No it can’t because a square is not a circle and their is no way for it to become a circle
The range is the highest number minus the lowest number
98 - 13 = 85 <==
Answer:
Original number 26.
Step-by-step explanation:
xy - two-digit number
1) x + y = 8
2) Original two-digit number can be written as
10*x + y
3) If the digits interchanged yx,
then the new number can be written as
10*y + x
4) Double the original number is
2*(10*x + y)
5) New number is 10 more than double the original number
(10*y + x) - (2*(10*x + y)) = 10
6) Now we have the system of 2 equations:
x + y = 8
(10*y + x) - (2*(10*x + y)) = 10 -----> 10y + x - (20x + 2y) = 10 ---> 8y - 19x = 10
x = 8 - y
8y - 19(8 - y) = 10
8y - 152 +19y = 10
27y = 162
y = 6
x = 8 - y = 8 - 6 = 2
x = 2
So, x =2, y = 6.
Original number 26.
Check:
Original number 26.
New number 62.
Double of the original number = 2*26= 52.
New number is 10 more than double the original number :
62 - 52 = 10 True
The given statement is proved by side-angle-side (SAS) theorem.
Yes, if two sides and the included angle of one triangle are congruent to the corresponding parts of another triangle the triangles are congruent.
The statement is proved by SAS theorem
<u>Side-Angle-Side (SAS) theorem: </u>
The triangles are congruent if two sides and the included angle of one triangle are equivalent to two sides and the included angle of another triangle.
Hence, The given statement is proved by side-angle-side (SAS) theorem.
To read more about Angles
brainly.com/question/22472034
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The given matrix equation is,
.
Multiplying the matrices with the scalars, the given equation becomes,
![\left[\begin{array}{cc}1.5x&9\\12&6\end{array}\right] +\left[\begin{array}{cc}y&4y\\3y&2y\end{array}\right] =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right] \\](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1.5x%269%5C%5C12%266%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dy%264y%5C%5C3y%262y%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dz%26z%5C%5C6z%262%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%20%20)
Adding the matrices,
![\left[\begin{array}{cc}1.5x+y&9+4y\\12+3y&6+2y\end{array}\right] =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right] \\](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1.5x%2By%269%2B4y%5C%5C12%2B3y%266%2B2y%5Cend%7Barray%7D%5Cright%5D%20%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dz%26z%5C%5C6z%262%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%20)
Matrix equality gives,

Solving the equations together,

We can see that the equations are not consistent.
There is no solution.