Using the Pythagorean Identity, we can find the value of cos θ when sin θ is given as √2/2. By applying the Pythagorean Identity, we can determine that cos θ is also √2/2.
The Pythagorean Identity is a fundamental trigonometric identity that relates the three basic trigonometric functions: sine (sin), cosine (cos), and tangent (tan). It states that sin²θ + cos²θ = 1.
Given that sin θ = √2/2, we can substitute this value into the Pythagorean Identity equation: (√2/2)² + cos²θ = 1.
Simplifying the equation, we have (2/4) + cos²θ = 1, which becomes 1/2 + cos²θ = 1.
Rearranging the equation, we get cos²θ = 1 - 1/2, which is equal to 1/2.
Taking the square root of both sides, we find that cos θ is equal to ±√(1/2). Since cosine is positive in the given context of the angle terminating, we take the positive square root and conclude that
cos θ = √2/2.
In summary, using the Pythagorean Identity, we determine that cos θ is also √2/2 when sin θ is given as √2/2 and the angle A terminates.
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What is the radius of this cone?
Answer:
It's 21.98 cm
Write the differential equation that governs the motion of the damped mass-spring system, and find the solution that satisfies the initial conditions specified. Units are mks; y is the damping coefficient, with units of kg/sec. m = 0.2 y = 2, and k = 5.8 Initial displacement is -1 and initial velocity is - 1 x+ x' + x = 0 X(t) = m = 0.1 y = 0.8, and k= 2.5 Initial displacement is 3 and initial velocity is -15 x+ x' + x = 0 x(t) =__.
We have c1 = 1 and c2 = -1.the final solution that satisfies the initial conditions is:
x(t) = e⁽⁻ᵗ⁾ * (cos(0.
the differential equation that governs the motion of a damped mass-spring system is given by:
mx'' + yx' + kx = 0
where m is the mass of the object, y is the damping coefficient (in kg/sec), k is the spring constant (in n/m), x is the displacement of the object from its equilibrium position, and t is time.
in this case, the values are m = 0.2 kg, y = 2 kg/sec, and k = 5.8 n/m.
to find the solution that satisfies the initial conditions, we are given the initial displacement and initial velocity. the initial conditions are:
x(0) = -1 (initial displacement)x'(0) = -1 (initial velocity)
to solve the differential equation and find x(t), we can use the method of solving second-order linear homogeneous differential equations. the characteristic equation corresponding to the given differential equation is:
mλ² + yλ + k = 0
substituting the values of m, y, and k, we have:
0.2λ² + 2λ + 5.8 = 0
using the quadratic formula, we can find the roots of this equation:
λ = (-2 ± √(2² - 4 * 0.2 * 5.8)) / (2 * 0.2)
= (-2 ± √(4 - 4 * 0.2 * 5.8)) / 0.4 = (-2 ± √(4 - 4.64)) / 0.4
= (-2 ± √(-0.64)) / 0.4 = (-2 ± √(-0.16)) / 0.4
since the discriminant is negative, the roots of the characteristic equation are complex conjugate. let's write them in the form λ = α ± βi:
λ = -1 ± 0.2i
using the form of the roots, the solution of the differential equation is:
x(t) = e⁽αᵗ⁾ * (c1cos(βt) + c2sin(βt))
where c1 and c2 are constants to be determined from the initial conditions.
plugging in the values of α and β, we have:
x(t) = e⁽⁻ᵗ⁾ * (c1cos(0.2t) + c2sin(0.2t))
to determine the values of c1 and c2, we can use the initial conditions:
x(0) = c1cos(0) + c2sin(0) = c1
-1 = c1
x'(t) = -e⁽⁻ᵗ⁾ * (c1cos(0.2t) + c2sin(0.2t)) + 0.2e⁽⁻ᵗ⁾ * (-c1sin(0.2t) + c2cos(0.2t))
x'(0) = -c1cos(0) + c2sin(0) = -c1-1 = -c1 2t) - sin(0.2t))
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R 110° S P In the diagram, O is the centre of the circle. P, Q, R and S are points on the circle. Angle ROP = 110° Calculate the size of angle RSP.
Answer:
angle RSP = 125°
Step-by-step explanation:
You want the measure of inscribed angle RSP that subtends long arc RQP when short arc RP is 110°.
Inscribed angleThe inscribed angle is half the measure of the arc it subtends. That long arc is the difference between 360° and the short arc RSP of 110°.
angle RSP = (1/2)(360° -110°) = 250°/2
angle RSP = 125°
__
Additional comment
Angle RSP is the supplement to inscribed angle RQP, which is half the measure of central angle ROP. 180° -(110°/2) = 125° = angle RSP, as above.
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Find the function f(x) such that f′(x) = 12x^2 − 4x − 7 y f(−1) = 2.
For a company that produces x number of units of an item per month, the marginal cost function is C′(x) = 0.8x+92. The
The slope of the line represented by the given parametric equations x = 8 + 2r and y = 1 is equal to the derivative dx/dr, which is 2.
To find the of the line represented by the given parametric equations, we slope can differentiate each equation with respect to the parameter.
Let's analyze the first set of parametric equations:
1. x = 8 + 2r
2. y = 1
Differentiating equation (1) with respect to r:
dx/dr = 2
This tells us that the rate of change of x with respect to the parameter r is 2.
Differentiating equation (2) with respect to r:
dy/dr = 0
This indicates that y is a constant and does not change with r.
Since the derivative of y with respect to r is zero, the value of y remains constant.
Hence, the slope of the line is 2.
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the midpoint of a class is the sum of its lower and upper limits divided by two. T/F
The midpoint of a class is indeed calculated by adding the lower and upper limits of the class and then dividing the sum by two. The statement is true.
This concept is commonly used in statistics and data analysis.
In statistical data, classes are often created to group data points within a range. Each class has a lower limit and an upper limit, defining the range of values it encompasses. The midpoint is a representative value within the class that provides a measure of its central tendency.
To calculate the midpoint, the lower and upper limits of the class are added together, resulting in the total range of the class. Dividing this sum by two gives the midpoint. It is important to note that the midpoint is not necessarily an actual data point; rather, it is a statistical measure used to represent the central value within a class.
For example, suppose we have a class with a lower limit of 10 and an upper limit of 20. Adding these values gives us 30, and dividing by two yields a midpoint of 15. This means that, for the purposes of analysis, we can consider the midpoint of the class as 15.
By calculating midpoints for each class in a data set, statisticians can summarize and analyze data in a more manageable and meaningful way. Midpoints help provide a sense of the central tendencies and distributions within the data, facilitating further statistical analysis and interpretation.
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Consider the integral [ 1 + x² If the integral is divergent, type an upper-case "D". Otherwise, evaluate the integral. 00 9x dx
Consider the integral -1 I -2e-21 dt If the integral is divergent, typ
For the first integral, we need to evaluate the integral of 9x with respect to x over the interval [0, 9].
The antiderivative of 9x is (9/2)x^2, so we can evaluate the definite integral as follows:
∫[0, 9] 9x dx = [ (9/2)x^2 ] [0, 9]
= (9/2)(9)^2 - (9/2)(0)^2
= (9/2)(81)
= 364.5
Therefore, the value of the integral is 364.5.
For the second integral, we need to evaluate the integral of -2e^(-21t) with respect to t. The antiderivative of -2e^(-21t) is (-2/21)e^(-21t), so we can evaluate the indefinite integral as follows:
∫ -2e^(-21t) dt = (-2/21)e^(-21t) + C
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7
For a certain drug, the rate of reaction in appropriate units is
given by R'(t) == +2, where t is measured in hours after the drug
is administered. Find the total reaction to the drug from t = 2 to
Solve the problem. For a certain drug, the rate of reaction in appropriate units is given by R'(t)=+, where t is measured in hours after the drug is administered. Find the total reaction to the drug f
The total reaction to the drug from t=2 to t=S is: 2S - 4 for a certain drug where the rate of reaction in appropriate units is given by R'(t) == +2, where t is measured in hours after the drug is administered.
The rate of reaction is given by R'(t) = +2, which means that the rate of reaction is constant and equal to 2 units per hour.
To find the total reaction over a certain time period, we need to integrate the rate of reaction with respect to time over that time period. The integral of a constant function is equal to the product of the constant and the interval over which it is being integrated.
To find the total reaction to the drug from t=2 to t=S, we need to integrate the rate of reaction function R'(t) with respect to t.
∫[R'(t)]dt = ∫[2]dt = 2t + C
To find the constant of integration, we can use the initial condition that at t=0, the reaction to the drug is 0.
So, 2(0) + C = 0
C = 0
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A circle has a radius of 27 cm. What is the area of the circle? If necessary, round your answer to the hundredths place.
Answer: The area of the circle is approximately 2290.22 cm^2 when rounded to the hundredths place.
When an archer shoots an arrow, the arrow follows a parabolic path through the
air. That parabola can be represented by a quadratic function, where x represents
the arrow's horizontal distance from the archer, and y represents the arrow's height
above the ground.
1. Philip releases an arrow from a shoulder height of 4 t
The arrow reaches a maximum height of 40 ft before
it begins to fall it lands on the ground about 160 ft
away from where he is standing. Philip models this
situation with a quadratic function. What is a zero of
the function? What is the yintercept of the graph
of the function? Construct a rough graph of the situation based on your
findings, where y represents the vertical height, in feet, and x represents the
horizontal distance, in feet.
2. Philip shoots an arrow from the top of a hill onto the ground. The flight of the
arrow can be modeled with the quadratic function y=-52² +15x+90, where
y represents the vertical height of the arrow, in meters, and x represents the
time, in seconds. Use factoring to find the positive zero of the function. What
does this zero represent? Then find the maximum value of the function by
completing the square. What does the maximum value represent? Explain
each step in your answer.
Answer:
a
Step-by-step explanation:
b
What is the name of the line of reflection that maps the pre-image to the image? y = x y-axis x-axis x = 1
The Coordinates of the pre-image and image or a diagram that shows the pre-image and image so that we can find the corresponding points and draw the line of reflection.
None of the given options (y = x, y-axis, x-axis, x = 1) is the name of the line of reflection that maps the pre-image to the image. The line of reflection is a line across which the pre-image is reflected or "flipped" to form the image. It is also known as the "mirror line" or "axis of reflection."
To find the line of reflection given a pre-image and its image, we can draw a line connecting each corresponding point of the pre-image and image, and then find the perpendicular bisector of each line segment. The intersection of these perpendicular bisectors is the line of reflection.
Alternatively, if we know the coordinates of the pre-image and image, we can use the midpoint formula to find the midpoint of each line segment connecting corresponding points, and then find the equation of the line that passes through these midpoints. This line will also be the line of reflection.To determine the line of reflection that maps the pre-image to the image, we need more information than what is given in the options provided. We need either the coordinates of the pre-image and image or a diagram that shows the pre-image and image so that we can find the corresponding points and draw the line of reflection.
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Answer:
Y- axis is the correct answer
Step-by-step explanation:
Correct on edge
pls help me quick erg
The distance Chyna ran is 7 1 / 2 kilometres.
How to find the distance Chyna ran?Ronaldo ran 9 / 10 fewer kilometres than Chyna ran. Ronaldo ran 6 3 / 5
kilometres
Therefore, the distance Chyna ran in kilometres can be calculated as follows:
Therefore,
Distance Ronaldo ran = 6 3 / 5 = 33 / 5 kilometres
Hence,
Distance Chyna ran = 33 / 5 + 9 / 10
Distance Chyna ran = 66 + 9 / 10
Distance Chyna ran = 75 / 10
Therefore,
Distance Chyna ran = 7 5 / 10
Distance Chyna ran = 7 1 / 2 kilometres
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Find a power series representation for the function and the radius of convergence R. Leave your answer in sigma notation f(x) = (1 +3.r)"
The power series representation of the function f(x) = (1 + 3x)^n is given by 1 + 3nx + 3n(n-1)x^2/2! + 3n(n-1)(n-2)x^3/3! + ..., and the radius of convergence is R = 1/3.
The power series representation of the function f(x) = (1 + 3x)^n can be obtained by using the binomial series expansion. The binomial series is given by:
(1 + x)^n = 1 + nx + n(n-1)x^2/2! + n(n-1)(n-2)x^3/3! + ...
In this case, we have f(x) = (1 + 3x)^n, where n is a constant. Substituting x with 3x, we get:
f(x) = (1 + 3x)^n = 1 + 3nx + 3n(n-1)x^2/2! + 3n(n-1)(n-2)x^3/3! + ...
This is the power series representation of the function f(x). The radius of convergence R for this power series can be found by considering the convergence properties of the series. In this case, the series will converge for values of x such that |3x| < 1, or |x| < 1/3. Therefore, the radius of convergence is R = 1/3.
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The sine of an obtuse angle is 0.53. Calculate the tangent of this angle to 4 decimal places.
The tangent of an obtuse angle with a sine of 0.53 is approximately 0.5592.
To find the tangent of an angle, we can use the relationship between the sine and tangent functions. The tangent of an angle is equal to the sine of the angle divided by the cosine of the angle. Since we are given the sine of the obtuse angle as 0.53, we need to find the cosine of the angle.
However, the sine alone does not provide enough information to determine the cosine and thus the angle itself. Therefore, it is not possible to calculate the exact value of the tangent based solely on the given information. However, if additional information or a specific angle measure is provided, we could determine the tangent using the trigonometric functions.
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A surface is a triangle with vertices A=(1, 0, 0), B=(0, 2, 0), and C=(0, 0, 3).
a) Find one parameterization →r(u,v)
for this triangle surface, remember lo include ranges of parameters u and v.
b) Use your parameterization to find one normal vector →N
for the surface.
c) Use formula Area = ∬ds
= ∬ |→N| dudv to find the area of the triangle.
Therefore, the area of the triangle is 3/2 square units.
a) One possible parameterization for the triangle surface is:
→r(u,v) = (1 - u - v)A + uB + vC, where 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1-u.
b) To find a normal vector →N, we can compute the cross product of two vectors lying on the triangle surface, for example:
→N = AB x AC = (-2, 1, 2)
Note that there are other possible choices for →N, since any non-zero scalar multiple of →N is also a valid normal vector for the triangle surface.
c) The area of the triangle can be computed as:
Area = ∬ds = ∬ |→N| dudv = |→N| ∬dudv
Since |→N| = sqrt(2^2 + 1^2 + 2^2) = 3, we have:
Area = 3 ∬dudv
Using the given ranges for u and v, we can evaluate the double integral as:
Area = 3 ∫0^1 ∫0^(1-u) dudv = 3 ∫0^1 (1-u)du = 3/2
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if triangle PRT∼triangle VST find PT.
The value of length PT is 25
What are similar triangles?Similar triangles are triangles that have the same shape, but their sizes may vary. The corresponding angles of similar triangles are congruent.
Also, the ratio of corresponding sides of similar triangles are equal.
Therefore;
6x-2/30 = 5x+13/36
36( 6x-2) = 30( 5x+13)
6( 6x -2) = 5( 5x+13)
36x -12 = 25x + 65
collecting like terms
36x-25x = 65+12
11x = 77
divide both sides by 11
x = 77/11
x = 7
Therefore,
PT = 5x+13
= 5× 7 + 13
= 12 +13
= 25
therefore the length PT is 25
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The histograms display the frequency of temperatures in two different locations in a 30-day period.
A graph with the x-axis labeled Temperature in Degrees, with intervals 60 to 69, 70 to 79, 80 to 89, 90 to 99, 100 to 109, 110 to 119. The y-axis is labeled Frequency and begins at 0 with tick marks every one unit up to 14. A shaded bar stops at 10 above 60 to 69, at 9 above 70 to 79, at 5 above 80 to 89, at 4 above 90 to 99, and at 2 above 100 to 109. There is no shaded bar above 110 to 119. The graph is titled Temps in Sunny Town.
A graph with the x-axis labeled Temperature in Degrees, with intervals 60 to 69, 70 to 79, 80 to 89, 90 to 99, 100 to 109, 110 to 119. The y-axis is labeled Frequency and begins at 0 with tick marks every one unit up to 16. A shaded bar stops at 2 above 60 to 69, at 4 above 70 to 79, at 12 above 80 to 89, at 6 above 90 to 99, at 4 above 100 to 109, and at 2 above 110 to 119. The graph is titled Temps in Desert Landing.
When comparing the data, which measure of variability should be used for both sets of data to determine the location with the most consistent temperature?
IQR, because Sunny Town is skewed
IQR, because Desert Landing is symmetric
Range, because Sunny Town is skewed
Range, because Desert Landing is symmetric
The measure of variability should be used for both sets of data to determine the location with the most consistent temperature is IQR (Interquartile Range).
The measure of variability should be used for both sets of data to determine the location with the most consistent temperatureThe reason for this is that IQR is less sensitive to extreme values or outliers, which can affect the range significantly. In the case of the histograms provided, it appears that Sunny Town is skewed and Desert Landing is roughly symmetric, which further supports the use of IQR as a measure of variability.
Therefore, we can calculate the IQR for both sets of data and compare them to determine which location has the most consistent temperature.
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What should be added to 15/16 to get 78/48
Answer:
First, find a common denominator for 16 and 48, which is 48.
Multiply the numerator and denominator of 15/16 by 3 to have a common denominator of 48:
(15/16) * (3/3) = 45/48
Now, subtract 45/48 from 78/48:
78/48 - 45/48 = (78 - 45) / 48 = 33/48
Therefore, the difference between 78/48 and 15/16 is 33/48.
So, to find the number that needs to be added to 15/16 to get 78/48, it is 33/48.
a⃗ =⟨−1,−4⟩ and b⃗ =⟨3,2⟩.
Represent a⃗ −b⃗ using the head-to-tail method.
Use the Vector tool to draw the vectors, complete the subtraction by the head-to-tail method, and draw a⃗ −b⃗ .
To use the Vector tool, select the initial point and then the terminal point.
The resultant of the vector subtraction of vector a and vector b is determined as ( -4, - 6).
What is the resultant of the vector addition?The resultant of the vector subtraction is calculated as follows;
The given vector expression;
a = ( -1, - 4) and b = (3, 2 )
Then the subtraction of the two vector is calculated as follows;
a - b = ( -1, -4 ) - (3, 2 )
To subtract each component of the vector, we will simply collect similar terms, for example, vectors in x direction will be subtracted and the vectors in y direction will be subtracted as well.
( -1, -4 ) - (3, 2 ) = ( -1 - 3, -4-2)
R = ( -4, - 6)
The vector diagram is in the image attached.
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Need help please help me
The given figure shows the Translation by 5 units left or right.
We have to given that;
Graph of transformation is shown in image.
Since, Translation is defined to slide a figure in any direction.
Reflection is defined as to flip a figure over a line.
Rotation is defined as to rotate a figure a certain degree around a point.
Dilation is defined as to enlarge or reduce a figure.
Hence, By graph, we get;
The given figure shows the Translation.
Because is translate by 5 units.
Thus, The given figure shows the Translation by 5 units left or right.
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Solve the initial value problem. dy dx =x*(y-4), y(0) = 6 The solution is (Type an implicit solution. Type an equation using x and y as the variables...
To solve the initial value problem dy/dx = x*(y-4) with the initial condition y(0) = 6, we can separate variables and integrate both sides.
Separating variables:
dy / (y - 4) = x dx
Integrating both sides:
∫ dy / (y - 4) = ∫ x dx
Taking the integral:
ln|y - 4| = (1/2) x^2 + C
To determine the constant of integration, we can use the initial condition y(0) = 6. Substituting x = 0 and y = 6 into the equation:
ln|6 - 4| = (1/2) (0)^2 + C
ln(2) = C
Therefore, the equation becomes:
ln|y - 4| = (1/2) x^2 + ln(2)
This is the implicit solution to the initial value problem.
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The market for smart thermostats has grown increasingly competitive. Producers of smart thermostats all rely on the same technology and face the same costs. The cost function for a smart thermostat producer is given by the following function:
c(y)= 3
y 3
−10y 2
+200y, where
y
stands for the number of smart thermostats produced and sold in a month. 1st attempt O See Hint Suppose the market demand for smart thermostats in any month is given by
Q D
=275−p.
In the long-run equilibrium, we would expect to find firms in the industry
Based on the given cost function and demand function, we can find the long-run equilibrium by setting the marginal cost equal to the marginal revenue.
The marginal cost is given by the derivative of the cost function:
MC(y) = 9y^2 - 20y + 200
The marginal revenue can be found by taking the derivative of the demand function and solving for p:
QD = 275 - p
p = 275 - QD
MR = d(QD)/dy * dp/dy = -1
Setting MC = MR:
9y^2 - 20y + 200 = -1
Simplifying and solving for y:
9y^2 - 20y + 201 = 0
Using the quadratic formula: y = (20 ± sqrt(400 - 4*9*201)) / 18
Since the discriminant is negative, there are no real solutions for y. This means there is no long-run equilibrium for this market.
In practical terms, this suggests that the market for smart thermostats may not be sustainable in the long run given the high cost structure and competitive pressures. Some firms may exit the market, while others may continue to innovate and differentiate their products to remain competitive.
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Solve for x. The triangles in each pair are similar
Answer: B). x=11
Step-by-step explanation:
[tex]\frac{48}{12} =\frac{3x+11}{11} \\[/tex]
Cross products
528=36x+132
Simplify for x
396=36x
11=x
I don't remember how to solve this I just don't understand it
Answer:
(5⁴)² = 5⁴ • 5²
Step-by-step explanation:
anything outside of the ( ) is times by whats in it. hope it helps!
Solve for x: 3^x-1=3^4
The solution for 'x' is equal to 5.
We are given,
[tex]3^{x-1}[/tex]=[tex]3^{4}[/tex]
We know that when two expressions have common bases, their exponents can be compared or equated to obtain the desired value. Here, since both the expressions have 3 as their bases, their exponents which are x - 1 and 4 can be compared to obtain the value of 'x'.
Comparing the exponents of the given expressions, we get,
⇒ x - 1 = 4
Transposing 1 to R.H.S.
⇒ x = 4 + 1
⇒ x = 5
Hence the solution to the given statement is 'x' =5.
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Answer:
x = 7
Step-by-step explanation:
[tex]\sf 3^x-1=3^4[/tex]
Use logarithms.
[tex]\sf log3^x-log1=log3^4\\\\\sf log3^x-log3^3=log3^4\\\\Add\:log3^3\:to\:both\:sides.\\\\\sf log3^x=log3^4+log3^3\\\\Put\: logarithm\:rules\\\\xlog3=4log3+3log3\\\\Divide\:both\:sides\:by\:log3\\\\x=\dfrac{4log3+3log3}{log3} \\\\x=7[/tex]
1 2 3 4 5 6 7 8 9 10 TIME REMAINING 49:37 Which is a diagonal through the interior of the cube? A cube. The top face has points A, B, D, C. The bottom face has points E, F, H, G. Side A H Side B E Side C H Side F G Mark this and return QUIKKKKK
A diagonal through the interior of a cube is a line segment that connects two opposite vertices of the cube. In this case, the cube has top face vertices A, B, D, and C, and bottom face vertices E, F, H, and G. There are four possible diagonals through the interior of the cube: AD, BC, AE, and CG.
- Diagonal AD connects vertex A from the top face to vertex D, also on the top face. This diagonal does not pass through the interior of the cube.
- Diagonal BC connects vertex B from the top face to vertex C, also on the top face. This diagonal does not pass through the interior of the cube.
- Diagonal AE connects vertex A from the top face to vertex E on the bottom face. This diagonal passes through the interior of the cube.
- Diagonal CG connects vertex C from the top face to vertex G on the bottom face. This diagonal passes through the interior of the cube.
Thus, the diagonals AE and CG are diagonals through the interior of the cube.
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WILL LEAVE POSITIVE RATE
Show that the following series diverges. Which condition of the Alternating Series Test is not satisfied? 1 3 4 Σ 1-16-1 k 2k+1 3 7 9 k=1 Let a > 0 represent the magnitude of the terms of the given s
The harmonic series Σ 1/k diverges, so by the Limit Comparison Test, Σ a_k also diverges.
The Alternating Series Test states that if a series has alternating terms that decrease in magnitude and approach zero, then the series converges.
However, in the given series, the terms do not decrease in magnitude. In fact, the terms increase as k increases. Therefore, the Alternating Series Test is not satisfied, and the series diverges.
Using the Limit Comparison Test,
Let b_k = 1/k. Then b_k > a_k for all k > 1, since 2k + 1 > k for k > 1. Also, lim (k → ∞) b_k / a_k = lim (k → ∞) k(2k + 1)/(k + 1)(k + 2) = 2/1 = 2, which is a finite nonzero number.
Therefore, both series either converge or diverge together. However, we know that the harmonic series Σ 1/k diverges, so by the Limit Comparison Test, Σ a_k also diverges.
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Please what’s the answer?
To solve the given equation the following steps are taken:
(a) given
(b) Substitution
(C) additive property
(d) division property
(e) symmetric property of equality
Hence option D is correct.
According to the given information,
9x - 5y = -8 , x = - 9 ⇒ this given expression
81 - 5y = -8 ⇒ it represents substitution property as value of x = -9 is substituted.
-5y = 73 ⇒ it represents property of addition.
y = -73/5 ⇒ it represents division property as it is divide by -5 both sides.
-73/5 = y ⇒ It represents symmetric property of equality.
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surface area . how do u do number 44
[tex] \boxed{\bf{Surface\:Area\:of\:Cone}} [/tex]
[tex] \begin{aligned}SA&=\pi \: r(s + r) \\&=3.14 \times 12 \times (26.8 + 12) \\ &=37.68 \times 38.8 \\&= \bold{ \underline{1461.984 \: ft ^{2} }} \end{aligned}[/tex]
[tex] \small{\blue{\mathfrak{That's\:it\: :)}}} [/tex]
Please answer the question as best as possible
Answer:
Quadrant 2
Step-by-step explanation:
In this question, we are given that [tex]\sin\theta[/tex] is positive, and [tex]\cos\theta[/tex] is negative.
This means that if we graph [tex]\theta[/tex] and find its point of intersection with the unit circle, the y-coordinate will be positive, and the x-coordinate will be negative.
If we look at the coordinate plane, the quadrant where y is positive and x is negative is Quadrant 2 (the upper left quadrant).
Why is it a good thing to run both a bivariate and a multivariate analysis when determining the linear relationship between two variables? What changes happen when other variables are introduced into the models?
Combining bivariate and multivariate analysis provides a more comprehensive view of the relationship between variables, which can inform better decision-making in a variety of fields.
Running both a bivariate and a multivariate analysis when determining the linear relationship between two variables is a good thing because it helps to better understand the relationship between the variables.
A bivariate analysis only looks at the relationship between two variables, while a multivariate analysis considers the impact of additional variables on the relationship.
When other variables are introduced into the models, the strength and direction of the relationship between the two original variables may change. This is because the new variables may be affecting one or both of the original variables.
Multivariate analysis can help identify and control for these effects, leading to a more accurate and reliable understanding of the relationship between the two variables.
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