A vertical angles and Angles 1 & 3
Answer:
d. converges, -25
Step-by-step explanation:
An infinite geometric series converges if the absolute value of the common ratio is less than 1.
Here, the common ratio is 4/5:
| 4/5 | = 4/5 < 1
So the series converges. The sum of an infinite geometric series is:
S = a₁ / (1 − r)
where a₁ is the first term and r is the common ratio.
Here, a₁ = -5 and r = 4/5:
S = -5 / (1 − 4/5)
S = -25
<h3>Answer: Choice D</h3>
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Explanation:
The long way to do this is to multiply all the fractions out by hand, or use a calculator to make shorter work of this.
The shortest way is to simply count how many negative signs each expression has.
The rule is: if there are an even number of negative signs, then the product will be positive. Otherwise, the product is negative.
For choice A, we have 3 negative signs. The result (whatever number it is) is negative. Choice B is a similar story. Choice C is also negative because we have 1 negative sign. Choices A through C have an odd number of negative signs.
Only choice D has an even number of negative signs. The two negatives multiply to cancel to a positive. The negative is like undoing the positive. So two negatives just undo each other. This is why the multiplied version of choice D will be some positive number.
Or you can think of it as opposites. If you are looking up (positive direction) and say "do the opposite" then you must look down (negative direction). Then if you say "do the opposite", then you must look back up in the positive direction.
Answer:
(2,0)
Step-by-step explanation:
Answer:
Step-by-step explanation:
If a point (x, y) is translated by 4 units horizontally right and 1 unit upwards, coordinates of the image point will be,
(x, y) → (x + 4. y + 1)
Therefore, vertices of the image triangle ABC will be,
A(2, 2) → A'(2+4, 2+1)
→ A'(6, 3)
B(5, -1) → B'(5+4, -1+1)
→ B'(9, 0)
C(1, -2) → C'(1+4, -2+1)
→ C'(5, -1)
Then reflected across y-axis.
Rule for the reflection across y-axis will be,
(x, y) → (-x, y)
A'(6, 3) → A"(-6, 3)
B'(9, 0) → B"(-9, 0)
C'(5, -1) → C"(-5, -1)
