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lapo4ka [179]
3 years ago
11

Joe is using the function E= 1.89x + 2.89y to minimize his expenses when selling

Mathematics
1 answer:
stepan [7]3 years ago
7 0

Answer:

(d)\ \$13.34

Step-by-step explanation:

Given

E = 1.89x + 2.89y -- Objective function

Constraints:

2x + y > 10

x + 2y > 8

x,y>0

Required

Minimum value of E

To do this, we apply graphical method

See attachment for plots of 2x + y > 10 and x + 2y > 8

From the attached plot, the point that satisfy x,y>0 is:

(x,y) = (4,2)

So, we have:

E = 1.89x + 2.89y

This gives:

E =1.89 * 4 + 2.89 * 2

E =7.56 + 5.78

E =13.34

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Suppose a city with population 800,000 has been growing at a rate of 5% per year. If this rate continues, find the population of
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2,340,208.5 or 2,340,209

Step-by-step explanation:

You could answer this by multiplying the current population of 800,000 by 1.05 (1.05 represents the annual growth rate of population) and do that 22 times.  But that would take a while.

we got 1.05 because the formula says that y = a( 1 + r ) power t

for a we will give it 1

so y = 1 ( 1 + 0.05) power t

it will give us 1.05

to do this faster, you would first calculate (1.05)22and then multiply this by 800,000.

So, 800,000 x (1.05)22 = 2,340,208.5 or 2,340,209

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2 years ago
Across which axis was point E reflected?
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Hope that helps! :D

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3 years ago
Write three decimals that are equivalent
Alexus [3.1K]

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Step-by-step explanation:

4 0
4 years ago
Estimate 1 13 cos(x2) dx 0 using the Trapezoidal Rule and the Midpoint Rule, each with n = 4. (Round your answers to six decimal
babunello [35]

Answer:

(a) 4.152698

(b) 3.215557

Step-by-step explanation:

(a)

\int\limits^{13}_1 {cos(x^2)} \, dx =M_n=$\sum_{n=1}^{\infty} f(m_i)\Delta x $

n=4, so :

Each subinterval has length :

\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3

Therefore the subintervals consist of:

[1,5], [5,9], [9,13]

Now, the midpoints of these subintervals are:

\frac{1+5}{2} =3\\\\\frac{5+9}{2} =7\\\\\frac{9+13}{2} =11

Hence:

M_4= 3*(cos(3^2))+3*(cos(7^2))+3*cos((11^2))\approx 4.152698

(b)

\int\limits^{13}_1 {cos(x^2)} \, dx =T_n=\frac{\Delta x}{2} (f(x_o)+2f(x_1)+2f(x_2)+...+2f(x_n_-_1)+f(x_n))

n=4, so :

Each subinterval has length :

\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3

Therefore the subintervals consist of:

[1,5], [5,9], [9,13]

The endpoints of the subintervals consist of:

5,9

Hence:

T_4= \frac{3}{2}(cos(1^2)+2*cos(5^2)+2*cos(9^2)+cos(13^2)) \approx 3.215557

8 0
3 years ago
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