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koban [17]
3 years ago
13

. Given PQ and point ???? that lies on PQ such that point ???? lies 7 / 9 of the length of PQ from point P along PQ:

Mathematics
1 answer:
Rudik [331]3 years ago
8 0

Answer:

Point R is closer to Q.

Step-by-step explanation:

Correct Question Statement:

Given PQ and point R that lies on PQ such that point ???? lies 7 / 9 of the length of PQ from point P along PQ:

a. Sketch the situation described.

b. Is point R closer to P or closer to R, and how do you know?

c. Use the given information to determine the following ratios:

i. PR: PQ

ii. RQ: PQ

iii. PR: RQ

iv. RQ: PR

d. If the coordinates of point P are (0, 0) and the coordinates of point R are (14, 21), what are the coordinates

of point Q.

ANSWER

a. PQ is the line as shown below:

P_______R__Q

Point R lies at 7/9 means if PQ is divided in 9 parts then R lies at 7th place.

b. point R is closer to Q as shown above.

c. Ratios:

i. PR:PQ = 7:9.

ii. RQ:PQ = 2:9

iii. PR:RQ = 7:2.

iv. RQ:PR = 2:7.

d. Suppose coordinates of Q are (x,y)

7/9 of x is 14 → (7/9)*x = 14 → x = \frac{14*9}{7} = 18

7/9 of y is 21 → (7/9)*y =21 → y =\frac{21*9}{7} =27

Thus, Q(18,27) are required coordinates.

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A pyramid with a square base is cut by a plane that is parallel to its base and is 2 units from the base. The surface area of th
icang [17]

Answer:

  6.83 units

Step-by-step explanation:

Let the height of the original pyramid be represented by h. Then the cut off top has a height of (h -2). The scale factor for the area is the square of the scale factor for height, so we have ...

  (height ratio)^2 = 1/2

  ((h -2)/h)^2 = 1/2

  (h -2)√2 = h . . . . . . square root; multiply by h√2

  h(√2 -1) = 2√2 . . . . add 2√2 -h

  h = (2√2)/(√2 -1) ≈ 6.8284 . . . units

The altitude of the original pyramid is about 6.83 units.

7 0
2 years ago
EXAMPLE 5 If f(x, y, z) = x sin(yz), (a) find the gradient of f and (b) find the directional derivative of f at (1, 2, 0) in the
solong [7]

Answer:

<h2>a) f =  sin(yz)i + xzcos(yz)j + xycos(yz)k</h2><h2>b) -2</h2>

Step-by-step explanation:

Given f(x, y, z) = x sin(yz), the formula for calculating the gradient of the function is expressed as ∇f(x, y, z) = fx(x, y, z)i+ fy(x, y, z)j+fz(x, y, z)k where;

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a) ∇f(x, y, z) = sin(yz)i + xzcos(yz)j + xycos(yz)k

The gradient of f =  sin(yz)i + xzcos(yz)j + xycos(yz)k

b) Directional derivative of f at (1,2,0) in the direction of v = i + 4j − k is expressed as ∇f(1, 2, 0)*v

∇f(1, 2, 0) = sin(2(0))i +1*0cos(2*0)j + 1*2cos(2*0)k

∇f(1, 2, 0) = sin0i +0cos(0)j + 2cos(0)k

∇f(1, 2, 0) = 0i +0j + 2k

Given v = i + 4j − k

∇f(1, 2, 0)*v (note that this is the dot product of the two vectors)

∇f(1, 2, 0)*v =  (0i +0j + 2k)*(i + 4j − k )

Given i.i = j.j = k.k =1 and i.j=j.i=j.k=k.j=i.k = 0

∇f(1, 2, 0)*v = 0(i.i)+4*0(j.j)+2(-1)k.k

∇f(1, 2, 0)*v = 0(1)+0(1)-2(1)

∇f(1, 2, 0)*v =0+0-2

∇f(1, 2, 0)*v= -2

 

Hence, the directional derivative of f at (1, 2, 0) in the direction of v = i + 4j − k is -2

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