What is the area of this composite shape?

Answer:

real and equal roots

Step-by-step explanation:

because at end of calculation,0=0

Answer:

0 = 2

No solution

Step-by-step explanation:

\(-2(t-4)=10-2t\\\\\mathrm{Expand\:}-2\left(t-4\right):\quad -2t+8\\\\-2t+8=10-2t\\\\\mathrm{Subtract\:}8\mathrm{\:from\:both\:sides}\\\\-2t+8-8=10-2t-8\\\\Simplify\\\\-2t=-2t+2\\\\\mathrm{Add\:}2t\mathrm{\:to\:both\:sides}\\\\-2t+2t=-2t+2+2t\\\\Simplify\\\\0=2\\\\\mathrm{The\:sides\:are\:not\:equal}\\\\\mathrm{No\:Solution}\)

Answer:

See below

Step-by-step explanation:

Complete the square

\(f(x)=x^2-25x\\\\f(x)+(\frac{-25}{2})^2=x^2-25x+(\frac{-25}{2})^2\\\\f(x)+156.25=x^2-25x+156.25\\\\f(x)+156.25=(x-12.5)^2\\\\f(x)=(x-12.5)^2-156.25\\\\f(x)=(x-\frac{25}{2})^2-\frac{625}{156}\)

Identify vertex

\((h,k)\rightarrow(-\frac{25}{2},-\frac{625}{156})\)

Answer:

Step-by-step explanation:

∫41 xh''(x)dx=41[x∫h''(x)dx-∫{1∫h''(x)dx}dx]+c

The value of \(\int\limits^{4}_{1} {x\cdot h''(x)} \, dx\) is 8.5. The choice that represent the best approximation is A.

This integral can be approximated by the following Riemann sum:

\(A = \Sigma\limits_{i=0}^{2} \left\{(x_{i+1}-x_{i})\cdot x_{i}\cdot h''(x_{i})+\frac{1}{2}\cdot (x_{i+1}-x_{i})\cdot [x_{i+1}\cdot h''(x_{i+1})-x_{i}\cdot h''(x_{i})] \right\}\)

\(A = \frac{1}{2} \cdot \Sigma\limits_{i=0}^{2} \left\{(x_{i+1}-x_{i})\cdot [x_{i+1}\cdot h''(x_{i+1})+x_{i}\cdot h''(x_{i})] \right\}\)

Then, the approximate value of the integral is:

\(A = \frac{1}{2}\cdot \{(2-1)\cdot [(2)\cdot 2+(1)\cdot (-5)]+(3-2)\cdot [(3)\cdot 1+(2)\cdot 2]+(4-1)\cdot [(4)\cdot 2+(3)\cdot 1]\}\)

\(A = 8.5\)

The value of \(\int\limits^{4}_{1} {x\cdot h''(x)} \, dx\) is 8.5. The choice that represent the best approximation is A. \(\blacksquare\)

The statement is incomplete and poorly formatted and table is missing. Complete statement is:

Selected values of the twice-differentiable function and its first and second derivatives are:

\(h(1) = 3\), \(h(2) = 6\), \(h(3) = 2\), \(h(4) = 10\)

\(h'(1) = 4\), \(h'(2) = -4\), \(h'(3) = 3\), \(h'(4) = 5\)

\(h''(1) = -5\), \(h''(2) = 2\), \(h''(3) = 1\), \(h''(4) = 2\)

Selected values of the twice-differentiable function \(h\) and its first and second derivatives are given in the table above. What is the value of \(\int\limits^{4}_{1} {x\cdot h''(x)} \, dx\)?

A. 9, B. 13, C. 23, D. 38

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Answer

9

Step-by-step explanation:

Divide 63 by 9 months

7 students per month

63 / 9 = 7

Answer:

a) P(x) = R(x) - C(x)

P(x) = 66x - 22x - 6600 =

         44x - 6600

b) P(2800) = 44(2800) - 6600 =

                   $116600

Answer:36

Step-by-step explanation:

f(x)=12-3*(-8)

=36

Answer: I’m really sorry if you get this wrong and I will feel so bad so please don’t take it from me because I’m only 15 and also doing summer school lol, but if I had to take a quick, random guess i’d say 4. Please wait until someone else responds or look it up! I don’t want to be the reason you get it wrong.

Answer:

Step-by-step explanation:

boody boody

The probability that on exactly one of these three occasions the voter approves of the mayor's work is given as follows:

0.189 = 18.9%.

The mass probability formula, giving the probability of x successes, is of:

\(P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}\)

\(C_{n,x} = \frac{n!}{x!(n-x)!}\)

The parameters are given by:

The values of these parameters in the context of this problem are given as follows:

n = 3, p = 0.7.

Then the probability of exactly one success is calculated as follows:

P(X = 1) = 3!/(1!2!) x 0.7 x (0.3)² = 0.189 = 18.9%.

The proportion of voters that approve the mayor's work is of 70%.

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Answer:

1.9685

Step-by-step explanation:

If you are thinking in inches and if you need to simplify its pretty easy.

To transform is to alter. So, changing any given geometric shape would be considered a geometric transformation.

What does one transformation mean?

A 180° rotation about the origin characterizes the sole transformation. Translation, rotation, reflection, and dilation are the four primary categories of transformation.

The term "Geometric" transformation refers to a group of picture transformations where the geometry of the image is modified without changing the actual pixel values. In general, various actions can be performed on it, but the actual pixel values will not change.

Students get the opportunity to consider fundamental mathematical concepts in novel ways thanks to geometric transformations (e.g., functions whose domain and range are R2). Students can view mathematics as a connected field by using geometric transformations as a framework.

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Δy=v-0.32 and dy = -0.32 .Δy and dy are both used to represent changes in the dependent variable y based on changes in the independent variable x.

Δy represents the change in y (the dependent variable) resulting from a specific change in x (the independent variable). In this case, y = x^4 + 7, x = -2, and Δx = dx = 0.01. Therefore, we need to calculate Δy and dy based on these values.

To calculate Δy, we substitute the given values into the derivative of the function and multiply it by Δx. The derivative of y = x^4 + 7 is dy/dx = 4x^3. Plugging in x = -2, we have dy/dx = 4(-2)^3 = -32. Now, we can calculate Δy by multiplying dy/dx with Δx: Δy = dy/dx * Δx = -32 * 0.01 = -0.32.

On the other hand, dy represents an infinitesimally small change in y due to an infinitesimally small change in x. It is calculated using the derivative of the function with respect to x. In this case, dy = dy/dx * dx = 4x^3 * dx = 4(-2)^3 * 0.01 = -0.32.

Therefore, both Δy and dy in this context have the same value of -0.32. They represent the change in y corresponding to the change in x, but Δy considers a specific change (Δx), while dy represents an infinitesimally small change (dx) based on the derivative of the function.

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Answer:

4

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pls rate ..

ANSWER

yes it is

you have 6 eggs right you remove 2 eggs from the 6 eggs which means you have 4 then the 2 you removed you fry them then the 2 eggs you fry you eat those same 2 eggs and then it leaves you with 4 eggs

Answer:

66

Step-by-step explanation:

The solution for b is y-mx in the equation y=mx+b.

The given equation is y=mx+b

y equal r=to m times of x plus b

We need to solve for b in the equation

To solve we have to isolate b from the equation

Subtract mx from both sides

y-mx=b

Hence, the solution for b is y-mx in the equation y=mx+b.

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Answer:

Step-by-step explanation:

Answer:

see explanation

Step-by-step explanation:

the error in the question is that

sec²x = \(\frac{1}{cos^2x}\) ≠ \(\frac{1}{sin^2x}\) , and

csc²x = \(\frac{1}{sin^2x}\) ≠ \(\frac{1}{cos^2x}\)

have been used incorrectly in simplifying the expression

then

\(\frac{tan^2x+1}{1+cot^2x}\)

= \(\frac{sec^2x}{csc^2x}\)

= \(\frac{\frac{1}{cos^2x} }{\frac{1}{sin^2x} }\)

= \(\frac{1}{cos^2x}\) × \(\frac{sin^2x}{1}\)

= \(\frac{sin^2x}{cos^2x}\)

= tan²x

Answer:

2.5 is the equal epression of 5/2

(a) The economic loading of unit 4 and B are 11.54 MW and 138.46 MW, respectively.

To find the economic loading of the two units when the total load to be supplied by the power station is 150 MW, we need to minimize the total fuel cost. Let x be the power generated by unit 4 and y be the power generated by unit B. Then, we have:

x + y = 150 (total load)

The total fuel cost C is given by:

C = F4(x) + Fb(y)

where F4(x) and Fb(y) are the fuel costs for units 4 and B, respectively. Using the given relations, we have:

F4(x) = 0.06x^2 + 11.4x

Fb(y) = 0.07y^2 + 10y

Substituting x = 150 - y, we get:

C(y) = 0.06(150-y)^2 + 11.4(150-y) + 0.07y^2 + 10y

Expanding and simplifying, we get:

C(y) = 0.013y^2 - 3.6y + 1710

To minimize C(y), we take its derivative with respect to y and set it equal to zero:

dC/dy = 0.026y - 3.6 = 0

y = 138.46 MW

Substituting y back into x = 150 - y, we get:

x = 11.54 MW

(b) The negative value that is  -1297.5 indicates that there is a net decrease in fuel cost per hour if the load is equally shared by the two units.

To find the net increase in fuel cost per hour if the load is equally shared by the two units, we need to calculate the fuel cost for each unit when they generate half of the total load (i.e., 75 MW). Using the given relations, we have:

F4(75) = 0.06(75)^2 + 11.4(75) = 1282.5

Fb(75) = 0.07(75)^2 + 10(75) = 1312.5

Therefore, the total fuel cost is:

C = F4(75) + Fb(75) = 2595

If the load is equally shared by the two units, each unit generates 75/2 = 37.5 MW. The fuel cost for each unit is:

F4(37.5) = 0.06(37.5)^2 + 11.4(37.5) = 641.25

Fb(37.5) = 0.07(37.5)^2 + 10(37.5) = 656.25

Therefore, the total fuel cost is:

C' = F4(37.5) + Fb(37.5) = 1297.5

The net increase in fuel cost per hour is:

ΔC = C' - C = 1297.5 - 2595 = -1297.5

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Answer:

x*2 = 4

x = 4^1/2

x=2

therefore the vertex of the graph will be equal to 2

Answer: 1,250,000

Step-by-step explanation:

Answer:

y=65 and x=25

Step-by-step explanation:

since 65 and x create a 90 degree angle, you subtract 90-65 to get the x (25.) and then since x and y ALSO make a 90 degree angle you subtract 90-25 to get y (65.)

Answer:

The answer is explained below

Step-by-step explanation:

Let x be the price of fertilizer A, y be the price of fertilizer B, and z be the price of fertilizer C

The first combination  costs $384 and consists of 6 liters of fertilizer A, 5 liters of fertilizer B, and 3 liters of fertilizer C. The first combination is given by the equation:

6X + 5Y + 3Z = 384

The second combination consists of 10 liters of A, 2 liters of B, and 6 liters of C, and it costs $516. The second combination is given by the equation:

10X + 2Y + 6Z = 516

The last combination consists of 4 liters of A, 8 liters of B, and 2 liters of C,  with a cost of $368. The last combination is given by the equation:

4X + 8Y + 2Z = 368

In Matrix form it can be represented as:

\(\left[\begin{array}{ccc}6&5&3\\10&2&6\\4&8&2\end{array}\right]\left[\begin{array}{c}X\\Y\\Z\end{array}\right]=\left[\begin{array}{c}384\\516\\368\end{array}\right]\)

\(\left[\begin{array}{c}X\\Y\\Z\end{array}\right]=\left[\begin{array}{ccc}6&5&3\\10&2&6\\4&8&2\end{array}\right]^{-1}\left[\begin{array}{c}384\\516\\368\end{array}\right]\\\\\left[\begin{array}{c}X\\Y\\Z\end{array}\right]=\left[\begin{array}{ccc}1.5714&-0.5&-0.857\\-0.142&0&0.2142\\-2.571&1&1.3571\end{array}\right]\left[\begin{array}{c}384\\516\\368\end{array}\right]\\\\\left[\begin{array}{c}X\\Y\\Z\end{array}\right]=\left[\begin{array}{c}30\\24\\28\end{array}\right]\)

Therefore:

X = $30, Y = $24, Z = $28

The price of fertilizer A = $30 per liter, The price of fertilizer B = $24 per liter and The price of fertilizer c = $28 per liter

Answer:

X = $30, Y = $24, Z = $28

Step-by-step explanation:

PLATO <3

The equation of the circle with center at (-7, 1) and radius of 11 is (x + 7)² + (y - 1)² = 121.

To find the equation of a circle with a given center and radius, we use the standard form equation of a circle:

(x - h)² + (y - k)² = r²

where (h, k) is the center of the circle and r is the radius.

In this case, the center is given as (-7, 1) and the radius is 11. So we substitute these values into the standard form equation and simplify:

(x - (-7))² + (y - 1)² = 11²

(x + 7)² + (y - 1)² = 121

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There are 18 teams in the lacrosse league in the third year. It's important to note that percentages are a way of representing fractions out of 100.

In the first year, the lacrosse league has 20 teams.

In the second year, the number of teams increases by 20%, which is 20/100 x 20 = 4. Therefore, the number of teams in the second year is 20 + 4 = 24.

In the third year, the number of teams decreases by 25% from the second year, which is 25/100 x 24 = 6. Hence, the number of teams in the third year is 24 - 6 = 18.

Therefore, there are 18 teams in the lacrosse league in the third year.

It's important to note that percentages are a way of representing fractions out of 100. In the second year, the 20% increase can also be written as a multiplication factor of 1.2 (1 + 20/100). Similarly, the 25% decrease in the third year can be written as a multiplication factor of 0.75 (1 - 25/100). Using these multiplication factors can make it easier to calculate the number of teams in each year.

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Using the fourth-order Runge-Kutta (RK-4) method with 10 segments, the numerical solution for the ordinary differential equation (ODE) y′sin(x) + ysin(x) = sin(2x) with the initial condition y(1) = 2 is found to be approximately y(0) ≈ 2.68051443.

The fourth-order Runge-Kutta (RK-4) method is a numerical technique commonly used to approximate solutions to ordinary differential equations. In this case, we are given the ODE y′sin(x) + ysin(x) = sin(2x) and the initial condition y(1) = 2, and we are tasked with finding the value of y(0) using RK-4 with 10 segments.

To apply the RK-4 method, we divide the interval [1, 0] into 10 equal segments. Starting from the initial condition, we iteratively compute the value of y at each segment using the RK-4 algorithm. At each step, we calculate the slopes at various points within the segment, taking into account the contributions from the given ODE. Finally, we update the value of y based on the weighted average of these slopes.    

By applying this procedure repeatedly for all the segments, we approximate the value of y(0) to be approximately 2.68051443 using the RK-4 method with 10 segments. This numerical solution provides an estimation for the value of y(0) based on the given ODE and initial condition.  

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Answer:

1st use distance formula for each side of the triangle

then use PT for checking if they are equal then u will get it as equal

then use area of triangle formula u will get the ans explanation:

Answer:

x = 4, -2 1/2

Step-by-step explanation:

factors:

(2x + 5) (x - 4)

set each factor equal to zero:

2x + 5 = 0

2x = -5

x = -5/2

x - 4 = 0

x = 4

Answer: 77

Took the test :)