend%7Barray%7D%5Cright%5D)
3.
we exchange rows three and four of the previous matrix and obtain the echelon form of the augmented matrix.
![\left[\begin{array}{cccc}1&2&3&1\\0&-3&-6&-3\\0&0&3&2\\0&0&0&0\\0&0&0&2\\0&0&0&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%263%261%5C%5C0%26-3%26-6%26-3%5C%5C0%260%263%262%5C%5C0%260%260%260%5C%5C0%260%260%262%5C%5C0%260%260%262%5Cend%7Barray%7D%5Cright%5D)
Since the only nonzero rows of the augmented matrix of the coefficient matrix are the first three, then the set
![\{\left[\begin{array}{c}1\\2\\3\end{array}\right],\left[\begin{array}{c}0\\-3\\-6\end{array}\right],\left[\begin{array}{c}0\\0\\3\end{array}\right] \}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C2%5C%5C3%5Cend%7Barray%7D%5Cright%5D%2C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C-3%5C%5C-6%5Cend%7Barray%7D%5Cright%5D%2C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C3%5Cend%7Barray%7D%5Cright%5D%20%5C%7D)
is a basis for Row (A2)
Now, observe that the last two rows of the echelon form of the augmented matrix have the last coordinate different of zero. Then, the system is inconsistent. This means that the system has no solutions.
