Ask question

Ask question

All categories

3 years ago

10

a boy flies a kite holding a 50 foot piece if string. the angle of elevation of the kite is 35. How high is the kite above the g

round.

1 answer:

Rus_ich [418]3 years ago

8 0

29'

x = height
cos(35˚) = 50/x
50/cos(35˚) = x
x = 28.6

You might be interested in

Which shaded grid represents 44%

Galina-37 [17]

Answer: third one

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A motor vehicle has a maximum efficiency of 39 mpg at a cruising speed of 40 mph. The efficiency drops at a rate of 0.1 mpg/mph

svp [43]

Answer:

38 mpg

Step-by-step explanation:

Initial efficiency = 39 mpg

Initial speed = 40 mph

Final speed = 50 mph

The efficiency drop between 40 mph and 50 mph is given by:

$E_d = 0.1 \frac{mpg}{mph}*\Delta V$

The total efficiency drop from 40 to 50 mph is:

$E_d = 0.1 \frac{mpg}{mph}*(50-40)mph\\E_d = 1\ mpg$

Therefore, the efficiency at 50 mph is:

$E_{50} = E_{40} -E_d\\E_{50} = 39 -1\\E_{50} = 38\ mpg$

5 0

3 years ago

What is 23/6 as a mixed number?

saveliy_v [14]

23/6 as a mixed fraction is 3(5/6)

7 0

3 years ago

Read 2 more answers

A product can be made in sizes huge, average and tiny which yield a net unit profit of $14, $10, and$5, respectively. Three cent

navik [9.2K]

Answer:

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃ to maximize

Subject to:

First center X₁₁ + X₂₁ + X₃₁ ≤ 550

Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

X₁₁ + X₁₂ + X₁₃ ≤ 710

X₂₁ + X₂₂ + X₂₃ ≤ 900

X₃₁ + X₃₂ + X₃₃ ≤ 350

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

Step-by-step explanation:

Let´s call Xij product size i produced in center j

According to this, we get the following set of variable

X₁₁ product size huge produced in center 1

X₁₂ product size huge produced in center 2

X₁₃ product size huge produced in center 3

X₂₁ product size average produced in center 1

X₂₂ product size average produced in center 2

X₂₃ product size average produced in center 3

X₃₁ product size-tiny produced in center 1

X₃₂ product size-tiny produced in center 2

X₃₃ product size-tiny produced in center 3

Then Objective function is

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Constrains

Center capacity

1.- First center X₁₁ + X₂₁ + X₃₁ ≤ 550

2.- Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

3.- Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

Water available

1.- 22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

2.- 22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

3.- 22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

Demand constrain

Product huge

X₁₁ + X₁₂ + X₁₃ ≤ 710

Product average

X₂₁ + X₂₂ + X₂₃ ≤ 900

Product tiny

X₃₁ + X₃₂ + X₃₃ ≤ 350

Fraction SP/CC must be the same

First and second centers fraction SP/CC

(X₁₁ + X ₂₁ + X₃₁)/ 11000 = (X₁₂ + X₂₂ + X₃₂)/ 2700

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

First and third centers fraction SP/CC

(X₁₁ + X ₂₁ + X₃₁)/ 11000 = ( X₁₃ + X₂₃ + X₃₃)/ 3400

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Subject to:

First center X₁₁ + X₂₁ + X₃₁ ≤ 550

Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

X₁₁ + X₁₂ + X₁₃ ≤ 710

X₂₁ + X₂₂ + X₂₃ ≤ 900

X₃₁ + X₃₂ + X₃₃ ≤ 350

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

6 0

3 years ago

The solutions to a certain quadratic equation are x = -4 and x = 3. Write the equation in standard form below.

elena-14-01-66 [18.8K]

$y=a(x-x_1)(x-x_2)\\\\x_1=-4;\ x_2=3\ therefore\\\\y=a(x+4)(x-3)\\\\y=a(x^2+x-12)\\\\\boxed{y=ax^2+ax-12a}\ where\ a\in\mathbb{R}-\{0\}$

5 0

3 years ago

Read 2 more answers