Answer:
Q1: 77°, Q2: 103°, Q3: 257°, Q4: 283°
Step-by-step explanation:
Note: Q1 = Quadrant 1
Q1: 77°
Q2: 180° - 77° = 103°
Q3: 180° + 77° = 257°
Q4: 360° - 77° = 283°
<span>here we can use Pythogoras' theorem.
in right angled triangles the square of the hypotenuse is equal to the sum of the squares of the other 2 sides.
hypotenuse is 19 cm. One side is 13 cm and we need to find the length of the third side.
19</span>²<span> = 13</span>²<span> + X</span>²<span>
X - length of the third side
361 = 169 + X</span>²<span>
X</span>²<span> = 361 - 169
X</span>²<span> = 192
X = 13.85 the length of third side rounded off to the nearest tenth is 13.9 cm</span>
Answer:
x=30, <A=60, <C=30
Step-by-step explanation:
A triangle always has its total angles added up to 180 degrees, so we can set up the following equation:
90+2x+x=180
90+3x=180
3x=90
x=30
Since x=30, this means <C is 30 degrees. For <A, it's twice as much as <C, so 2*30=60.
So x=30, <A=60, <C=30
Hope this helped!
Answer:
Step-by-step explanation:
x
2
+
x
−
6
=
(
x
+
3
)
(
x
−
2
)
x
2
−
3
x
−
4
=
(
x
−
4
)
(
x
+
1
)
Each of the linear factors occurs precisely once, so the sign of the given rational expression will change at each of the points where one of the linear factors is zero. That is at:
x
=
−
3
,
−
1
,
2
,
4
Note that when
x
is large, the
x
2
terms will dominate the values of the numerator and denominator, making both positive.
Hence the sign of the value of the rational expression in each of the intervals
(
−
∞
,
−
3
)
,
(
−
3
,
−
1
)
,
(
−
1
,
2
)
,
(
2
,
4
)
and
(
4
,
∞
)
follows the pattern
+
−
+
−
+
. Hence the intervals
(
−
3
,
−
1
)
and
(
2
,
4
)
are both part of the solution set.
When
x
=
−
1
or
x
=
4
, the denominator is zero so the rational expression is undefined. Since the numerator is non-zero at those values, the function will have vertical asymptotes at those points (and not satisfy the inequality).
When
x
=
−
3
or
x
=
2
, the numerator is zero and the denominator is non-zero. So the function will be zero and satisfy the inequality at those points.
Hence the solution is:
x
∈
[
−
3
,
−
1
)
∪
[
2
,
4
)
graph{(x^2+x-6)/(x^2-3x-4) [-10, 10, -5, 5]}
