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Ipatiy [6.2K]
2 years ago
14

contractor wishes to build 9 houses, each different in design. In how many ways can he place these houses on a street if 6 lots

are on one side of the street and 3 lots are on the opposite side?
Mathematics
1 answer:
Tju [1.3M]2 years ago
5 0

The houses can be placed in 362,880 ways.

<u>Step-by-step explanation:</u>

The 9 houses are each in different design.

The each lot can place any of the 9 houses.

  • The 1st lot can place anyone house of all the 9 houses.
  • The 2nd lot can place one of remaining 8 houses.
  • The 3rd lot can place one of remaining 7 houses.

Similarly, the process gets repeated until the last house is placed on a lot.

<u>From the above steps, it can be determined that :</u>

The number of ways to place the 9 houses in 9 lots = 9!

⇒ 9×8×7×6×5×4×3×2×1

⇒ 362880 ways.

Therefore, the houses can be placed in 362880 ways.

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Triangle ABC and triangle CDE are similar right triangles. Which proportion can be used to show that the slope of AC is equal to
shusha [124]

Answer:

Option C is the correct option.

Step-by-step explanation:

Considering the triangle ABC, the slope of the line CA is given by \frac{AB}{BC} = \frac{6 - 3}{-3 - (- 1)}

Again, considering the triangle CDE, the slope of the line EC is given by

\frac{CD}{DE} = \frac{3 - (- 3)}{- 1 - 3 }

Since CA and EC represents the same straight line so, we can write

\frac{6 - 3}{-3 - (- 1)} = \frac{3 - (- 3)}{- 1 - 3 }

Therefore, option C is the correct option. (Answer)

8 0
2 years ago
Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y
zalisa [80]

Answer:

y has a finite solution for any value y_0 ≠ 0.

Step-by-step explanation:

Given the differential equation

y' + y³ = 0

We can rewrite this as

dy/dx + y³ = 0

Multiplying through by dx

dy + y³dx = 0

Divide through by y³, we have

dy/y³ + dx = 0

dy/y³ = -dx

Integrating both sides

-1/(2y²) = - x + c

Multiplying through by -1, we have

1/(2y²) = x + C (Where C = -c)

Applying the initial condition y(0) = y_0, put x = 0, and y = y_0

1/(2y_0²) = 0 + C

C = 1/(2y_0²)

So

1/(2y²) = x + 1/(2y_0²)

2y² = 1/[x + 1/(2y_0²)]

y² = 1/[2x + 1/(y_0²)]

y = 1/[2x + 1/(y_0²)]½

This is the required solution to the initial value problem.

The interval of the solution depends on the value of y_0. There are infinitely many solutions for y_0 assumes a real number.

For y_0 = 0, the solution has an expression 1/0, which makes the solution infinite.

With this, y has a finite solution for any value y_0 ≠ 0.

8 0
2 years ago
Ramona's backyard is fenced and represented by parallelogram YARD with measures in meters. She installed a 12-meters fence to se
Alisiya [41]
The fencing line x is the height of a rectangle triangle of base = y, hypothenuse of 9 m, so we use Pythagoras theorem to solve:

hyp^2 = height^2 + base^2
9^2 = x^2 + y^2
x^2 = 81 - y^2


we can see that x is also the height of another rectangle triangle of base = 15 - y, hypothenuse of 12 m, so we use Pythagoras theorem to solve:
hyp^2 = height^2 + base^2
12^2 = x^2 + (15 - y)^2

lets expand:
144 = x^2 + 225 - 30y + y^2

substitute x^2 from the first equation in the last:
144 = 81 - y^2 + 225 - 30y + y^2
144 = 81 + 225 - 30y
30y = -144 + 81 + 225
y = 5.4 m

substitute in the fence equation:
x^2 = 81 - y^2
x^2 = 81 - 5.4^2
x = 7.2 m that is the length of the fence



3 0
2 years ago
Ethan bought 4 packages of pencils. After he gave 8 pencils to his friends, he had 40 pencils
LiRa [457]

He had 40 pencils left after he gave away 8, so originally he had 40 + 8 pencils, which is 48.

Now, he bought 4 packages, which had a total of 48 pencils, so divide 48 by 4, which is 12. He had 12 pencils in each package.

To determine the solution arithmetically, first add 8 to 40, then divide 48 by 4.

To determine the solution algebraically, set up and solve the equation 40 = 4x - 8.

Each package contained 12 pencils.

Hope this helps

8 0
2 years ago
Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 6,000 pounds and the
solniwko [45]

Answer:

(5953.52,6046.49)

Step-by-step explanation:

We are given the following in the question:

Mean, \mu = 6,000 pounds

Sample size, n = 40

Alpha, α = 0.05

Standard deviation, σ = 150 pounds

95% Confidence interval:

\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}

Putting the values, we get,

z_{critical}\text{ at}~\alpha_{0.05} = 1.96

6000 \pm 1.96(\dfrac{150}{\sqrt{40}} )\\\\ = 6000 \pm 46.4854=\\(5953.5146,6046.4854)\approx (5953.52,6046.49)

are the limits within which 95% of the sample means occur.

6 0
1 year ago
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