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KIM [24]
3 years ago
14

Help please Find the surface area, including the floor, of his tent.

Mathematics
1 answer:
natita [175]3 years ago
3 0

Answer:

52.8 m. sq.

Step-by-step explanation:

2.6 x 1.5 + 1.5 (3) = 7.8 / 2 = 3.9 x 2 = 7.8 (Triangles)

3 x 5 = 15 x 2 = 30 (Side Rectangles)

5 x 1.5 + 1.5 (3) = 15

15 + 30 + 7.8 = 52.8 m. sq.

Hope this helps!

If something is wrong, please let me know.

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If a giraffe is 9 feet tall, what is its height in decimeters?​
Dmitry [639]

Answer:

The height of the giraffe is 27.432 decimeters.

Step-by-step explanation:

To convert 1 foot to decimeters you would multiply by 3.048.

We need to multiply 9 feet by 3.048 which would give you the answer of 27.432 decimeters.

6 0
2 years ago
Dan can run 1000 meters in 5 minutes how many can he run in 2 minutes
Rudik [331]

Answer:

800 m

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
What is the magnitude of -3 +4i?
Goshia [24]

Answer:

we have

-3+4i

let magnitude be |z|

|z|=\sqrt{( - 3) {}^{2}  + 4 {}^{2} }

|z|=5 unit

6 0
3 years ago
A cable television compan is laying cable in an area with underground utilities. Two subdivisions are located on opposite sides
Schach [20]

Answer:

Step-by-step explanation:

Given:

Point Q = Q is on the north bank 1200 m east of P $40/m to lay cable underground and $80/m way to lay the cable.

Lets denote T be a point on north bank, therefore, point T is t m east and 100m north of P.

Note that 0 < t < 1200

Distance from P to T using Pythagoras theorem:

PT^2 = PQ^2 + QT^2

PT = √(t²+100²) = √(t²+10,000)

Distance from T to Q:

TQ = 1200 - t

From above, Cable from P to T is underwater, and costs $80/m to lay.

Cable from T to Q is underground, and costs $40/m to lay.

Total cost of the cable:

C(t) = 80√(t²+10,000) + 40(1200 - t)

Cost is at its minimum: when dC/dt = C'(t) = 0 and d^2C/dt^2 = C''(t) > 0

dC/dt = C'(t)

= 80 * 1/2 (t² + 10,000)^(-1/2) × 2t + 40 × (-1)

dC/dt = C'(t)

= 80t/√(t² + 10,000) - 40 = 0

80t/√(t² + 10,000) = 40

80t = 40√(t² + 10,000)

2t = √(t² + 10,000)

square both sides:

4t² = t² + 10,000

3t² = 10,000

t² = 10,000/3

t = 100/√3

≈ 57.735

d^2C/dt^2 =

C''(t) = [80√(t² + 10,000) - 80t × t/√(t²+10,000)] /√(t²+10,000)

C''(t) = [(80(t²+10,000) - 80t²) / √(t²+10,000)] /√(t²+10,000)

C''(t) = 800,000 / (t² + 10,000)^(3/2)

C''(t) > 0 for all t

Therefore C(t) is minimized at t = 100/√3

C(100/√3) = 80√(10,000/3 + 10,000) + 40 × (1200 - 100/√3)

= 80 √(40,000/3) + 48,000 - 4000/√3

= 16000/√3 + 48,000 - 4000/√3

= 48,000 + 12,000/√3

= 48,000 + 4,000√3

= 4000 (12 + √3)

≈ 54,928.20

Minimum cost is $54,928.20

This occurs when cable is laid underwater from point P to point T(57.735 m east and 100m north of P) and then underground from point T to point Q (1200 - 57.735)

= 1142.265 m

5 0
3 years ago
What is the answer to X^2-61=20
Reika [66]
X^2 -61 =20
X^2= 81x =9 x=-9
7 0
3 years ago
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