Answer:
21 girls.
Step-by-step explanation:
50% of 42 students are girls.
So 42/2 = 21.
Therefore 21 of the students are girls.
<em>Hope I helped</em>
Answer:
The correct option is parallelogram ABCD is a rhombus, because the diagonal bisects two angles
Step-by-step explanation:
In triangle ABD:
∠B = ∠D
Thus AB=AC by the property of opposite sides of equal angles are equal
In triangle CBD
∠B = ∠D
Thus CB=CD by the property of opposite sides of equal angles are equal
Thus all four sides of quadrilateral ABCD are equal
And diagonal BD bisects the angles
So, it is a rhombus
Therefore the correct option is parallelogram ABCD is a rhombus, because the diagonal bisects two angles....
Each bus was filled with 54 students
Answer:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Let's find the answer.
Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.
Equations:
Eq. 1 : x1 + 2x2 + 5x3 = 5
Eq. 2 : x1 + x2 + x3 = 6
E1. 3 : 4x1 + 6x2 + 5x3 = 7
Coefficients for x1 ; x2 ; x3
From eq. 1 : 1 ; 2 ; 5
From eq. 2 : 1 ; 1 ; 1
From eq. 3 : 4 ; 6 ; 5
So matrix A is:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D)
And the vector of vriables (X) is:
![\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D)
Now we can find the resulting vector (B) using the 'resulting values' from each equation:
![\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
In conclusion, AX=B is:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
9514 1404 393
Answer:
sum = ∑[n=1,5] 4^(n-4)
Step-by-step explanation:
First of all, you need to be able to describe the n-th term.
Here, we have ...
term #: 1, 2, 3, 4, 5
value: 4^-3, 4^-2, 4^-1, 4^0, 4^1
That is, the exponent of 4 is 4 less than the term number. So, the n-th term is 4^(n-4). The sum of the 5 terms shown is then ...
