Find the sum of the arithmetic series where n = 50, a = 9, and t50 = 331.

The equation of the line in the slope-intercept form is: y = -2.00 x + 10.

The slope and provided point are both in the equation for the straight line.

The generic point (x, y) must satisfy the equation if we have a non-vertical land in a that passes through any point(x1, y1) has gradient m.

y-y₁ = m(x-x₁)

It is the necessary equation for a line in the form of a point-slope.

That gift card to the Coffee Café has been provided. = $10

Medium coffee = $2.00

customers

The quantity of coffees a consumer can purchase can be represented using a linear equation.

Slope formula

m = (8 - 10)/(1 - 0)

m = -2

Now consider the line's point-slope shape.

y-y₁ = m(x-x₁)

( y - 10) = -2.00 ( x - 0)

y = -2.00 x + 10

The slope: m = - 2.00

The equation of the line: y = -2.00 x + 10

The y-intercept is 10

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Step-by-step explanation:

this is the answer in the picture

Answer:

9

Step-by-step explanation:

Answer: 9

Step-by-step explanation:

Square Rooting is basically going down to the origins of the number, trying to figure out what times (x) would give you your current number. In order it get 81, you would need  to do 9*9, therefore making the square root 9

Answer: In decimal form, 14.85

Answer in fraction form: 297/20

Step-by-step explanation:

Answer:

2x - 3y = 6

\(x = 3 + \frac{3y}{2}\)

\(y = 2 + \frac{2x}{3}\)

4y - x = 12

\(x = 12 + 4y\\y = 3 + \frac{x}{4}\)

Step-by-step explanation:

Divide both sides of the equation to "solve for x and y."

The domain of the function is Dοmain = [0, 25218]

Range = [0, 45, 90, .....,1134810]

In mathematics, the domain of a function is the set of all possible input values (often referred to as the independent variable).

In this case, the maximum capacity of the stadium is 25218 people, so the domain of the function is [0, 25218], including 0 for the case of no attendance.

As we knοw fοr the given questiοn :

• Dοmain will be the number οf peοple whο will be frοm 0 tο 25218

• Range will be the amοunt οf mοney οr revenue which will be [45×0, 45×1, 45×2, ........45×25218]

Sο,

Dοmain = [0, 25218]

Range = [0, 45, 90, .....,1134810]

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Answer:

\( \boxed{ \bold{ \huge{ \boxed{ \sf{10 {x}^{2} + 2x - 9}}}}}\)

Step-by-step explanation:

\( \sf{3 {x}^{2} + 2x - 5 + ( - 4 + 7 {x}^{2} )}\)

When there is a ( + ) in front of an expression in parentheses, there is no need to change the sign of each term.

That means , the expression remains the same.

⇒\( \sf{3 {x}^{2} + 2x - 5 - 4 + 7 {x}^{2} }\)

Collect like terms

⇒\( \sf{ {10x}^{2} + 2x - 5 - 4}\)

Calculate

⇒\( \sf{10 {x}^{2} + 2x - 9}\)

Hope I helped!

Best regards!!

Answer:

He rode farther on Sunday

135/7 - 139/8 = 107/56 or 1 51/56 miles farther

Answer:D

Step-by-step explanation: I got a hundred on my test.

The values of the variable based on the equation include:

a. x = 32

b. x = 55

c. y = -18

d. m = -70

e. p = 3

f. t = 7

g. x = -11

h. n = -25

i. u = 16

j. v = 27

k. x = -9

l. w = 64

m. y = -78

n. y = 36

o. r = 5

It should be noted that an equation simply means a formula that can be used to express the equality of two expressions.

a. 1/8x = 4

x = 4 × 8

x = 32

b. 1/5x = 11

x = 5 × 11

x = 55

c. 1/9y = -2

y = -2 × 9

y = -18

d. 1/2m = -35

m = 2 × -35

m = -70

e. 6p = 18

p = 18/6

p = 3

f. 12t = 84

t = 84/12

t = 7

g. 3x = -33

x = -33/3

x = -11

h. -4n = 100

n = 100/-4

n = -25

i. -3u = -48

u = -48/-3

u = 16

j. 2v = 54

v = 54 / 2

v = 27

k. -72 = 8x

x = -72 / 8

x = -9

l. 1/4w = 16

w = 16 × 4

w = 64

m. 13 = -1/6y

y = 13 × -6

y = -78

n. -18 = -1/2y

y = -18 × -2

y = 36

o. 1/2r = 5/2

r = 5/2 × 2/1

r = 5

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The general solution of the partial differential equation is:

U(x, t) = Σ [Aₙ*sin((nπ/3)x)]*e^(-(nπ/3)²t)

The partial differential equation (PDE) is given as:

Uxx = (1/2)Ut with the boundary and initial conditions as 0< X <3 with U(0,t) = U(3, t)=0, U(0, t) = 5sin(4πx)

Assume that the solution can be written as a product of two functions:

U(x, t) = X(x)T(t)

Substituting this into the PDE, we have:

X''(x)T(t) = (1/2)X(x)T'(t)

Dividing both sides by X(x)T(t), we get:

(X''(x))/X(x) = (1/2)(T'(t))/T(t)

Since the left side only depends on x and the right side only depends on t, both sides must be equal to a constant, denoted as -λ²:

(X''(x))/X(x) = -λ²

(1/2)(T'(t))/T(t) = -λ²

Simplifying the second equation, we have:

T'(t)/T(t) = -2λ²

Solving the second equation, we find:

T(t) = Ce^(-2λ²t)

Applying the boundary condition U(0, t) = 0, we have:

U(0, t) = X(0)T(t) = 0

Since T(t) ≠ 0, we must have X(0) = 0.

Applying the boundary condition U(3, t) = 0, we have:

U(3, t) = X(3)T(t) = 0

Again, since T(t) ≠ 0, we must have X(3) = 0.

Therefore, we can conclude that X(x) must satisfy the following boundary value problem:

X''(x)/X(x) = -λ²

X(0) = 0

X(3) = 0

The general solution to this ordinary differential equation is given by:

X(x) = Asin(λx) + Bcos(λx)

Applying the initial condition U(x, 0) = 5*sin(4πx), we have:

U(x, 0) = X(x)T(0) = X(x)C

Comparing this with the given initial condition, we can conclude that T(0) = C = 5.

Therefore, the complete solution for U(x, t) is given by:

U(x, t) = Σ [Aₙsin(λₙx) + Bₙcos(λₙx)]*e^(-2(λₙ)²t)

where:

Σ represents the summation over all values of n

λₙ are the eigenvalues obtained from solving the boundary value problem for X(x).

To find the eigenvalues λₙ, we substitute the boundary conditions into the general solution for X(x):

X(0) = 0: Aₙsin(0) + Bₙcos(0) = 0

X(3) = 0: Aₙsin(3λₙ) + Bₙcos(3λₙ) = 0

From the first equation, we have Bₙ = 0.

From the second equation, we have Aₙ*sin(3λₙ) = 0. Since Aₙ ≠ 0, we must have sin(3λₙ) = 0.

This implies that 3λₙ = nπ, where n is an integer.

Therefore, λₙ = (nπ)/3.

Substituting the eigenvalues into the general solution, we have:

U(x, t) = Σ [Aₙ*sin((nπ/3)x)]*e^(-(nπ/3)²t)

where Aₙ are the coefficients that can be determined from the initial condition.

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Answer:

B)65 g/m2

Step-by-step explanation:

I did it on usa test prep

Answer:

C

Step-by-step explanation:

I did it on USATestPrep

The graph of the function \(f(x) = (9/4)x^2\) is a symmetric upward-opening parabola.

Overall, the graph represents a quadratic function with a positive leading coefficient, resulting in an upward-opening parabolic curve. The graph has been attached.

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Answer:

Domain: [-3,3)

Range: [-4,5]

Step-by-step explanation:

Domain is all the x's on the graph (or if you have an equation, all the x's allowed to be used in the equation)

Here the graph exists from -3 to 3, but not including 3. The square brackets means "including" and regular parenthesis means "not including".

The range is all the y-values on the graph (or if you have an equation, its all the y-values that can be produced by the equation) This graph goes from -4 to 5, I wrote it to include -4 and 5. If you think that curve is floating over the -4 and purposefully and absolutely missing it then the range should be (-4,5] But it looks like it touches -4 in the image.

Answer:

1

Step-by-step explanation:

Use the coordinates of the points to determine the slope

slope = Δy/Δx = (8-7)/(5-4) = 1

The value of x is -1 which makes the model true

The equation from the given model will be 5x+6=1

We have to find the value of x

5x+6=1

Subtract 6 from both sides

5x=-5

Divide both sides by 5

x=-1

Hence, the value of x is -1 which makes the model true

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Answer:

m<2 = 100

Step-by-step explanation:

136 <abc Given

136-197= 39 = <1

so 136 - 39 = 197=<2

Answer:

collect like terms

Step-by-step explanation:

collect all the x numbers and the power of 3 numbers to simplify

An example of a boundary value problem is the Sturm-Liouville problem, which entails determining the eigenvalues and eigenfunctions of a differential equation that complies with specific boundary requirements.

The general formula for the Sturm-Liouville problem's solution in x is y(x) = a * f(x) + b * g(x), where a and b are constants and f(x) and g(x) are the differential equation's eigenfunctions. When the differential equation and boundary conditions are solved, the eigenvalues and eigenfunctions are discovered.

For instance, if the differential equation has the following form: -y" + q(x)y = lambda* w(x)y where y is the dependent variable, y" is the second derivative of y, q(x) and w(x) are functions of x, and lambda is the eigenvalue, the boundary conditions can be of the following

form: where L is the length of the interval on which the differential equation is defined, y(0) = 0, and y(L) = 0.

The general solution in x can be expressed in the form: y(x) = a * f(x) + b * g(x), where a and b are constants and f(x) and g(x) are the eigenfunctions of the differential equation. The eigenvalues and eigenfunctions can be discovered by solving this differential equation and the boundary conditions.

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Answer:

136

Step-by-step explanation:

it adds for every time.

Answer:

The  AILP for management to decide  how many classes should be scheduled for each subject daily is

The objective function is mathematically represented as

       \(M  =  P_c  *  a  + P_w  *  b\)

=>    \(M  = 720  a  + 300  b\)

Now the first constraints to this functions is

     \(t_c  *  x  + t_w * y  \le t_a\)

=>  \( 7.5  x  + 3 y  \le 56\)

Another constraints to this function is

     \(k  x  +  u  y  \le  N\)

=>    \(6  x  +  2  y  \le  100\)

Here x and  y are the number of classes    

Step-by-step explanation:

From the question we are told that

   The time required for each ITC class is  \(t_c  = 7.5 \ hours\)

   The profit of each ITC class is  \(P_c  =  \$ 720\)

   The time require for  CWP  class is  \(t_w  =  3 \  hours\)

    The profit for   CWP  class is  \(P_w  =  \$ 300\)

   The total time available  is \(t_a  =  56 \  hours / day\)

   the maximum number of trainee that can be accommodated in a daily basis is   \(N  =  100\)

   The class size limit for  ITC  is  \(k  =6\)

    The class size limit for CWP is  \(u  =12\)

Generally the aim of Hi-Tech is to maximize profit

So the objective function will be a function that maximizes profit

Generally the objective function is mathematically represented as

       \(M  =  P_c  *  a  + P_w  *  b\)

=>    \(M  = 720  a  + 300  b\)

Now the constraints to this functions are

     \(t_c  *  x  + t_w * y  \le t_a\)

=>  \( 7.5  x  + 3 y  \le 56\)

Another constraints to this function is

     \(k  x  +  u  y  \le  N\)

=>    \(6  x  +  2  y  \le  100\)

Here x and  y are the number of classes

Answer:

1. The integral of \(\(\cos(x) e^{\sin(x)} dx\)\) is \(\(e^{\sin(x)} + C\)\), where \(\(C\)\) is the constant of integration.

2. The integral of \(\(\cos^2(x) dx\)\) can be rewritten using the double-angle formula as \(\(\frac{x}{2} + \frac{\sin(2x)}{4} + C\).\)

3. The integral of \(\(\frac{\cos(\sqrt{x})}{\sqrt{x}} dx\)\) is \(\(2\sin(\sqrt{x}) + C\)\).

4. The integral of \(\(\frac{1}{\sqrt[3]{1-7x}} dx\)\)is \(\(-\frac{3(1 - 7x)^{2/3}}{14} + C\)\).

5. The integral of \(\(\frac{1}{2+5x} dx\)\) is \(\(\frac{\ln|2+5x|}{5} + C\)\).

Step-by-step explanation:

Let's break down each integral:

1. Integral of \(\(\cos(x) e^{\sin(x)} dx\)\)

  This is a case of a simple substitution. We can let \(\(u = \sin(x)\)\), then \(\(du = \cos(x) dx\)\). Substituting these into the integral, we get \(\(\int e^u du\)\), which is simply \(\(e^u + C\)\). Substituting back for \(\(u\)\), we get \(\(e^{\sin(x)} + C\)\).

2. Integral of \(\(\cos^2(x) dx\)\)

  The double-angle formula is used here. We know that \(\(\cos^2(x) = \frac{1 + \cos(2x)}{2}\)\). Substituting this into the integral, we get \(\(\int \frac{1 + \cos(2x)}{2} dx\)\), which can be separated into two simpler integrals: \(\(\frac{1}{2} \int dx + \frac{1}{2} \int \cos(2x) dx\)\). The integral of \(\(dx\) is \(x\)\), and the integral of \(\(\cos(2x)\)\) is \(\(\frac{1}{2}\sin(2x)\)\). So, the result is \(\(\frac{x}{2} + \frac{\sin(2x)}{4} + C\)\).

3. Integral of \(\frac{\cos(\sqrt{x})}{\sqrt{x}} dx\)

  This is another case of simple substitution. We can let \(\(u = \sqrt{x}\)\), then \(\(du = \frac{1}{2\sqrt{x}} dx\), or \(2 du = \frac{dx}{\sqrt{x}}\)\). Substituting these into the integral, we get \(\(2 \int \cos(u) du\), which is \(2\sin(u) + C\)\). Substituting back for \(\(u\)\), we get \(\(2\sin(\sqrt{x}) + C\)\).

4. Integral of \(\(\frac{1}{\sqrt[3]{1-7x}} dx\)\)

  Here, we can let \(\(u = 1 - 7x\)\), then \(\(du = -7 dx\)\), or \(\(-\frac{1}{7} du = dx\)\). Substituting these into the integral, we get \(\(-\frac{1}{7} \int u^{-1/3} du\)\), which is \(\(-\frac{3}{7}u^{2/3} + C\)\). Substituting back for \(\(u\)\), we get \(\(-\frac{3(1 - 7x)^{2/3}}{14} + C\)\).

5. Integral of \(\(\frac{1}{2+5x} dx\)\)

  This is a standard form of integral that results in a natural logarithm. The integral of \(\(\frac{1}{a+bx} dx\)\) is \(\(\frac{1}{b} \ln |a+bx| + C\)\). So, the result is \(\(\frac{\ln|2+5x|}{5} + C\)\).


Hope This Helps!

Given that the triangle is a right triangle, the value of sin (n) is √5/5.

Hence, option D)√5/5 is the correct answer.

Trigonometry is simply a branch in mathematics that looks into the relationships between angles and side lengths of a triangle.

Given the data in the question;

To determine the value of sine n, we will use one of six functions of an angle commonly used in trigonometry.

Sine of angle n = Opposite of angle n / Hypotenuse

sin(n) = Opposite / Hypotenuse

Plug in the given values and simplify

sin(n) = Opposite / Hypotenuse

sin (n) = 2 / 2√5

Next, rationalize the denominator.

Cancel out the common factor 2.

sin (n) = 2 / 2√5

sin (n) = 1 /√5

Multiply both the numerator and the denominator by √5.

sin (n)  = (1×√5) / (√5 × √5)

sin (n)  = √5/5

Given that the triangle is a right triangle, the value of sin (n) is √5/5.

Hence, option D)√5/5 is the correct answer.

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Given that 16 families went on a trip and the cost of the trip was Rs. 2,16,352.The amount paid by each family is to be determined by unitary method Hence each family paid Rs.13522

Now, let's solve this by using the method of unitary method. To find the cost of 1 family trip, we will divide the total cost of the trip by the number of families.2,16,352 / 16 = 13,522 So, the cost of the trip per family is Rs. 13,522.Hence, each family paid Rs. 13,522 for the trip.

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Answer:

Step-by-step explanation

1. The total cost of the trip for all 16 families is Rs 2,16,352.
2. To find out how much each family paid, we need to divide the total cost by the number of families: Rs 2,16,352 ÷ 16.
3. When we do the division, we get the result: Rs 13,522.

Now let's check if this result is correct:

1. If each family paid Rs 13,522 for the trip, then the total cost for all 16 families would be: 16 × Rs 13,522 = Rs 2,16,352.
2. This is exactly the same as the total cost given in the problem statement.

So we have shown that each family paid **Rs 13,522** for the trip

Answer:

Basic facts:

1foot =12 inch

Step-by-step explanation:

Vidal's treehouse was 6 feet which are 72 inches

Mateo's treehouse was 48 inches

72-48=24

24/12=2      

2 feet and 0 inches

√89 sana makatulong ito sa inyo