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Ilya [14]
3 years ago
7

Help me with this please!!!!!!

Mathematics
1 answer:
Gnoma [55]3 years ago
3 0

Answer:

False

Step-by-step explanation:

If the point (-4,3) is the center of a circle and that circles perimeter contains the point (4,9) there it no way that the point (2,-5) could be in that circle

Hope this is right heh, I'm not really good with these type of problems. byeee <3

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What is the perimeter of this figure
Serjik [45]

Answer:

40

Step-by-step explanation:

So far, the Lengths we have are

4, 4, 8, and 4

for the bottom length, add 8 and 4

For the side length, and 4 and 4

so the lenghts are 4, 4, 8, 4, 8, and 12

4+4+8+4+8+12=40

5 0
3 years ago
-4 , 1 2 units down would be
Artyom0805 [142]
-4,12 if on a coordinate plane would be 4 to the left and twelve down(if not what your looking for for the answer than sorry)
7 0
3 years ago
I'm struggling with this, and would like someone to please help me with this. Thank you!
rodikova [14]
What is it? Put the picture up
4 0
3 years ago
Which is the value of this expression when j=-2 and k=-1? <br><br> ( jk ^-2 / j^-1 k^-3 ) ^3
KengaRu [80]

Answer:

Step-by-step explanation:

No idea

6 0
2 years ago
How do I solve this limits problem?
motikmotik

Each piece of this function is continuous on its respective domain (because all polynomials are continuous functions), meaning

• 2 - x exists for all x < -1

• x exists for all -1 ≤ x < 1

• (x - 1)² exists for all x ≥ 1

So this really just leaves the points where the pieces are split up, i.e. x = -1 and x = 1. At both of these points, the two-sided limit exists as long as the one-side limits from both sides exist and are equal to one another.

At x = -1, as I said in my comment, you have

\displaystyle \lim_{x\to-1^-}f(x) = \lim_{x\to-1}(2-x) = 2-(-1) = 3

while

\displaystyle \lim_{x\to-1^+}f(x) = \lim_{x\to-1}x = -1

But -1 ≠ 3, so the two-sided limit

\displaystyle\lim_{x\to-1}f(x)

does not exist. So a = -1 is one of the points you would list.

At x = 1, we have

\displaystyle \lim_{x\to1^-}f(x) = \lim_{x\to1}x = 1

while

\displaystyle \lim_{x\to1^+}f(x) = \lim_{x\to1}(x-1)^2 = (1-1)^2 = 0

and again the one-sided limits don't match, so this two-sided limit also does not exist, making a = 1 the other answer.

6 0
3 years ago
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