It’s simple. go to desmos and put in both equations. wherever the lines intercept put those two points in (x,y)
' n ' the 'n'th term
1 4
2 2
3 0
4 -2 .
In general, the 'n'th term of the sequence is A(n) = 6 - 2n .
Answer:
No, it would not be an equilateral triangle.
Step-by-step explanation:
An equilaterial triangle has all three sides equal. Raising the roof will increse the two vertical sides, but if the bottom is not widened, it will reamin the original lenght. Only two sides are now equal. No longer equilateral /
Part A:
For the first figure it can be seen that the orientation of the pre-image (<span>∆RST) is the same as that of the image (∆R'S'T').
The image is obtained from the pre-image by shifting the pre-image some units down.
Therefore, the </span><span>single transformation that transforms ∆RST to ∆R'S'T' is a translation.
Part B:</span>
<span>
For the second figure it can be seen that the orientation of the pre-image (<span>∆RST) is the same as that of the image (∆R'S'T').
The image is obtained from the pre-image by shifting the pre-image some units down and some units to the right.
Therefore, the </span><span>single transformation that transforms ∆RST to ∆R'S'T' is a translation.
Part C:
</span></span><span>
For the third figure it can be seen that the orientation of the pre-image (<span>∆RST) is not the same as that of the image (∆R'S'T').
The image is obtained from the pre-image by rotating the pre-image some 180 degrees.
Therefore, the </span><span>single transformation that transforms ∆RST to ∆R'S'T' is a rotation.
Part D:
</span></span><span>
For the fourth figure it can be seen that the orientation of the pre-image (<span>∆RST) is not the same as that of the image (∆R'S'T').
The image is obtained from the pre-image by refrecting the pre-image.
Therefore, the </span><span>single transformation that transforms ∆RST to ∆R'S'T' is a reflection.</span></span>