Find the maximum and the minimum value of the following objective​ function, and the value of x and y at which they occur. The function F=2x+16y subject to 5x+3y≤37, 3x+5y≤35, x≥0, y≥0
The maximum value of the objective function is ___ when x=___ and y=___

Answer:

[0.25, 2]

Step-by-step explanation:

We have

4t² ≤ 9t-2

subtract 9t-2 from both sides to make this a quadratic

4t²-9t+2 ≤ 0

To solve this, we can solve for 4t²-9t+2=0 and do some guess and check to find which values result in the function being less than 0.

4t²-9t+2=0

We can see that -8 and -1 add up to -9, the coefficient of t, and 4 (the coefficient of t²) and 2 multiply to 8, which is also equal to -8 * -1. Therefore, we can write this as

4t²-8t-t+2=0

4t(t-2)-1(t-2)=0

(4t-1)(t-2)=0

Our zeros are thus t=2 and t = 1/4. Using these zeros, we can set up three zones: t < 1/4, 1/4<t<2, and t>2. We can take one random value from each of these zones and see if it fits the criteria of

4t²-9t+2 ≤ 0

For t<1/4, we can plug in 0. 4(0)²-9(0) + 2 = 2 >0 , so this is not correct

For 1/4<t<2, we can plug 1 in. 4(1)²-9(1) +2 = -3 <0, so this is correct

For t > 2, we can plug 5 in. 4(5)²-9(5) + 2 = 57 > 0, so this is not correct.

Therefore, for 4t^2 ≤ 9t-2 , which can also be written as 4t²-9t+2 ≤ 0, when t is between 1/4 and 2, the inequality is correct. Furthermore, as the sides are equal when t= 1/4 and t=2, this can be written as [0.25, 2]

Answer:

{-6,0,2,4}

Step-by-step explanation:

Answer:

AB = 5.6 cm

Step-by-step explanation:

Reference angle (θ) = 62°

Hypotenuse = 12 cm

Adjacent = AB

Apply the trigonometric ratio formula, CAH, which is:

Cos θ = Adj/Hyp

Plug in the values

Cos 62° = AB/12

12*Cos 62° = AB

5.63365876 = AB

AB = 5.6 cm (1 decimal place)

Answer: hello there here are your answers:

8) B multiplication property of zero

9) C additive identify

1) 8

2) -2a-7

Step-by-step explanation:

1) [tex]\frac{3+u}8^{2} \\u=5 \\so\\ \frac{3+5}8^{2} \\3+5=8^{2} \\8^{2} =\\64 \\\\ 64/8=\\\\\\8 \\there.\\\\\\[/tex]

2)[tex]-2(a-7)\\\\-2(a)(-7)\\\\=-2a+14\\\\\\there[/tex]

Answer:

the answer of r is 8 i hope it will help

Answer:

79.5, 110.5, 141.5, 172.5, 203.5, 234.5

Step-by-step explanation:

Given

The attached distribution

Required

The lower class limits

To do this, we simply subtract 0.5 from the lower interval

From the attached distribution, the lower intervals are:

80.0, 111.0, 142.0, 173,0 .......

So, the lower class limits are:

[tex]80.0-0.5 = 79.5[/tex]

[tex]111.0-0.5 = 110.5[/tex]

[tex]142.0-0.5 = 141.5[/tex]

[tex]173.0-0.5 = 172.5[/tex]

[tex]204.0-0.5 = 203.5[/tex]

[tex]235.0-0.5 = 234.5[/tex]

Answer:

[tex]X=6.67ft/s[/tex]

Step-by-step explanation:

From the question we are told that:

Height of pole [tex]H_p=15[/tex]

Height  of man [tex]h_m=6ft[/tex]

Speed of Man [tex]\triangle a =4ft/s[/tex]

Distance from pole [tex]d=35ft[/tex]

Let

Distance from pole to man=a

Distance from man to shadow =b

Therefore

 [tex]\frac{a+b}{15}=\frac{b}{6}[/tex]

 [tex]6a+6b=15y[/tex]

 [tex]2a=3b[/tex]

Generally the equation for change in velocity is mathematically given by

 [tex]2(\triangle a)=3(\triangle b )[/tex]

 [tex]2*4=3(\triangle b)[/tex]

 [tex]\triangle a=\frac{8}{3}[/tex]

Since

The speed of the shadow is given as

 [tex]X=\triangle b+\triangle a[/tex]

 [tex]X=4+8/3[/tex]

 [tex]X=6.67ft/s[/tex]

Answer:

in a square all sides are equal so x has to equal

128

Hope This Helps!!!

In cylindrical coordinates, W is the set of points

W = {(r, θ, z) : 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π and -2 ≤ z ≤ 5}

(A) Then the integral of f(x, y, z) over W is

[tex]\displaystyle\iiint_W(x^2+y^2+z^2)\,\mathrm dV = \int_0^{2\pi}\int_0^2\int_{-2}^5r(r^2+z^2)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta[/tex]

(B)

[tex]\displaystyle \int_0^{2\pi}\int_0^2\int_{-2}^5r(r^2+z^2)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta = 2\pi \int_0^2\int_{-2}^5(r^3+rz^2)\,\mathrm dz\,\mathrm dr \\\\\\= 2\pi \int_0^2\left(zr^3+\frac13rz^3\right)\bigg|_{z=-2}^{z=5}\,\mathrm dr \\\\\\= 2\pi \int_0^2\left(\frac{133}3r+7r^3\right)\,\mathrm dr \\\\\\= 2\pi \left(\frac{133}6r^2+\frac74r^4\right)\bigg|_{r=0}^{r=2} \\\\\\= 2\pi \left(\frac{110}3\right) = \boxed{\frac{220\pi}3}[/tex]

9514 1404 393

Answer:

  y -9

Step-by-step explanation:

From 4 years ago until 5 years from now, John will age 9 years. That is, his age 4 years ago is 9 years less than it will be in 5 years.

John's age 4 years ago is y-9 years.

Answer:

The expected number of winning balls that Heather draws is 0.3.

Step-by-step explanation:

The balls are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Expected value of the hypergeometric distribution:

The expected value is given by:

[tex]E(X) = \frac{nk}{N}[/tex]

Expected number of blue and green balls:

40 balls, which means that [tex]N = 40[/tex]

2 are chosen, which means that [tex]n = 2[/tex]

25 are blue, which means that [tex]k = 25[/tex]

So

[tex]E(X) = \frac{nk}{N} = \frac{25(2)}{40} = 1.25[/tex]

1.25 balls are expected to be blue and 2 - 1.25 = 0.75 green.

Of the blue balls, 12% are winning.

Of the green balls, 20% are winning.

Calculate the expected number of winning balls that Heather draws.

[tex]E_w = 1.25*0.12 + 0.75*0.2 = 0.3[/tex]

The expected number of winning balls that Heather draws is 0.3.

Answer:

x = 16, or if you didn't want the value for x,

2x + 3 = 35

Step-by-step explanation:

Three more: +3

Twice a number: 2x

Combined:

2x + 3 = 35.

Get rid of the 3 by subtracting it from both sides:

2x = 32

Get rid of the 2 by dividing it from both sides:

x = 16

Answer:

The number is 16.

Step-by-step explanation:

Let the unknown number be x.

Now we translate the sentence into an equation piece by piece.

Three more than twice a number is 35.

2x

Three more than twice a number is 35.

2x + 3

Three more than twice a number is 35.

2x + 3 = 35

Now we solve the equation.

Subtract 3 from both sides.

2x = 32

Divide both sides by 2.

x = 16

Answer: The number is 16.

P.S. Notice that x was a variable that was introduced solely to solve the problem. The original problem is a word problem, not an equation, and has no x in it. The correct answer makes no reference to x since x was used to solve the equation but is not part of the given problem. The person asking the question has no idea what x is. He just wants a number as an answer.

Answer:

11 hours is right answer i hope it will help you

Answer:

C. 17 units

Step-by-step explanation:

Surface area of rectangular prism is given as:

A = 2lw + 2lh + 2wh

A = 930 square units

l = 12 units

h = 9 units

w = ? (We're to find the width)

Plug in the value into the formula

930 = 2*12*w + 2*12*9 + 2*w*9

930 = 24w + 216 + 18w

Add like terms

930 - 216 = 42w

714 = 42w

Divide both sides by 42

714/42 = 42w/42

17 = w

w = 17 units

Answer:

The answer is "47.5354".

Step-by-step explanation:

In the given graph it is a half-circle and a triangle.

So, the diameter of the circle is 6.2 so the radius is 3.1

[tex]\text{Area of a circle}= \pi r^2\\\\\text{Area of a triangle}= \frac{1}{2} b h\\\\[/tex]

Calculating the total area of the shape:

[tex]= \pi r^2+\frac{1}{2} \times b\times h\\\\ = 3.14 \times 3.1^2+\frac{1}{2} \times 6.2 \times 5.6\\\\ = 3.14 \times 3.1^2+\frac{1}{2} \times 6.2 \times 5.6\\\\=3.14 \times 9.61+\frac{1}{2} \times 34.72\\\\=3.14 \times 9.61+\frac{1}{2} \times 34.72\\\\= 30.1754+17.36\\\\=47.5354\\\\[/tex]

Step-by-step explanation:

it will help u

Answer:

[tex]\bar x = 107.11[/tex]

[tex]\sigma_x = 31.07[/tex]

Step-by-step explanation:

See comment for complete question

Given

[tex]x: 97\ 178\ 129\ 90\ 75\ 94\ 116\ 100\ 85[/tex]

Solving (a): The sample mean

This is calculated using:

[tex]\bar x = \frac{\sum x}{n}[/tex]

So, we have:

[tex]\bar x = \frac{97+ 178+ 129+ 90+ 75+ 94+ 116+ 100+ 85}{9}[/tex]

[tex]\bar x = \frac{964}{9}[/tex]

[tex]\bar x = 107.11[/tex]

Solving (b): The sample standard deviation

This is calculated as:

[tex]\sigma_x = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}[/tex]

So, we have:

[tex]\sigma_x = \sqrt{\frac{(97 - 107.11)^2 +.............+ (85- 107.11)^2 }{9-1}}[/tex]

[tex]\sigma_x = \sqrt{\frac{7720.8889}{8}}[/tex]

[tex]\sigma_x = \sqrt{965.1111125}[/tex]

[tex]\sigma_x = 31.07[/tex]

The work is simply the dot product of the force and displacement (which I assume are given in Newtons and meters, respectively):

W = F • d

W = (5i + 3j - 2k) N • ((4i - j + 3k) m - (j + k) m)

W = (5i + 3j - 2k) • (4i - 2j + 2k) Nm

W = (20 - 6 - 4) Nm

W = 10 J

Answer:

10

Step-by-step explanation:

Sorry. I needed to answer this question to get access.

Answer:

I think you have a type.. "the seventh number must be a 1"

there are no 1's in the original set of numbers

Step-by-step explanation:

Answer:

Attached images

It was just easier for me this way.

Let me know in comments if you have questions.

Step-by-step explanation:

Answer:

Step 1

Because the number in front of the bracket is 1 and it is also affected by the negative sign(-),5 is supposed to be negative not positive because (negative by positive is negative)

And since the first step has an error in it,the remaining steps would also be wrong.

(Q.1)

[tex]y = \dfrac{C - e^x}{2x} \implies y' = \dfrac{-2xe^x-2C+2e^x}{4x^2} = \dfrac{-xe^x-C+e^x}{2x^2}[/tex]

Then substituting into the DE gives

[tex]\dfrac{-xe^x-C+e^x}{2x^2} = -\dfrac{2\left(\dfrac{C-e^x}{2x}\right) + e^x}{2x}[/tex]

[tex]\dfrac{-xe^x-C+e^x}{2x^2} = -\dfrac{C-e^x + xe^x}{2x^2}[/tex]

[tex]\dfrac{-xe^x-C+e^x}{2x^2} = \dfrac{-C+e^x - xe^x}{2x^2}[/tex]

and both sides match, so y is indeed a valid solution.

(Q.2)

[tex]\ln\left(y^x\right)\dfrac{\mathrm dy}{\mathrm dx} = 3x^2y[/tex]

This DE is separable, since you can write [tex]\ln\left(y^x\right)=x\ln(y)[/tex]. So you have

[tex]x\ln(y)\dfrac{\mathrm dy}{\mathrm dx} = 3x^2y[/tex]

[tex]\dfrac{\ln(y)}y\,\mathrm dy = 3x\,\mathrm dx[/tex]

Integrate both sides (on the left, the numerator suggests a substitution):

[tex]\dfrac12 \ln^2(y) = \dfrac32 x^2 + C[/tex]

Given y (2) = e ³, we find

[tex]\dfrac12 \ln^2(e^3) = 6 + C[/tex]

[tex]C = \dfrac12 \times3^2 - 6 = -\dfrac32[/tex]

so that the particular solution is

[tex]\dfrac12 \ln^2(y) = \dfrac32 x^2 - \dfrac32[/tex]

[tex]\ln(y) = \pm\sqrt{3x^2 - 3}[/tex]

[tex]\boxed{y = e^{\pm\sqrt{3x^2-3}}}[/tex]

(Q.3) I believe I've already covered in another question you posted.

Answer:

hey hi mate

hope you like it

plz mark it as brainliest

Answer:

main age = total age/total people

if Main age is = 27

[tex]27 = \frac{ \times }{5} [/tex]

and x = 135

Total age is = 135

then main age is 35

[tex][35 = \frac{y}{6} [/tex]

and y = 210

first main age - second main age = age of the person participating

210 - 135 = 75

HAVE A NİCE DAY

Step-by-step explanation:

GREETINGS FROM TURKEY

[tex]\frac{13}{28}[/tex]

Given:

Blue marbles: 3

Reb marbles: 5

Total marbles: 8

Two marbles are selected at random, one after the other with replacement.

Getting the same colour of marbles from the selection means the two marbles are both red or both blue.

(a) Probability of getting 2 marbles being red in colour

i. Probability of picking a red at the first selection:

Number of red marbles ÷ Total number of marbles

=> 5 ÷ 8 = [tex]\frac{5}{8}[/tex]

ii. Probability of picking a red at the second selection:

Number of remaining red marbles ÷ Total number of remaining marbles

Since after the first pick, the marble is not replaced, the remaining red marbles is 4 while the total number of remaining marbles is 7

=> 4 ÷ 7 = [tex]\frac{4}{7}[/tex]

iii. The probability of getting both marbles being red is the product of i and ii above. i.e

[tex]\frac{5}{8}[/tex] x [tex]\frac{4}{7}[/tex] = [tex]\frac{5}{14}[/tex]

(b) Probability of getting 2 marbles being blue in colour

i. Probability of picking a blue at the first selection:

Number of blue marbles ÷ Total number of marbles

=> 3 ÷ 8 = [tex]\frac{3}{8}[/tex]

ii. Probability of picking a blue at the second selection:

Number of remaining blue marbles ÷ Total number of remaining marbles

Since after the first pick, the marble is not replaced, the remaining blue marbles is 2 while the total number of remaining marbles is 7

=> 2 ÷ 7 = [tex]\frac{2}{7}[/tex]

iii. The probability of getting both marbles being blue is the product of i and ii above. i.e

[tex]\frac{3}{8}[/tex] x [tex]\frac{2}{7}[/tex] = [tex]\frac{3}{28}[/tex]

(c) Probability of getting 2 marbles of the same colour.

The probability of getting 2 marbles of same colour is the sum of the probability of getting both marbles of red colour and the probability of getting both marbles as blue colour. i.e The sum of a(iii) and b(iii)

[tex]\frac{5}{14}[/tex] + [tex]\frac{3}{28}[/tex] = [tex]\frac{13}{28}[/tex]

The probability of getting 2 of the same colour is [tex]\frac{13}{28}[/tex]

im struggling with the same one