Answer:
The correct option is second one i.e 24 units.
Therefore the height of the triangle is
Step-by-step explanation:
Given:
An equilateral triangle has all sides equal.
ΔMNO is an Equilateral Triangle with sides measuring,
NR is perpendicular bisector to MO such that
.NR ⊥ Bisector.
To Find:
Height of the triangle = NR = ?
Solution :
Now we have a right angled triangle NRM at ∠R =90°,
So by applying Pythagoras theorem we get
Substituting the values we get
Therefore the height of the triangle is
Answer:What was done from the first to the second equation was that the fraction was simplified. 126 and 32 have a common factors of 2.
126/32=(63*2)/(16*2)
The 2s in the top and bottom can cancel out, leaving the fraction 63/16.
In addition, since there are x terms on the top and bottom, they cancelled out as well.
x/x^3=1/x^2
This leaves an x^2 term on the bottom.
Thus, if a is 16, and b is 2, you will have an equivalent form of the fraction.
Step-by-step explanation:
Answer:
The dimensions that minimize the surface are:
Wide: 1.65 yd
Long: 3.30 yd
Height: 2.20 yd
Step-by-step explanation:
We have a rectangular base, that its twice as long as it is wide.
It must hold 12 yd^3 of debris.
We have to minimize the surface area, subjet to the restriction of volume (12 yd^3).
The surface is equal to:
The volume restriction is:
If we replace h in the surface equation, we have:
To optimize, we derive and equal to zero:
![dS/dw=36(-1)w^{-2} + 8w=0\\\\36w^{-2}=8w\\\\w^3=36/8=4.5\\\\w=\sqrt[3]{4.5} =1.65](https://tex.z-dn.net/?f=dS%2Fdw%3D36%28-1%29w%5E%7B-2%7D%20%2B%208w%3D0%5C%5C%5C%5C36w%5E%7B-2%7D%3D8w%5C%5C%5C%5Cw%5E3%3D36%2F8%3D4.5%5C%5C%5C%5Cw%3D%5Csqrt%5B3%5D%7B4.5%7D%20%3D1.65)
Then, the height h is:
The dimensions that minimize the surface are:
Wide: 1.65 yd
Long: 3.30 yd
Height: 2.20 yd
Answer: t=9
Step-by-step explanation:
It is given in the question that,
Line QS bisects angle PQR. Solve for x and find the measure of angle PQR.
And
Since QS bisects angle PQR, therefore
Substituting the values, we will get
