Answer:
Step-by-step explanation:
You gave very clear instructions on how to draw this graph, so that's what I did. What you need to remember in particular with limits is that you do not care in the least what happens AT the x value of 2, only what happens as it is being approached. Because we are asked the limit as x is approaching from the left and the right, this is a one-sided limit question. In order for the limit to exist as x approaches 2 (NOT from the left or the right), the limit would have to agree from the left and the right, and this one doesn't. Having said that there is "a horizontal line crossing the y-axis at 5 that ends at the open point (2, 5)..." is a limit approaching x from the left. Therefore,
Having also said there is "...another horizontal line starting at the open point (2, -3) and continues to the right..." is a limit approaching x from the right. Therefore,
The closed point at (2, 1) is where x IS, and remember that we don't care about what happens AT x. So disregard this point in limits.
Cosine is the second one the first one is Sine and the last one is tangent
Answer:
its B. -3
Step-by-step explanation:
Remove 3 negative tiles .
Can I get brainliest??????
Answer:
Step-by-step explanation:
Using the section formula, if a point (x,y) divides the line joining the points (x
1
,y
1
) and (x
2
,y
2
) in the ratio m:n, then
(x,y)=(
m+n
mx
2
+nx
1
,
m+n
my
2
+ny
1
)
The vertices of the triangle are given to be (x
1
,y
1
),(x
2
,y
2
) and (x
3
,y
3
). Let these vertices be A,B and C respectively.
Then the coordinates of the point P that divides AB in l:k will be
(
l+k
lx
2
+kx
1
,
l+k
ly
2
+ky
1
)
The coordinates of point which divides PC in m:k+l will be
⎩
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎧
m+k+l
mx
3
+(k+l)
(l+k)
lx
2
+kx
1
,
m+k+l
my
3
+(k+l)
(l+k)
ly
2
+ky
1
⎭
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎫
⇒(
m+k+l
kx
1
+lx
2
+mx
3
,
m+k+l
ky
1
+ly
2
+my
3
)
Answer:
a
Step-by-step explanation: