Helppp me with this ,I will mark brainlest

Chad has a rope that is 8 yards long. How many pieces of rope measuring 4/7 of a yard can he divide his rope into?

statuscvo [17]

Answer:

B

Step-by-step explanation:

Your answer is 28 because 4/7 turn it to a decimal and then divide 8 from it and that is how you get your answer 28.

Hope this helps:)

Pls mark brainlist

6 0

3 years ago

Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,

posledela

Answer:

number of multiplies is n!
n=10, 3.6 ms
n=15, 21.8 min
n=20, 77.09 yr
n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

mpy[n] = n·mp[n-1]

mpy[2] = 2

So, ...

mpy[n] = n! . . . n ≥ 2

__

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0

3 years ago

Multiplicative inverse of fraction 4/12

docker41 [41]

Answer:

12/4

hope it helps!

Step-by-step explanation:

A multiplicative inverse is a reciprocal... A reciprocal is one of a pair of numbers that when multiplied with another number equals the number 1.

So the inverse of 4/12 is 12/4

Because 4/12*12/4= 1

5 0

3 years ago

Read 2 more answers

A heavy rope, 50 ft long, weighs 0.6 lb/ft and hangs over the edge of a building 120 ft high. Approximate the required work by a

Anastasy [175]

Answer:

Exercise (a)

The work done in pulling the rope to the top of the building is 750 lb·ft

Exercise (b)

The work done in pulling half the rope to the top of the building is 562.5 lb·ft

Step-by-step explanation:

Exercise (a)

The given parameters of the rope are;

The length of the rope = 50 ft.

The weight of the rope = 0.6 lb/ft.

The height of the building = 120 ft.

We have;

The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;

ΔW₁ = 0.6Δx·x

The work done for the second half, ΔW₂, is given as follows;

ΔW₂ = 0.6Δx·x + 25×0.6 × 25 = 0.6Δx·x + 375

The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375

∴ We have;

W = $2 \times \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= 2 \times \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 750$

The work done in pulling the rope to the top of the building, W = 750 lb·ft

Exercise (b)

The work done in pulling half the rope is given by W₂ as follows;

$W_2 = \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 562.5$

The work done in pulling half the rope, W₂ = 562.5 lb·ft

6 0

2 years ago

-2(6+s) Greater than or equal to -15 - 2s

Eddi Din [679]

Answer:

I'm leaning mostly towards C Because there is a solution, None of them are real numbers and if they were, S would = 5 Especially positive S

~ Zachary

7 0

3 years ago