Answer:
Step-by-step explanation:
Given:
The given mixed fractions are:
Both the mixed fractions have the whole part same as -5. Now, in order to compare the two, we have to compare their fractional parts 
and 
.
As the denominators of both the fractions are same and equal to 8.
For same denominators of two positive fractions, the numerator with a greater value has a larger fraction.
For same denominators of two negative fractions, the numerator with a smaller value has a larger fraction.
Here, both the fractions are negative, so the numerator that is smaller has the larger fraction.
Here, 2 is less than 7. So, the fraction 
is greater than that of 
.
Therefore,
I think you need to use the SoHCaHToA method here.
S means sin
C means cos
T means tan
O means opposite
H means hypotenuse
A means adjacent
The hypotenuse is the longest side in a right angled triangle.
The adjacent is the side between the angle and the right angle.
The opposite is the side that is left.
If you have the value of the opposite cross all of the o's out. The same with the other sides. If only one type of letter is crossed out, cross out the type of letter that stands for the length you want to find. The part of the word with two crosses tells you what function to use (sin, cos or tan)
In this example, the function used will be sin.
sinx°=o÷h
cosx°=a÷h
tanx°=o÷a
since we are using sin, the formula we will use is :
sinx°=o÷h
Then you plug in what you know.
sin30°=30÷h
h=30÷sin30°
h=60ft
If there was a case when you needed to work out an angle, inverse sin, or cos, or tan would be used.
Answer:
Area of trapezium = 4.4132 R²
Step-by-step explanation:
Given, MNPK is a trapezoid
MN = PK and ∠NMK = 65°
OT = R.
⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).
Now, sum of interior angles in a quadrilateral of 4 sides = 360°.
⇒ x + x + 65° + 65° = 360°
⇒ x = 115°.
Here, NS is a tangent to the circle and ∠NSO = 90°
consider triangle NOS;
line joining O and N bisects the angle ∠MNP
⇒ ∠ONS = 
= 57.5°
Now, tan(57.5°) = 
⇒ 1.5697 = 
⇒ SN = 0.637 R
⇒ NP = 2×SN = 2× 0.637 R = 1.274 R
Now, draw a line parallel to ST from N to line MK
let the intersection point be Q.
⇒ NQ = 2R
Consider triangle NQM,
tan(∠NMQ) = 
⇒ tan65° =
⇒ QM = 
QM = 0.9326 R .
⇒ MT = MQ + QT
= 0.9326 R + 0.637 R (as QT = SN)
⇒ MT = 1.5696 R
⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R
Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).
⇒ A = (
) × (ST)
= (
) × 2 R
= 4.4132 R²
⇒ Area of trapezium = 4.4132 R²
98.07. its a longer number but thats it for short.
I got 6.49204 but you will need to round so D 6.5
I added all of the numbers up
