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suter [353]
2 years ago
9

A summer camp cookout is planned for the campers and their families. There is room for 200 people. Each adult costs $4, and each

camper costs $3. There is a maximum budget of $750. Write the system of inequalities to represent this real-world scenario, where x is the number of adults and y is the number of campers.
A.
x + y \leqslant 200 \\ 4x + 3y \leqslant 750
B.
x + y \leqslant 750 \\ 4x + 3y \leqslant 200
C.
x + y \leqslant 200 \\ 3x + 4y \leqslant 750
D.
x + y \leqslant 750 \\ 3x + 4y \leqslant 200
​
Mathematics
2 answers:
Sati [7]2 years ago
5 0

Answer:

A.\\\\x+y\leq200\\\\4x+3y\leq750

Step-by-step explanation:

x - number of adults

y - number of campers

<em>The room for 200 people</em>: x + y ≤ 200

<em>Each adult costs $4, and each camper costs $3</em>: 4x and 3y

<em>A maximum budget of $750</em>: 4x + 3y ≤ 750

Bess [88]2 years ago
4 0

Answer:

A)  

x + y ≤ 200

4x + 3y ≤ 750

Step-by-step explanation:

i just took the test

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You are a bus driver and are scheduled to depart from Terminal A at 9:18 a.M. And arrive at Terminal B at 10:03 a.M. You arrive
Vesna [10]

Answer:

9 minutes

Step-by-step explanation:

departure terminal A : 9:18 a.M

arrive terminal B : 10:03 a.M

time between B and A : 60 minutes (1 hour) - 18 minutes + 3 minutes = 45 minutes

arrive to a stop at : 9:58 a.M

time between A and stop : 58 minutes - 18 minutes = 40 minutes

5 mores tops including B

a stop - 1 stop next -2 stops next -3 stops next -4 stops next -B

average time between stops: 2 minutes

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a stop 40 minutes + (2 minutes to stop 1 + 1 minute in stop 1 )+ (2 minutes to stop 2 + 1 minute in stop 2 )+ (2 minute to stop 3 + 1 minute in stop 3 )+ (2 minutes to stop 4 + 1 minute in stop 4 )+ (2 minutes to stop B)

40 + 2 + 1 + 2 + 1 + 2+ 1 + 2+1 +2=54 minutes

real time between terminal B and A : 54 minutes

schedule time between terminal B and A : 45 minutes

minutes past scheduled time to arrive at Terminal B: 54 minutes -45 minutes=

9 minutes

8 0
2 years ago
In triangle XYZ, XY = 13, YZ=20, and XZ=25. What is the measure of angle Z to the nearest degree?
anastassius [24]
We can use Law of Cosines to solve for the angle of Z. The solution is shown below:
cos C=(a²+b²-c²)/2ab
cos Z = (yz² + xz² - xy² )/2*yz*xz
cos Z = (20² + 25  - 13²)/2*20*25
cos Z = 856 / 1000
Z=31.13°

The answer is angle 31.13°. 
4 0
2 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina in the shape of an isosceles right triangle with equal sides of length a if
Katyanochek1 [597]

Answer:

Ix = Iy =  a^4 / 12

Ixy = a^4 / 24

Step-by-step explanation:

Solution:-

- Sketch the right angled isosceles triangle as shown in the attachment.

- The density (ρ) is a multivariable function of both coordinates (x and y).

                             ρ ( x, y ) = k*(x^2 + y^2)

Where,  k: coefficient of proportionality.

- The lamina and density are symmetrical across the line y = x.  (see attachment). The center of mass must lie on this line.

- The coordinates of centroid ( xcm and ycm) are given by:

                            x_c_m = \frac{M_y}{m} = \frac{\int \int {x*p(x,y)} \, dA }{\int \int {p(x,y)} \, dA } \\\\y_c_m = \frac{M_x}{m} = \frac{\int \int {y*p(x,y)} \, dA }{\int \int {p(x,y)} \, dA } \\\\m = mass = \int \int {p(x,y)} \, dA

- The coordinates of xcm = ycm, they lie on line y = x.

- Calculate the mass of lamina (m):

                            m = mass = \int \int {p(x,y)} \, dA = \int\limits^0_a \int\limits_0 {k*(x^2 + y^2)} \, dy.dx \\\\m = k\int\limits^a_0 {(x^2y + \frac{y^3}{3}) } \, dx|\limits^a^-^x_0 = k\int\limits^a_0 {(\frac{-4x^3}{3}  + 2ax^2-ax^2+\frac{a^3}{3}) } \, dx\\\\m = k* [ \frac{-x^4}{3} + \frac{2ax^3}{3} - \frac{a^2x^2}{2} +\frac{a^3x}{3}] | \limits^a_0   \\\\m = k* [ \frac{-a^4}{3} + \frac{2a^4}{3} - \frac{a^4}{2} +\frac{a^4}{3}] = \frac{ka^4}{6}

- Calculate the Moment (My):

                     M_y = \int \int {x*p(x,y)} \, dA = \int\limits^0_a \int\limits_0 {k*x*(x^2 + y^2)} \, dy.dx \\\\M_y = k\int\limits^a_0 x*{(x^2y + \frac{y^3}{3}) } \, dx|\limits^a^-^x_0 = k\int\limits^a_0 x*{(\frac{-4x^3}{3}  + 2ax^2-ax^2+\frac{a^3}{3}) } \, dx\\\\M_y = k* [ \frac{-4x^5}{15} + \frac{ax^4}{2} - \frac{a^2x^3}{3} +\frac{a^3x^2}{6}] | \limits^a_0   \\\\M_y = k* [ \frac{-4a^5}{15} + \frac{a^5}{2} - \frac{a^5}{3} +\frac{a^5}{6}] = \frac{ka^5}{15}

- Calculate ( xcm = ycm ):

               xcm = ycm = ( ka^5 /15 ) / ( ka^4/6) = 2a/5        

- Now using the relations for Ix, Iy and I: We have:

                   Ix = bh^3 / 12

                   Iy = hb^3 / 12

Where,        h = b = a .... (Right angle isosceles)

                   Ix = Iy =  a^4 / 12

                   Ixy = b^2h^2 / 24

                   Ixy = a^4 / 24

                                           

8 0
2 years ago
A rectangular garden is to be constructed using a rock wall as one side of the garden and wire fencing for the other three sides
givi [52]

Answer:

x  =  28 m

y  =  14  m

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Step-by-step explanation:

Rectangular garden    A (r ) =  x * y

Let´s call x the side of the rectangle to be constructed with a rock wall, then only one x side of the rectangle will be fencing with wire.

the perimeter of the rectangle is  p  =  2*x  +  2*y    ( but in this particular case only one side x will be fencing with wire

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A(y ) =  56*y  -  2*y²

Tacking derivatives on both sides of the equation we get:

A´(y )  =  56  - 4 * y        A´(y) = 0     56  -  4*y  =  0    4*y  =  56

y =  14 m

and x  =  56  - 2*y    =  56 - 28  = 28 m

Then dimensions of the garden:

x  =  28 m

y  =  14  m

A(max)  =  392 m²

How do we know that the area we found is a local maximum??

We find the second derivative

A´´(y)  = - 4     A´´(y)  <  0   then the function A(y) has a local maximum at y = 14 m

4 0
1 year ago
A large dump truck will empty its full load of stone onto a small plot of land for storage. The rock forms a cone that is 9 feet
Troyanec [42]
48 square feet is the area of thr base.
3 0
2 years ago
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