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azamat
3 years ago
12

Besides 26 and 1, what is one factor of 26?

Mathematics
2 answers:
iogann1982 [59]3 years ago
8 0
One factor of twenty six is thirteen
Ainat [17]3 years ago
5 0
13 and 2, I think that is the only one left.

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I need help. please show your work!
vodomira [7]
X =  10 * 3^n
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8 0
3 years ago
Read 2 more answers
I need help please!!!!!!!
k0ka [10]

Answer:

1 roll of ribbon, 1 package of buttons, and 3 packages of beads

Step-by-step explanation:

Use the information you found about how much each makes to answer the question.

Lynette will make 6 decorations.

Since 1 roll makes 6, she needs 1 roll.

Since 1 package of buttons makes 6, she needs 1 package.

Since 1 package of beads makes 2, she needs 3 packages.

5 0
3 years ago
There are 5 types of fruit in a basket. There are 3 apples, 2 bananas, 4 oranges, 5 kiwi, and 1 passionfruit. You ask a friend t
creativ13 [48]

Answer:

Five out of fifteen or one out of three

Step-by-step explanation:

Add up all of the fruit in the basket - 15

Number of kiwi - 5

If you simplify 5/15 you get:

1/3

3 0
2 years ago
16 is 40% of what number?
saw5 [17]
Sixteen is 40% of 40
7 0
3 years ago
Read 2 more answers
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
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