Suppose that each student at a university has one of 4 expected graduation years and one of 21 majors. How many students must be enrolled to guarantee 2 graduations in the same year and major?

Answer:

[tex] {x}^{ - \frac{5}{3} } \\ \frac{1}{ {x}^{ \frac{5}{3} } } [/tex]

Answer:

The answer is the square root of 8, or 2root2.

Step-by-step explanation:

Use the distance formula.

d = sqrt((-3-(-5))^2 + (-3-(-1)))

d = sqrt(2^2 + 2^2)

d = sqrt(4 + 4)

d = sqrt8

d = sqrt4 * sqrt2

d = 2 * sqrt2

Answer :

Let in Triangle abc,

a=90,

b = (we have to find)

c = (we have to find)

c + 144=180 (linear pair)

c = 180 - 144

c = 36

a + b + c = 180 (Angle sum property)

90 + b + 36 = 180

126 + b = 180

b = 180-126

b = 54

Answer:

b) 8

Step-by-step explanation:

the angles are a linear pair meaning that they have a sum of 180. So 22x+4+180. Solve that and x=8

Answer:

B)     8

Step-by-step explanation:

(13x + 1) + (9x + 3) = 180

Combine like terms

22x + 4 = 180

Subtract 4 from both sides

22x = 176

Divide both sides by 22

x = 8

Answer:

5(n + 3)

Step-by-step explanation:

Factor out 5

5(n + 3)

Answer:

5(n + 3)

should I also give u an explanation

Answer:

Line A equation = y=-9/8x+69/4

Line B equation = y=-10/9x+131/9

Step-by-step explanation:

Answer:

area is 1/2bh

in an euqilateral triangle, all sides are equal so base and height are 12

1/2 x 12 x 12

1/2 x 144= 72 (B)

Answer: Choice C. [tex]36\sqrt{3}[/tex]

This is the same as writing 36*sqrt(3)

===================================================

Work Shown:

x = 12 = side length of equilateral triangle

A = area of equilateral triangle

A = 0.25*sqrt(3)*x^2

A = 0.25*sqrt(3)*12^2

A = 0.25*sqrt(3)*144

A = 0.25*144*sqrt(3)

A = 36*sqrt(3) .... answer is choice C

Side note: the area approximates to 62.354 square units.

Answer:

Step-by-step explanation:

[tex]log\ \frac{1}{x} = - log \ x \\\\log_a \ a = 1\\\\log \ a^x = x \ log \ a[/tex]

(iii)

[tex]m = log_2 \ \frac{1}{32}\\\\m = log_2 \ \ 32^{-1}\\\\m = - 1 \ log_2 32\\\\m = - 1 \ log_2 \ 2^5\\\\m = -1 \times 5 \ log_2 \ 2\\\\m = -5 \ log_2 \ 2\\\\m = -5 \times 1 = -5[/tex]

(iv)

[tex]log_a \ m = 0\\\\m = a^0 = 1[/tex]                       [tex][ \ log _a \ 1 = 0 \ ][/tex]

(vii)

[tex]log_{32} \ 8 = m\\\\8 = 32^m\\\\8^{ \frac{1}{m}} = 32 \\\\(2^{3 })^ { \frac{1}{m}} = 2^{5}\\\\2^{\frac{3}{m}} = 2^5 \\\\\frac{3}{m} = 5\\\\3 = 5 \times m \\\\m = \frac{3}{5}[/tex]

(viii)

[tex]log_5 \ ( 2m + 5 ) = 3\\\\(2m +5 ) = 5^3\\\\2m + 5 = 125\\\\2m = 125 - 5 \\\\2m = 120 \\\\m = 60[/tex]

(x)

[tex]log_{m^3} \ 64 = \frac{2}{3}\\\\64 = (m^3)^{ \frac{2}{3}}\\\\64 = m^{ 3 \times \frac{2}{3}}\\\\64 = m ^2\\\\\sqrt{64} = m \\\\8 = m[/tex]

Answer:

first you have to find the number of ways 3 men can be chosen, then the number of ways 2 women can be chosen, and then you need to multiply these numbers together to get the number of ways because multiplication will show the total arrangements possibilities. use combination since order does not matter.

number ways for 2 out of 10 women total: 10 choose 2= 45

number of ways for 3 out of 9 men total: 9 choose 3= 84

84x45= 3780

3780 total ways

number of ways

Answer:

[tex]h(t)[/tex] is likely a quadratic function.

Based on values in the table, domain of [tex]h(t)[/tex] : [tex]\lbrace 0,\, 1,\, 2,\, 3,\, 4,\, 5,\, 6,\, 7,\, 8\rbrace[/tex]; range of [tex]h(t)\![/tex]: [tex]\lbrace 0,\, 35.1,\, 60.1\, 75.2,\, 80.3,\, 75.3,\, 60.2,\, 35.0 \rbrace[/tex].

Step-by-step explanation:

By the power rule, [tex]h(t)[/tex] is a quadratic function if and only if its first derivative, [tex]h^\prime(t)[/tex], is linear.

In other words, [tex]h(t)[/tex] is quadratic if and only if [tex]h^\prime(t)[/tex] is of the form [tex]a\, x + b[/tex] for some constants [tex]a[/tex] and [tex]b[/tex]. Tables of differences of [tex]h(t)\![/tex] could help approximate  whether [tex]h^\prime(t)\![/tex] is indeed linear.

Make sure that values of [tex]t[/tex] in the first row of the table are equally spaced. Calculate the change in [tex]h(t)[/tex] over each interval:

Consecutive changes to the value of [tex]h(t)[/tex] appears to resemble a line with slope [tex](-10)[/tex] within a margin of [tex]0.2[/tex]. Hence, it is likely that [tex]h(t)\![/tex] is indeed a quadratic function of [tex]t[/tex].

The domain of a function is the set of input values that it accepts. For the [tex]h(t)[/tex] of this question, the domain of [tex]h(t)\![/tex] is the set of values that [tex]t[/tex] could take. These are listed in the first row of this table.

On the other hand, the range of a function is the set of values that it outputs. For the [tex]h(t)[/tex] of this question, these are the values in the second row of the table.

Since both the domain and range of a function are sets, their members are supposed to be unique. For example, the number "[tex]0[/tex]" appears twice in the second row of this table: one for [tex]t = 0[/tex] and the other for [tex]t = 8[/tex]. However, since the range of [tex]h(t)[/tex] is a set, it should include the number [tex]0\![/tex] only once.

Answer:

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

Step-by-step explanation:

                                Profit $    mach. 1      mach. 2

Product 1     ( x₁ )       30             0.5              1

Product 2    ( x₂ )       50             2                  1

Product 3    ( x₃ )       20             0.75             0.5

Machinne 1 require  2 operators

Machine   2 require  1  operator

Amaximum of  100 hours of labor available

Then Objective Function:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Constraints:

1.-Machine 1 hours available  40

In machine 1    L-H  we will need

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

2.-Machine 2   hours available  40

1*x₁  +  1*x₂   + 0.5*x₃   ≤  40

3.-Labor-hours available   100

Machine 1     2*( 0.5*x₁ +  2*x₂  +  0.75*x₃ )

Machine  2       x₁   +   x₂   +  0.5*x₃  

Total labor-hours   :  

2*x₁  +  5*x₂  +  2*x₃  ≤  100

4.- Production requirement:

x₁  ≤  0.5 *( x₁ +  x₂  +  x₃ )     or   0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

5.-Production requirement:

x₃  ≥  0,2 * ( x₁  +  x₂   +  x₃ )  or    -0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

General constraints:

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

The model is:

z  =  30*x₁  +  50*x₂  +  20*x₃      to maximize

Subject to:

0.5*x₁  +  2*x₂  + 0.75*x₃  ≤  40

1*x₁  +  1*x₂   + 0.5*x₃       ≤  40

2*x₁  +  5*x₂  +  2*x₃        ≤  100

0.5*x₁  -  0.5*x₂  -  0.5*x₃  ≤ 0

-0.2*x₁  - 0.2*x₂ + 0.8*x₃   ≥  0

x₁  ≥   0       x₂    ≥   0       x₃     ≥   0           all integers

After 6 iterations with the help of the on-line solver AtomZmaths we find

z (max)  =  1250 $

x₁  = 25    x₂  =  0   x₃  =  25

Answer:

-1

Step-by-step explanation:

Answer:

the answer is 6%

hdhxbxcbxbxszznzj

Answer:

Problem 20)

[tex]\displaystyle \frac{dy}{dx}=(\cos x)^x\left(\ln \cos x-x\tan x\right)[/tex]

Problem 21)

A)

The velocity function is:

[tex]\displaystyle v(t) =2\pi(\cos(2\pi t)-\sin(\pi t))[/tex]

The acceleration function is:

[tex]\displaystyle a(t)=-2\pi^2(2\sin(2\pi t)+\cos(\pi t))[/tex]

B)

[tex]s(0)=2\text{, }v(0) = 2\pi \text{ m/s}\text{, and } a(0) = -2\pi^2\text{ m/s$^2$}[/tex]

Step-by-step explanation:

Problem 20)

We want to differentiate the equation:

[tex]\displaystyle y=\left(\cos x\right)^x[/tex]

We can take the natural log of both sides. This yields:

[tex]\displaystyle \ln y = \ln((\cos x)^x)[/tex]

Since ln(aᵇ) = bln(a):

[tex]\displaystyle \ln y =x\ln \cos x[/tex]

Take the derivative of both sides with respect to x:

[tex]\displaystyle \frac{d}{dx}\left[\ln y \right]=\frac{d}{dx}\left[x \ln \cos x\right][/tex]

Implicitly differentiate the left and use the product rule on the right. Therefore:

[tex]\displaystyle \frac{1}{y}\frac{dy}{dx}=\ln \cos x+x\left(\frac{1}{\cos x}\cdot -\sin(x)\right)[/tex]

Simplify:

[tex]\displaystyle \frac{1}{y}\frac{dy}{dx}=\ln \cos x-\frac{x\sin x}{\cos x}[/tex]

Simplify and multiply both sides by y:

[tex]\displaystyle \frac{dy}{dx}=y\left(\ln \cos x-x \tan x\right)[/tex]

Since y = (cos x)ˣ:

[tex]\displaystyle \frac{dy}{dx}=(\cos x)^x\left(\ln \cos x-x\tan x\right)[/tex]

Problem 21)

We are given the position function of a particle:

[tex]\displaystyle s(t)= \sin (2\pi t)+2\cos(\pi t)[/tex]

A)

Recall that the velocity function is the derivative of the position function. Hence:

[tex]\displaystyle v(t)=s'(t)=\frac{d}{dt}[\sin(2\pi t)+2\cos(\pi t)][/tex]

Differentiate:

[tex]\displaystyle \begin{aligned} v(t) &= 2\pi \cos(2\pi t)-2\pi \sin(\pi t)\\&=2\pi(\cos(2\pi t)-\sin(\pi t))\end{aligned}[/tex]

The acceleration function is the derivative of the velocity function. Hence:

[tex]\displaystyle a(t)=v'(t)=\frac{d}{dt}[2\pi(\cos(2\pi t)-\sin(\pi t))][/tex]

Differentiate:

[tex]\displaystyle \begin{aligned} a(t)&=2\pi[-2\pi\sin(2\pi t)-\pi\cos(\pi t)]\\&=-2\pi^2(2\sin(2\pi t)+\cos(\pi t))\end{aligned}[/tex]

B)

The position at t = 0 will be:

[tex]\displaystyle \begin{aligned} s(0)&=\sin(2\pi(0))+2\cos(\pi(0))\\&=\sin(0)+2\cos(0)\\&=(1)+2(1)\\&=2\end{aligned}[/tex]

The velocity at t = 0 will be:

[tex]\displaystyle \begin{aligned} v(0)&=2\pi(\cos(2\pi (0)-\sin(\pi(0))\\&=2\pi(\cos(0)-\sin(0))\\&=2\pi((1)-(0))\\&=2\pi \text{ m/s}\end{aligned}[/tex]

And the acceleration at t = 0 will be:

[tex]\displaystyle \begin{aligned} a(0) &= -2\pi ^2(2\sin(2\pi(0))+\cos(\pi(0)) \\ & = -2\pi ^2(2\sin(0)+\cos(0)) \\ &= -2\pi ^2(2(0)+(1)) \\ &= -2\pi^2(1) \\ &= -2\pi^2\text{ m/s$^2$} \end{aligned}[/tex]

-4,-5,0,1,2,2,4,5

Hope it works perfectly...

Answer:

The elevation of Lucas Oil Field is of 715 feet.

Step-by-step explanation:

Elevation of Sports Authority Field:

Of 5280 ft.

Compared to Lucas Oil Field:

275ft than seven times the Lucas Oil Field elevation of x.

Equation:

[tex]275 + 7x = 5280[/tex]

What is the elevation of Lucas Oil Field?

Solving the above equation:

[tex]7x = 5280 - 275[/tex]

[tex]7x = 5005[/tex]

[tex]x = \frac{5005}{7}[/tex]

[tex]x = 715[/tex]

The elevation of Lucas Oil Field is of 715 feet.

Answer:

since it's the lhs we are concerned about, i. e., Y axis, so it must be either up or down. now look at the question, it says 8x2 - (1), it means one lower value of y i. e., 1 unit down

Answer:It is shifted 1 unit down.

Step-by-step explanation:

first and last one i think

Answer:

[tex]4 \pi {?}^{2} [/tex]

hope it is helpful to you

9514 1404 393

Answer:

  a.  m = 2; y intercept = (0, 4); y = 2 x + 4

Step-by-step explanation:

The slope can be found using the formula ...

  m = (y2 -y1)/(x2 -x1)

  m = (18 -4)/(7 -0) = 14/7

  m = 2

The y-intercept is the first given point: (0, 4).

The equation for the line in slope-intercept form is

  y = mx + b

Here, we have m=2 and b=4, so the equation is ...

  y = 2x +4

[tex]\implies {\blue {\boxed {\boxed {\purple {\sf { \: (x - 5)(x - 4) }}}}}}[/tex]

[tex]\large\mathfrak{{\pmb{\underline{\blue{Step-by-step\:explanation}}{\blue{:}}}}}[/tex]

[tex] {x}^{2} - 9x + 20[/tex]

[tex] = {x}^{2} - 4x - 5x + 20[/tex]

Taking "[tex]x[/tex]" as common from first two terms and "[tex]5[/tex]" from last two terms, we have

[tex] = x(x - 4) - 5(x - 4)[/tex]

Taking the factor [tex](x-4)[/tex] as common,

[tex] = (x - 5)(x - 4)[/tex]

[tex]\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35♛}}}}}[/tex]

Answer:

The value of k is 14.

Step-by-step explanation:

(x - 2) is a factor of the polynomial

[tex]x - 2 = 0 \rightarrow x = 2[/tex]

This means that [tex]p(2) = 0[/tex]

p(x) = 2x² + 3x - k​

[tex]p(2) = 2(2)^2 + 3(2) - k[/tex]

[tex]0 = 8 + 6 - k[/tex]

[tex]14 - k = 0[/tex]

[tex]k = 14[/tex]

The value of k is 14.

Answer:

3 - 2i

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Algebra II

Step-by-step explanation:

Step 1: Define

Identify

(1 + 3i) + (2 - 5i)

Step 2: Simplify

Answer:

U'/U

Step-by-step explanation:

Answer:

[tex]60x+75[/tex]

Step-by-step explanation:

We want to find the total amount of money (in cents) of the expression:

[tex]\text{$x$ nickels, $(x+3)$ quarters, and $3x$ dimes}[/tex]

Since each nickel is worth five cents, each quarter 25 cents, and each dime ten cents, we can write that:

[tex]\displaystyle \text{Total}=5(x)+25(x+3)+10(3x)[/tex]

Simplify. Distribute:

[tex]T=5x+25x+75+30x[/tex]

Combine like terms. Therefore, the total amount of money (in cents) is represented by:

[tex]T=60x+75[/tex]

Answer:

Well it all started by drawing some equilateral triangles so that they made a regular hexagon: hexagon from unit length triangles. Then we ...

Answer:

8.) 8,780 toffees can be stored in 122 such bags.

9.) 2,460 people are required to control the working of 88 such stations.

Step-by-step explanation:

QUESTION 8:

We know that there are 72 toffees per bag

So if we want to know how many toffees can be stored in 122 such bags:

72 x 122 = 8,784

ROUND IT TO THE NEAREST TENS:

Look at the digit to the right of the tens place: 8,784–>4. 4 and below means that the digit will be repkaced by 0 and the tens place remains the same.

So 8,784 rounded to tens is 8,780

Tye answer for question 9 is

2,460 (do the same thing)

Answer:

?

Step-by-step explanation:

i cant not explian that