Complete Question
Which of the following statements are true and which are false? Give reasons for your answers.
(i) Only one line can pass through a single point.
(ii) There are an infinite number of lines which pass through two distinct points.
(iii) A line can be produced indefinitely on both sides.
(iv) If two circles are equal, then their radii are equal.
(v) if AB=PQ and PQ=XY, then AB=XY.
A (i),(ii) - True
(iii),(iv),(v)-False
B(i),(ii),(iii) -True
(iv),(v)-False
C (i),(ii) -False
(iii),(iv),(v)-True
D (i),(ii),(iii) -False
(iv),(v)-True
Answer:
The correct option is C
Step-by-step explanation:
i is false because several lines can pass through a single point
ii is false because only one line can pass through two distinct points
iii is true because you can extend a line from both points (start and end points )
iv is true because when two equal circle are placed together and radius is trace we will discover that they are equal
v is true because from Euclid's First Axiom , if a= c and c = d the a = d
Answer:
3 units right, 4 down = (x+3, y-4)
3 units left, 4 up = (x-3, y+4)
4 units right, 3 down = (x+4, y-3)
Step-by-step explanation:
Actually, the answer is not A, if you're saying A is the first choice above. That's incorrect. You will need to use the Geometric mean for right triangles here to figure out what the value of a is. We will use this form:
. We have a value for YZ of 3; side a is XZ. That means in order to solve this we need WZ, which we can find using pythagorean's theorem. 3^2 + 4^2 = c^2 and 9 + 16 = c^2 and c = 5. Now we fill in accordingly:
. Cross-multiply to get 3XZ=25 and side XZ is
. XZ is 25/3 and YZ is 3, so 25/3 - 3 = XY. That means that XY (side a) = 16/3 or 5 1/3, choice B, or the second one down.
If you take the derivative of your equation, you get:
2′″−″−′+′=0
2
y
′
y
″
−
x
y
″
−
y
′
+
y
′
=
0
or
″(2′−)=0.
y
″
(
2
y
′
−
x
)
=
0.
Let =′
v
=
y
′
and we have ′(2−)=0,
v
′
(
2
v
−
x
)
=
0
,
so either ′=0
v
′
=
0
and =
v
=
c
or =/2
v
=
x
/
2
.
Then ′=
y
′
=
c
and so =+
y
=
c
x
+
d
or ′=/2
y
′
=
x
/
2
and =2/4.
y
=
x
2
/
4.
Plugging the first into the original equation gives =−2
d
=
−
c
2
. So there are two solutions =−2
y
=
c
x
−
c
2
for some constant
c
and =2/4
y
=
x
2
/
4
. I don't know if this is all the solutions
