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Tpy6a [65]
2 years ago
5

Can someone please help me with this question

Mathematics
1 answer:
Travka [436]2 years ago
6 0

Answer:

88 in^2

Step-by-step explanation:

2(2 x 6 + 4 x 6 + 4 x 2)

2(wl+hl+hw)

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6^2-49•1/7<br><br><br> PLEASE HELP <br><br><br> 6^2-49 ( times) 1/7
Masja [62]

The answer is 29 I put it in my scientific calculator

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3 years ago
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Kendra ordered a set of blue and brown pins. She received 100 pins, and 30% of them were blue. How many blue pins did Kendra rec
natka813 [3]

Answer:

30 blue pins

Step-by-step explanation:

30% is the same as 30 out of 100, so this can be written in this fraction: \frac{30}{100}

now, you should multiply this fraction by the total number of pins she ordered:

100 x \frac{30}{100}

to solve this, you can simplify by dividing 100 by 100, this equals 1

30 times 1 is 30

there are 30 blue pins

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2 years ago
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Write a story that would use the following equation 12r = 156 to satisfy the story.
Eddi Din [679]
12r=156

Answer r=13

it will help you
4 0
3 years ago
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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
Find the value of x.
Vikentia [17]
The interior angles of a quadrilateral shape add up to 360 so 116+93+45+x = 360 with x equaling 106
7 0
2 years ago
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