A 42 24÷16 =1.5 1.5×28 =42 b i think its 78.4 im not sure
well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus
Answer:
it's 13 meters taller
Step-by-step explanation:
both meters are not the same which means it can't be another number
Answer:
The point-slope formula states:
(
y
−
y
1
)
=
m
(
x
−
x
1
)
Where
m
is the slope and
(
x
1
y
1
)
is a point the line passes through.
Substituting the slope and values from the point in the problem gives:
(
y
−
−
1
)
=
3
5
(
x
−
−
3
)
(
y
+
1
)
=
3
5
(
x
+
3
)
If you want the equation in the somewhat more familiar slope-intercept form we can solve this equation for
y
. The slope-intercept form of a linear equation is:
y
=
m
x
+
b
Where
m
is the slope and
b
is the y-intercept value.
y
+
1
=
(
3
5
⋅
x
)
+
(
3
5
⋅
3
)
y
+
1
=
3
5
x
+
9
5
y
+
1
−
1
=
3
5
x
+
9
5
−
1
y
+
0
=
3
5
x
+
9
5
−
5
5
y
=
3
5
x
+
4
5
The point-slope formula states:
(
y
−
y
1
)
=
m
(
x
−
x
1
)
Where
m
is the slope and
(
x
1
y
1
)
is a point the line passes through.
Substituting the slope and values from the point in the problem gives:
(
y
−
−
1
)
=
3
5
(
x
−
−
3
)
(
y
+
1
)
=
3
5
(
x
+
3
)
If you want the equation in the somewhat more familiar slope-intercept form we can solve this equation for
y
. The slope-intercept form of a linear equation is:
y
=
m
x
+
b
Where
m
is the slope and
b
is the y-intercept value.
y
+
1
=
(
3
5
⋅
x
)
+
(
3
5
⋅
3
)
y
+
1
=
3
5
x
+
9
5
y
+
1
−
1
=
3
5
x
+
9
5
−
1
y
+
0
=
3
5
x
+
9
5
−
5
5
y
=
3
5
x
+
4
5
Step-by-step explanation:
0 and 1 are neither prime nor composite. A prime is any number greater than 1 that has just 1 and itself as factors. Primes can only start at x > 1
When that happens (when you start with numbers greater than one) p^2 is a composite consisting of 2 primes, so any composite will obey the law that he number will have at least 3 factors making it up -- in this case p p^2 and 1.
So the answer to the question by definition is that 0 numbers can have the property of both p and p^2 to be prime.
