Answer:
Adjacent angles. x = 55
Step-by-step explanation:
Adjacent angles are angles that are right next to each other, sharing one side.
Since that is a right angle (90º):
X + 35 = 90
X + (35-35) = 90 - 35
X = 55
Hope this helps! Have a great day!
Answer:
y = 2
Step-by-step explanation:
2 + 8y= 9y
2 = 9y - 8y
2= y
What is the SECRET to football BETTING ??
Answer:
Median = 19.5
Quartile for Q1 = 14
Quartile for Q3 = 23.5
Step-by-step explanation:
Our data is
22 15 13 25 11 19 30 17 20 20
Let us first arrange the data in ascending order
11 13 15 17 19 20 20 22 25 30
As there are total 10 terms in the data , the median will be the average of the 5th and th 6th term
5th entity = 19
6th entity = 20
Average
= ![\frac{19+20}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B19%2B20%7D%7B2%7D)
Average
= ![\frac{39}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B39%7D%7B2%7D)
Median = 19.5
The Formula to find the Quartile is
Quartile For Q1
term
n=10
term
term
2.75 term
This means that Quartile for Q1 is average of 2nd and 3rd term
2nd term = 13 , third term = 15
Average = ![\rac{13+15}{2}](https://tex.z-dn.net/?f=%5Crac%7B13%2B15%7D%7B2%7D)
= 14
Quartile For Q3
term
n=10
term
term
= 8.25
Hence the Quartile for Q3 will be the average of the 8th and 9th term
8th term = 22
9th term = 25
Average= ![\frac{22+25}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B22%2B25%7D%7B2%7D)
= 23.5
Hence Quartile for Q1 = 14
Quartile for Q3 = 23.5
Answer:
![\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24](https://tex.z-dn.net/?f=%5Ctext%7BExact%3A%20%7DAB%3D25%5Csqrt%7B6%7D%2C%5C%5C%5Ctext%7BRounded%3A%20%7DAB%5Capprox%2061.24)
Step-by-step explanation:
We can use the Law of Sines to find segment AD, which happens to be a leg of
and the hypotenuse of
.
The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:
![\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}](https://tex.z-dn.net/?f=%5Cfrac%7Ba%7D%7B%5Csin%20%5Calpha%7D%3D%5Cfrac%7Bb%7D%7B%5Csin%20%5Cbeta%7D%3D%5Cfrac%7Bc%7D%7B%5Csin%20%5Cgamma%7D)
Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is
. The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:
![\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}](https://tex.z-dn.net/?f=%5Cangle%20CAD%2B%5Cangle%20ACD%2B%5Cangle%20CDA%3D180%5E%7B%5Ccirc%7D%2C%5C%5C%5Cangle%20CAD%2B60%5E%7B%5Ccirc%7D%2B75%5E%7B%5Ccirc%7D%3D180%5E%7B%5Ccirc%7D%2C%5C%5C%5Cangle%20CAD%3D180%5E%7B%5Ccirc%7D-75%5E%7B%5Ccirc%7D-60%5E%7B%5Ccirc%7D%2C%5C%5C%5Cangle%20CAD%3D45%5E%7B%5Ccirc%7D)
Now use this value in the Law of Sines to find AD:
![\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}](https://tex.z-dn.net/?f=%5Cfrac%7BAD%7D%7B%5Csin%2060%5E%7B%5Ccirc%7D%7D%3D%5Cfrac%7B100%7D%7B%5Csin%2045%5E%7B%5Ccirc%7D%7D%2C%5C%5C%5C%5CAD%3D%5Csin%2060%5E%7B%5Ccirc%7D%5Ccdot%20%5Cfrac%7B100%7D%7B%5Csin%2045%5E%7B%5Ccirc%7D%7D)
Recall that
and
:
![AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}](https://tex.z-dn.net/?f=AD%3D%5Cfrac%7B%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5Ccdot%20100%7D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%7D%2C%5C%5C%5C%5CAD%3D%5Cfrac%7B50%5Csqrt%7B3%7D%7D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%7D%2C%5C%5C%5C%5CAD%3D50%5Csqrt%7B3%7D%5Ccdot%20%5Cfrac%7B2%7D%7B%5Csqrt%7B2%7D%7D%2C%5C%5C%5C%5CAD%3D%5Cfrac%7B100%5Csqrt%7B3%7D%7D%7B%5Csqrt%7B2%7D%7D%5Ccdot%5Cfrac%7B%20%5Csqrt%7B2%7D%7D%7B%5Csqrt%7B2%7D%7D%3D%5Cfrac%7B100%5Csqrt%7B6%7D%7D%7B2%7D%3D%7B50%5Csqrt%7B6%7D%7D)
Now that we have the length of AD, we can find the length of AB. The right triangle
is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio
, where
is the side opposite to the 30 degree angle and
is the length of the hypotenuse.
Since AD is the hypotenuse, it must represent
in this ratio and since AB is the side opposite to the 30 degree angle, it must represent
in this ratio (Derive from basic trig for a right triangle and
).
Therefore, AB must be exactly half of AD:
![AB=\frac{1}{2}AD,\\AB=\frac{1}{2}\cdot 50\sqrt{6},\\AB=\frac{50\sqrt{6}}{2}=\boxed{25\sqrt{6}}\approx 61.24](https://tex.z-dn.net/?f=AB%3D%5Cfrac%7B1%7D%7B2%7DAD%2C%5C%5CAB%3D%5Cfrac%7B1%7D%7B2%7D%5Ccdot%2050%5Csqrt%7B6%7D%2C%5C%5CAB%3D%5Cfrac%7B50%5Csqrt%7B6%7D%7D%7B2%7D%3D%5Cboxed%7B25%5Csqrt%7B6%7D%7D%5Capprox%2061.24)