All categories

3 years ago

If the (Z) clock (O) is (O) one (M) 1:10 what will it be in 34 hours and 89 minutes?

2 answers:

6 0

if the (Z) clock (O) is (O) one (M) 1:10 what will it be in 34 hours and 89 minutes?
Explain in 1 to 3 sentences.
yes or no.
12:39

Mama L [17]3 years ago

5 0

Answer: 12:39 AM

Step-by-step explanation: it would be 12:39 because 34 hours is 2 days with a leftover 10 hours. The 89 minutes include an hour so it becomes 11 hours and 29 minutes, add this number to 1:10 and there you go

You might be interested in

Help I will be marking brainliest!!

lora16 [44]

Answer:

Given: base = 10.4ft, height = 12.5

Area of octagon = 8(1/2 × b × h)

8(1/2 × 10.4 × 12.5)
520ft²

Volume of pool = 520ft² × 3ft

1560ft³

Now, 1 cubic ft takes 7.5 gallons to fill.

Therefore, 1560 cubic ft takes,

1560 × 7.5
11700

So, Correct choice - [D] 11,700.

5 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

Plz help and explain

vovikov84 [41]

To solve this problem, we need to first find the dimensions of the side of the blue and purple squares.

We're given that the purple (smaller) square has a side length of x inches.
We are also given that the blue band has a width of 5 inches.
Since the blue band surrounds the purple square on both sides, the length of the blue square is x+2(5)=x+10 inches.

The net area of the band is therefore the difference of the area of the blue square and the purple square, namely take out the area of the purple square from the blue.

Therefore
Area of band
$=(x+10)^2-x^2$ [recall $(a+b)^2=a^2+2ab+b^2$
$=x^2+20x+100-x^2$
$=20x+100$ or 20(x+5) if you wish.

8 0

3 years ago

Rosa has 55 books and gives 5 of them to her friend. How many does she have left?

Alecsey [184]

Answer:

55-5

50 books she has left

7 0

3 years ago

The yearbook committee polled 80 randomly selected students from a class of 320 ninth graders to see if they would be willing to

jek_recluse [69]

Answer: