simplify the expression 6²+4x3-18 divided by 6

Answer:

8/ln(9)

Step-by-step explanation:

Let u=cos(x). Then du=-sin(x) dx.

If x=0, then u=cos(0)=1.

If x=pi/2, then ucos(pi/2)=0.

The intgral is rewriten as

Integrate((9^u (-du) , from u=1 to u=0)

-9^u/ln(9) |u=1..0)

Upperlimit goes in first no matter if it is tinier in value:

(-9^0/ln(9)--9^1/ln(9))

-1/ln(9)+9/ln(9)

8/ln(9)

Answer:

x=-\(\frac{3+-\sqrt{41} }{8}\)

Step-by-step explanation:

We can use the quadratic formula here. We then plug in the values.

The parallelogram bounded by the lines 2x + 3y = 1, 2x + 3y - 3 3x - 2y = 0, 3x - 2y = 4,0 ≠ -4, there is no intersection point between these two lines.

The double integral ∫∫Ώ (x + y)² dxdy over the region Ώ, which is the parallelogram bounded by the lines 2x + 3y = 1, 2x + 3y - 3 = 0, 3x - 2y = 0, and 3x - 2y = 4,  to find the limits of integration for x and y.

To determine the limits of integration,  the intersection points of the given lines.

The intersection of the lines 2x + 3y = 1 and 2x + 3y - 3 = 0:

Subtracting the second equation from the first equation,

(2x + 3y) - (2x + 3y - 3) = 1 - 0

3 = 1

Since 3 ≠ 1, there is no intersection point between these two lines.

find the intersection of the lines 2x + 3y = 1 and 3x - 2y = 0:

Solving the system of equations,

2x + 3y = 1 ...(1)

3x - 2y = 0 ...(2)

Multiplying equation (1) by 3 and equation (2) by 2,

6x + 9y = 3 ...(3)

6x - 4y = 0 ...(4)

Subtracting equation (4) from equation (3),

(6x + 9y) - (6x - 4y) = 3 - 0

13y = 3

Simplifying,

y = 3/13

Substituting this value of y into equation (2),  solve for x:

3x - 2(3/13) = 0

3x = 6/13

x = 2/13

Therefore, the intersection point of the lines 2x + 3y = 1 and 3x - 2y = 0 is (x, y) = (2/13, 3/13).

the intersection of the lines 3x - 2y = 0 and 3x - 2y = 4:

Subtracting the second equation from the first equation,

(3x - 2y) - (3x - 2y) = 0 - 4

0 = -4

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Answer:

y = -15x + 50

Step-by-step explanation:

the pattern in the x side is +1 and the pattern on the y side is -15 so it is 15/1 which = m in y = mx + b

then you multiple -15 times whatever x and add 50 to get y like

y = -15 x 2 + 50 = 20

For the interval estimation of μ when σ is known and the sample is large, the proper distribution to use is Normal distribution.

What is normal distribution?

A probability distribution that is symmetric about the mean is the normal distribution, sometimes referred to as the Gaussian distribution. It demonstrates that data that are close to the mean occur more frequently than data that are far from the mean. The normal distribution appears as a "bell curve" on a graph.

The random variables whose values can locate any unknown value within a defined range are those that follow the normal distribution. for instance, determining the school's student population's height. Here, the distribution may take into account any value, but it will be constrained to, say, a range of 0 to 6 feet.

Hence, For the interval estimation of μ when σ is known and the sample is large, the proper distribution to use is, Normal distribution.

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Answer: 6

Step-by-step explanation:

If her lunch weighs a pound, and her gym clothes weigh two pounds, then she must have six books in her backpack.

1 + 2 + 6 = 9

Answer: mean: 32.375 median: 30 mode: 30 range: 39

Step-by-step explanation:

The exact values of all angles in the interval [0, 360°) that satisfy the given equations are:

data = 45°, 135°, 315°.

To solve the given trigonometric equations, we will consider each equation separately.

tan²(data) - 1 = 0:

First, let's rewrite tan²(data) as (sin(data)/cos(data))²:

(sin(data)/cos(data))² - 1 = 0

Now, we can factor the equation:

(sin²(data) - cos²(data)) / cos²(data) = 0

Using the Pythagorean identity sin²(data) + cos²(data) = 1, we can substitute sin²(data) with 1 - cos²(data):

((1 - cos²(data)) - cos²(data)) / cos²(data) = 0

Simplifying further:

1 - 2cos²(data) = 0

Rearranging the equation:

2cos²(data) - 1 = 0

Now, we solve for cos(data):

cos²(data) = 1/2

cos(data) = ± √(1/2)

cos(data) = ± 1/√2

cos(data) = ± 1/√2 * √2/√2

cos(data) = ± √2/2

From the unit circle, we know that cos(data) = √2/2 corresponds to angles 45° and 315° in the interval [0, 360°). Therefore, the solutions for data are:

data = 45° and data = 315°.

cos(data)sin(data) = 1:

Since cos(data) ≠ 0 (otherwise the equation wouldn't hold), we can divide both sides by cos(data):

sin(data) = 1/cos(data)

sin(data) = 1/√2

From the unit circle, we know that sin(data) = 1/√2 corresponds to angles 45° and 135° in the interval [0, 360°). Therefore, the solutions for data are:

data = 45° and data = 135°.

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Answer: angle 2=40

Step-by-step explanation:

angle 1 equal to 40

angle A= 180-40=140

angle D is the same as angle A, both will equal 40

angle 2 is the same as angle 1 so angle 2 will equal to 40

Answer:

x = 64

Step-by-step explanation:

The given polygon is a quadrilateral — a closed figure with 4 sides. We know that the measures of the interior angles of a quadrilateral sum to 360º (e.g., the measures of a rectangle's four angles: 90º + 90º + 90º + 90º = 360º). Using this information, we can construct an equation with xº and the rest of the angle measures given in the diagram.

xº + 100º + 130º + 66º = 360º

To solve this equation, we can subtract (100 + 130 + 66)º from both sides.

xº = 360º - (100 + 130 + 66)º

xº = 360º - 296º

xº = 64º

x = 64

Answer:

m<2 = 65

m<3 = 115

m<4 = 65

m<5 = 115

m<6 = 65

m<7 = 115

m<8 = 65

Step-by-step explanation:

Answer:

The number of miles increases as time increases

The number of hours causes a change in the number of miles ridden

The Variable h is the independent variable

Step-by-step explanation:

Answer:

y = -3

Step-by-step explanation:

y=2cos(x/4)−3

range of cos α : -1 ≤ cos α ≤ 1

-2 ≤ 2cos(x/4) ≤ 2

-5 ≤ 2cos(x/4)−3 ≤ -1       (-2-3= -5 and 2-3= -1) ....Range ( 2 x Amplitude)

midline: midline of y = -1 and y = -5        y = -3

The equatiion where C is the constant of integration.To evaluate the indefinite integral ∫(3e^x + 4x^5 - x^3/4 + 1)dx, we can integrate each term separately.

∫3e^x dx = 3∫e^x dx = 3e^x + C₁

∫4x^5 dx = 4∫x^5 dx = 4 * (1/6)x^6 + C₂ = (2/3)x^6 + C₂

∫-x^3/4 dx = (-1/4)∫x^3 dx = (-1/4) * (1/4)x^4 + C₃ = (-1/16)x^4 + C₃

∫1 dx = x + C₄

Now, we can combine these results to obtain the final answer:

∫(3e^x + 4x^5 - x^3/4 + 1)dx = 3e^x + (2/3)x^6 - (1/16)x^4 + x + C

Therefore, the indefinite integral of (3e^x + 4x^5 - x^3/4 + 1)dx is:

∫(3e^x + 4x^5 - x^3/4 + 1)dx = 3e^x + (2/3)x^6 - (1/16)x^4 + x + C

where C is the constant of integration.

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Answer:

8

Step-by-step explanation:

Plug in the value of k into your expression and solve.

40-4k^3

40-4(2)^3

40-4(8)

40-32

8

Step-by-step explanation:(4/100x)+(6/100y)=178---------(i)

2х+3y=8900-----------(i)

x+y=3600---------(ii)

where x is phone

y is pen

The system of equations that describes the situation are 4x + 6y = 17800 and x + y = 3600.

A linear equation is an algebraic equation of degree one. In general, the variable or the variables(in the case of a linear equation in two variables) the variables are x and y.

Let the cost of 1 phone is x dollars and cost of 1 computer is y dollars.

Cameron earns a 4% commission on the total dollar amount of all phone sales he makes and earns a 6% commission on all computer sales.

∴ (4/100)x + (6/100)y = 178

4x + 6y = 17800...(i)

Cameron made a total of $3600 in sales.

∴ x + y = 3600...(ii)

Now to solve the equations we'll multiply eqn(ii) by 4 or 6 to cancel out x or y, and then we'll substitute the numerical value of x or y in any of these equations.

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Answer:

(4+√142)/(3) or x=(4-√142) /3

Step-by-step explanation:

(4+√142)/(3) or x=4-√142 /3

The multiplicative inverse property of the questions that we have here are:

This is the term that is used in Mathematics to show that we are to multiply a fraction by the inverse of that fraction.

A good example of this would be the form that is written as:

1/a would have the inverse property written as a/1

such that 1/a *a/1 = 1

In the same way,

we have

9/7 * 7 / 9 = 1

4/3 * 3/4 = 1

5 1/3 = 16 / 3 * 3/ 16 = 1

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Adjust the parameters as needed, such as the step size (h) and the final x-value (xn), and run the code to obtain the solution for y(x).

The resulting plot will show the solution curve.

To solve the given set of differential equations using the Runge-Kutta method in MATLAB, we need to convert the second-order differential equations into a system of first-order differential equations.

Let's define new variables:

y = y(x)

z = dz/dx

Now, we have the following system of first-order differential equations:

dy/dx = z (1)

dz/dx = -k/(2y) (2)

To apply the Runge-Kutta method, we need to discretize the domain of x. Let's assume a step size h for the discretization. We'll start at x = 0 and proceed until x = infinite.

The general formula for the fourth-order Runge-Kutta method is as follows:

k₁ = h f(xn, yn, zn)

k₂ = h f(xn + h/2, yn + k₁/2, zn + l₁/2)

k₃ = h f(xn + h/2, yn + k₂/2, zn + l₂/2)

k₄ = h f(xn + h, yn + k₃, zn + l₃)

yn+1 = yn + (k₁ + 2k₂ + 2k₃ + k₄)/6

zn+1 = zn + (l₁ + 2l₂ + 2l₃ + l₄)/6

where f(x, y, z) represents the right-hand side of equations (1) and (2).

We can now write the MATLAB code to solve the differential equations using the Runge-Kutta method:

function [x, y, z] = rungeKuttaMethod()

   % Parameters

   k = 1;              % Constant k

   h = 0.01;           % Step size

   x0 = 0;             % Initial x

   xn = 10;            % Final x (adjust as needed)

   n = (xn - x0) / h;  % Number of steps

   % Initialize arrays

   x = zeros(1, n+1);

   y = zeros(1, n+1);

   z = zeros(1, n+1);

   % Initial conditions

   x(1) = x0;

   y(1) = 0;

   z(1) = 0;

   % Runge-Kutta method

   for i = 1:n

       k1 = h * f(x(i), y(i), z(i));

       l1 = h * g(x(i), y(i));

       k2 = h * f(x(i) + h/2, y(i) + k1/2, z(i) + l1/2);

       l2 = h * g(x(i) + h/2, y(i) + k1/2);

       k3 = h * f(x(i) + h/2, y(i) + k2/2, z(i) + l2/2);

       l3 = h * g(x(i) + h/2, y(i) + k2/2);

       k4 = h * f(x(i) + h, y(i) + k3, z(i) + l3);

       l4 = h * g(x(i) + h, y(i) + k3);

       y(i+1) = y(i) + (k1 + 2*k2 + 2*k3 + k4) / 6;

       z(i+1) = z(i) + (l1 + 2*l2 + 2*l3 + l4) / 6;

       x(i+1) = x(i) + h;

   end

   % Plotting

   plot(x, y);

   xlabel('x');

   ylabel('y');

   title('Solution y(x)');

end

function dydx = f(x, y, z)

   dydx = z;

end

function dzdx = g(x, y)

   dzdx = -k / (2*y);

end

% Call the function to solve the differential equations

[x, y, z] = rungeKuttaMethod();

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The p-value for study B is 0.98.

With the same set of data, two studies were conducted.

Study A was a two-sided test while study B was a one-sided test.

The p-value of the test in study A is 0.040.

If the test statistic from your sample has a negative value, the p-value for a two-sided test is equal to twice the p-value for the lower-tailed p-value. If the test statistic from your sample has a positive value, the p-value is two times that of the upper-tailed p-value.

So, in conclusion:

The one-tail p-value equals one minus half the two-tailed value.

The two-tail p-value is twice the one-tail p-value.

As study A is a two-sided test with p-value = 0.04.

The p-value of study B will be:

p = 1 - ( 1/2 ) × 0.04

p = 1 - 0.02

p = 0.98

The p-value for study B is 0.8.

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The calculated concentration of [PO43-] at equilibrium is approximately 7.28 x 10^20 M.

To determine the concentration of [PO43-] at equilibrium, we need to consider the reaction between Pb(NO3)2 and Na3PO4. The balanced equation for the reaction is:

3Pb(NO3)2 + 2Na3PO4 -> Pb3(PO4)2 + 6NaNO3

Given the initial volumes and concentrations of the solutions, we can calculate the moles of Pb(NO3)2 and Na3PO4 used in the reaction. From there, we can determine the moles of Pb3(PO4)2 formed and the resulting concentration of [PO43-] at equilibrium.

Using the volumes and concentrations provided (75 ml of 0.80 M Pb(NO3)2 and 10 ml of 0.10 M Na3PO4), we find that the moles of Pb(NO3)2 used is 0.06 mol and the moles of Na3PO4 used is 0.001 mol. Since the stoichiometric ratio between Pb3(PO4)2 and PO43- is 2:1, we can calculate that 0.03 mol of Pb3(PO4)2 is formed.

To determine the concentration of [PO43-] at equilibrium, we divide the moles of Pb3(PO4)2 formed by the total volume of the solution (85 ml), resulting in a concentration of approximately 7.28 x 10^20 M.

Therefore, at equilibrium, the concentration of [PO43-] is approximately 7.28 x 10^20 M. This value is obtained by considering the stoichiometry of the reaction and the initial concentrations and volumes of the solutions.

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Answer:

32 million / x million year 3 * 100%

Step-by-step explanation:

Since they mention that the total value of azco sales remain the same as year 2, therefore the earnings for year 2 and 3 are the same.

Which means that the percentage of sales profits would be 32 million dollars divided by the profits of year 3, which we do not know because the statement does not tell them but they tell us that they do not change with respect to year 2.

Therefore the percentage would be:

32 million / x million year 3 * 100%

Answer:

10

Step-by-step explanation:

Answer:

10 km

Step-by-step explanation:

total no of km covered=400

total no of trips=40

distance in each trip= 400/40

=10km

Therefore, we have shown that \(3xy + 4 \)is odd when x and y are both odd integers.

To show that 3xy + 4 is odd, we will consider the fact that x and y are both odd integers since they are not even.

Let x = 2a + 1 and y = 2b + 1, where a and b are integers.

Now, we can substitute these expressions into the given equation:

\(3xy + 4 = 3(2a + 1)(2b + 1) + 4 \)

= \( 3(4ab + 2a + 2b + 1) + 4 \)

= \( 12ab + 6a + 6b + 3 + 4 \)

=  \(2(6ab + 3a + 3b) + 7 \)

The term (6ab + 3a + 3b) is an integer, let's call it c. Thus, our equation becomes:

2c + 7

Since 2c is even, and 7 is odd, the sum (2c + 7) is odd.

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The union of events A and B, denoted by A∪B, contains all outcomes that are in A or B. Option B

In probability theory, events A and B represent sets of outcomes from a given experiment or sample space. The union of two events A and B is the set of all outcomes that belong to either A or B or both. It is formed by combining the elements from both sets without repetition.

Mathematically, the union of events A and B is defined as:

A∪B = {x : x ∈ A or x ∈ B}

This means that any outcome x that is in event A or event B (or both) will be included in the union of A and B.

To illustrate this concept, consider the following example:

Event A: Rolling an even number on a fair six-sided die

A = {2, 4, 6}

Event B: Rolling a number greater than 4 on a fair six-sided die

B = {5, 6}

The union of A and B, denoted by A∪B, will contain all outcomes that are in A or B:

A∪B = {2, 4, 5, 6}

Therefore, option B is the correct choice. The union of events A and B contains all outcomes that are in A or B.

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Answer:

19

Step-by-step explanation:

(8*19+5*18)/(8+5)=242/13=18 8/13

round

19

(a) the Riemann sum using left endpoints and n equals 4 is L₄ = 1620.

(b) the Riemann sum using right endpoints and n equals 4 is R₄ = 2916.

What is riemann sum?

A Riemann sum is a method used in calculus to approximate the area under a curve or the value of an integral. It involves dividing the region under the curve into smaller subintervals and approximating the area of each subinterval using a specific rule.

To find the Riemann sum using left endpoints and n equals 4, we need to divide the interval [2, 14] into four equal subintervals. The width of each subinterval, denoted as Δx, is calculated as (b - a) / n, where b is the upper limit (14) and a is the lower limit (2).

So, Δx = (14 - 2) / 4 = 12 / 4 = 3.

Now, we evaluate the function at the left endpoint of each subinterval and multiply it by the width (Δx).

The left endpoints for n = 4 are: x₀ = 2, x₁ = 5, x₂ = 8, x₃ = 11.

The Riemann sum using left endpoints is given by:

L₄ = f(x₀) Δx + f(x₁) Δx + f(x₂) Δx + f(x₃) Δx.

Now, let's substitute the given function f(x) = 2x² + 4x + 2 into the Riemann sum equation:

L₄ = (2x₀² + 4x₀ + 2) Δx + (2x₁² + 4x₁ + 2) Δx + (2x₂² + 4x₂ + 2) Δx + (2x₃² + 4x₃ + 2) Δx.

L₄ = (2(2)² + 4(2) + 2) (3) + (2(5)² + 4(5) + 2) (3) + (2(8)² + 4(8) + 2) (3) + (2(11)² + 4(11) + 2) (3).

L₄ = (8 + 8 + 2) (3) + (50 + 20 + 2) (3) + (128 + 32 + 2) (3) + (242 + 44 + 2) (3).

L₄ = (18)(3) + (72)(3) + (162)(3) + (288)(3).

L₄ = 54 + 216 + 486 + 864.

L₄ = 1620.

Therefore, the Riemann sum using left endpoints and n equals 4 is L₄ = 1620.

To find the Riemann sum using right endpoints and n equals 4, the process is similar. However, this time we evaluate the function at the right endpoint of each subinterval.

The right endpoints for n = 4 are: x₁ = 5, x₂ = 8, x₃ = 11, x₄ = 14.

The Riemann sum using right endpoints is given by:

R₄ = f(x₁) Δx + f(x₂) Δx + f(x₃) Δx + f(x₄) Δx.

Substituting the function into the Riemann sum equation:

R₄ = (2x₁² + 4x₁ + 2) Δx + (2x₂² + 4x₂ + 2) Δx + (2x₃² + 4x₃ + 2) Δx + (2x₄² + 4x₄ + 2) Δx.

R₄ = (2(5)² + 4(5) + 2) (3) + (2(8)² + 4(8) + 2) (3) + (2(11)² + 4(11) + 2) (3) + (2(14)² + 4(14) + 2) (3).

R₄ = (50 + 20 + 2) (3) + (128 + 32 + 2) (3) + (242 + 44 + 2) (3) + (392 + 56 + 2) (3).

R₄ = (72)(3) + (162)(3) + (288)(3) + (450)(3).

R₄ = 216 + 486 + 864 + 1350.

R₄ = 2916.

Therefore, the Riemann sum using right endpoints and n equals 4 is R₄ = 2916.

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Answer:

Step-by-step explanation:

When doing the calculations it is seen that the ordered pair that would not be a solution would be number 2: (4,9) because if the X is 4 then the Y would be 13 and not 9.

The answers after calculations↓

(0,5)

(4,13)

(-2,1)

(2,9)