What is the slope of a line that passes through (–4,–13) and (19,11)?a. 23/24
b. 24/23
c. -23/24
d. - 24/23
Answers
Answer 1
Answer:
for points (x1,y1) and (x2,y2)
slope=(y1-y2)/(x1-x2)
(-4,-13)
(19,11)
x1=-4
y1=-13
x2=19
y2=11
slope=(-13-11)/(-4-19)=-24/-23=24/23
B is answer
slope=(y1-y2)/(x1-x2)
(-4,-13)
(19,11)
x1=-4
y1=-13
x2=19
y2=11
slope=(-13-11)/(-4-19)=-24/-23=24/23
B is answer
Answer 2
Answer:
B. %7D%7B19-(-4)%7D%20%3D%20%5Cfrac%7B24%7D%7B23%7D)
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has center
P
(2, 0) and radius 20. Circle
Q
has center
Q
(0, 4) and radius 2.
(a)
Describe the rule for translating center
Q
onto center
P
.
(b)
Determine the scale factor for dilating circle
Q
so that it has the same radius as circle
P.
(c)
Are circles
P
and
Q
similar? Explain your answer.
Answers
Siras ! Don't try to picture this all in your head !
You'll wear out your brain.
You MUST sketch it on a piece of paper.
Draw an x-axis and a y-axis, then draw the two circles.
I'm drawing myself a picture right now, and I'm
supposed to be some kind of a genius.
a). In order to move Circle-Q so that both centers are
at the same point, you need to move the center of Q
4 units down and 2 units to the right.
When you do that, you'll have the little circle inside the
big circle, with their centers both at the same place.
b). The radius of Q is 2.
The radius of P is 20.
What do you have to multiply 2 by, in order to get 20 ?
THAT's the scale factor to dilate Q so that it has the same
radius as P.
When you do that, suddenly it'll look like you only have one circle
on the paper ... they'll both have the same radius and their centers are
at the same place, so you can't tell them apart.
c). All circles are similar !
I went online (you could easily go there too). I searched the question
"Are circles similar ?" and a lot of interesting stuff came up. (you could
do that too). I saw a lot of ways to prove that all circles are similar.
The best one says:
Two figures are "similar" if you can make one of them
exactly fit on top of the other one (make them congruent)
with translations and dilations.
You just did that with P and Q !
-- Translation is moving them around.
You moved Q and put the centers of both circles at the same place.
-- Dilation is blowing it up or blowing it down, so its size changes
but its shape doesn't change.
You blew Q up so that it had the same radius as P.
Then the two circles exactly fit over/under each other.
So the two circles are similar.
You'll wear out your brain.
You MUST sketch it on a piece of paper.
Draw an x-axis and a y-axis, then draw the two circles.
I'm drawing myself a picture right now, and I'm
supposed to be some kind of a genius.
a). In order to move Circle-Q so that both centers are
at the same point, you need to move the center of Q
4 units down and 2 units to the right.
When you do that, you'll have the little circle inside the
big circle, with their centers both at the same place.
b). The radius of Q is 2.
The radius of P is 20.
What do you have to multiply 2 by, in order to get 20 ?
THAT's the scale factor to dilate Q so that it has the same
radius as P.
When you do that, suddenly it'll look like you only have one circle
on the paper ... they'll both have the same radius and their centers are
at the same place, so you can't tell them apart.
c). All circles are similar !
I went online (you could easily go there too). I searched the question
"Are circles similar ?" and a lot of interesting stuff came up. (you could
do that too). I saw a lot of ways to prove that all circles are similar.
The best one says:
Two figures are "similar" if you can make one of them
exactly fit on top of the other one (make them congruent)
with translations and dilations.
You just did that with P and Q !
-- Translation is moving them around.
You moved Q and put the centers of both circles at the same place.
-- Dilation is blowing it up or blowing it down, so its size changes
but its shape doesn't change.
You blew Q up so that it had the same radius as P.
Then the two circles exactly fit over/under each other.
So the two circles are similar.
2 sinx cosx - 1 + 2 sin2x
2 sin x cos2x - 1 + 2 sin2x
2 sin x cos2x - sin x + 1 - 2 sin2x
2 sin x cos2x - 1 - 2 sin2x
Answers
Note that
sin 2x = 2 sin(x) cos(x)
cos 2x = cos²x - sin²x = 2cos²x - 1 = 1 - 2sin²x
Therefore
sin 2x - cos 2x
= 2 sin(x) cos(x) - (1 - 2sin²x)
= 2 sin(x) cos(x) - 1 + 2sin²x
Answer: 2 sin(x) cos(x) - 1 + 2 sin²x
sin 2x = 2 sin(x) cos(x)
cos 2x = cos²x - sin²x = 2cos²x - 1 = 1 - 2sin²x
Therefore
sin 2x - cos 2x
= 2 sin(x) cos(x) - (1 - 2sin²x)
= 2 sin(x) cos(x) - 1 + 2sin²x
Answer: 2 sin(x) cos(x) - 1 + 2 sin²x
Answers
10 5/12 written as an improper fraction is equivalent to 125/12
10 5/12
10 times the denominator 12=120
plus the numerator 5=125
=125 over the original denominator 12
125/12
Hope this helped
10 times the denominator 12=120
plus the numerator 5=125
=125 over the original denominator 12
125/12
Hope this helped