Answer:
B. 
Step-by-step explanation:
Since 3.74 is rounded to the hundredths place, it is not the ideal exact solution for m^2 = 14.
To solve for m you would have to square root m and 14, which gives you
. This is the best answer for the problem.
Answer:
62.4 square centimeters
Step-by-step explanation:
The picture of the question in the attached figure
we know that
The surface area of the triangular pyramid is equal to the area of the triangular base plus the area of its three lateral triangular faces
In this problem the triangles are equilateral, that means, the
surface area is equal to the area of four congruent equilateral triangles
so
![A=4[\frac{1}{2}(b)(h)]](https://tex.z-dn.net/?f=A%3D4%5B%5Cfrac%7B1%7D%7B2%7D%28b%29%28h%29%5D)
we have

substitute
![A=4[\frac{1}{2}(6)(5.2)]=62.4\ cm^2](https://tex.z-dn.net/?f=A%3D4%5B%5Cfrac%7B1%7D%7B2%7D%286%29%285.2%29%5D%3D62.4%5C%20cm%5E2)
We know that
the angle MOP is equal to arc MNP because is a central angle
the answer is
angle MOP=126°
Answer:
r = i + j + (-2/3)(3i - j)
Step-by-step explanation:
Vector Equation of a line - r = a + kb ; where r is the resultant vector of adding vector a and vector b and k is a constant
if a = i + j ; b = t(3i - j) and r = -i +s(j)
for this to be true all the vector components must be equal
summing i 's
i + 3ti = -i; then t = -2/3
j - tj = sj; then s = 1-t; substitue t; s=1+2/3 = 5/3
so r = i + j + (-2/3)(3i - j) which will symplify to -i + 5/3j
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