Not entirely, it could be an isoceles triangle (a triangle which all sides are different lengths) but there is still a possibility.
Almost 45 minutes. I’m just guessing no
Answer:
x = 4
Step-by-step explanation:
If you do not want to read the explanation go the the next part in bold
Ok so the shorter leg of the triangle is a radius of the circle. ( This is indicated by the point in the middle of the circle.) for the hypotenuse we are given one part of the length (2) The other part also happens to be a radius ( remember that a radius is a line that starts from the center of the triangle to any point of the circumference. ) Also remember that the radius is equal to 3. That being said the hypotenuse = 2+3 which equals 5.
This triangle then happens to be a right triangle. ( A triangle formed by a tangent line and a radius is a right triangle.) This means that we can use the Pythagorean theorem to solve for x.
Below here is where the work is shown
a² + b² = c². where a and b = legs and c = hypotenuse. We are given the hypotenuse (5) and a leg (3) So we plug in what we are given and solve for the missing information. 5² = 3² + b²
5² = 25
3² = 9
we would then have 25 = 9 + b²
Next we subtract 9 from each side
25 - 9 = 16
9 - 9 cancels out
Now we have 16 = b²
finally we want to get rid of the ²
To do so we take the square root of each side

we're left with b = 4 which means that x = 4
Answer:
a = 3, b = 0, c = 0, d = -2
Step-by-step explanation:
<em>To find the reflection Multiply the matrices</em>
∵ The dimension of the first matrix is 2 × 2
∵ The dimension of the second matrix is 2 × 3
<em>1. Multiply the first row of the 1st matrix by each column in the second matrix add the products of each column to get the first row in the 3rd matrix.</em>
2. Multiply the second row of the 1st matrix by each column in the second matrix add the products of each column to get the second row of the 3rd matrix
×
= ![\left[\begin{array}{ccc}(1*0+0*0)&(1*3+0*0)&(1*0+0*2)\\(0*0+-1*0)&(0*3+-1*0)&(0*0+-1*2)\end{array}\right]=\left[\begin{array}{ccc}0&3&0\\0&0&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%281%2A0%2B0%2A0%29%26%281%2A3%2B0%2A0%29%26%281%2A0%2B0%2A2%29%5C%5C%280%2A0%2B-1%2A0%29%26%280%2A3%2B-1%2A0%29%26%280%2A0%2B-1%2A2%29%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%263%260%5C%5C0%260%26-2%5Cend%7Barray%7D%5Cright%5D)
Compare the elements in the answer with the third matrix to find the values of a, b, c, and d
∴ a = 3
∴ b = 0
∴ c = 0
∴ d = -2
the answer is in the picture I mentioned as the answer ok ◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉◉‿◉