giving brainliest do 2 and 3

Answer:

-15,3   2,-11   -3,-7    12,4

The equation which has a value greater than 7,245 as required in the task content is; Choice D; three-halves times 7,245 equals = 10,867.5.

It follows from the task content that the value greater than 7,245 among the answer choices is to be determined.

First the value of each of the expressions can be evaluated as follows;

A. one-third times 7,245 equals = (1/3) × 7,245 = 2415.

B. two-thirds times 7,245 equals = (2/3) × 7,245 = 4,830.

C. three-thirds times 7,245 equals = (3/3) × 7,245 = 7,245

D. three-halves times 7,245 equals = (3/2) × 7,245 = 10,867.5

Ultimately, the required number which is greater than 7,245 is; Choice D; three-halves times 7,245 equals = 10,867.5.

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Answer:

To find -134 + 53, add the ones place (4+3=7) and the tens place (5+3=8), giving you -81.

Answer: -81.

Answer:

-81

Step-by-step explanation:

You subtract the absolute values and take the sign of the larger absolute value.

Absolute value is the distance from zero.  This will be a positive number

134 - 53 = 81

The sign will be negative because the sign of the higher absolute value number is negative.  134 has a larger absolute value than 53 and it is negative, so our answer is negative.

Helping in the name of Jesus.

Answer:

Division to get 28

Step-by-step explanation:

Answer:

2

Step-by-step explanation:

Answer:

7, 7, 60, 60, 60

Step-by-step explanation:

When you draw the circles thhey all hhave radius AB so AB=AC=BC =7

and since all sides are congruent you have a equilateral treangle so all angles are equal to 180/3=60

We will use the following properties of Laplace transform:

L{δ(t-a)} = e^(-as)

L{y'(t)} = sY(s) - y(0)

L{y''(t)} = s^2Y(s) - sy(0) - y'(0)

Applying Laplace transform to both sides of the differential equation and using the above properties, we get:

s^2Y(s) - sy(0) - y'(0) + 4(sY(s) - y(0)) + 5Y(s) = e^(-2πs)

Simplifying and solving for Y(s), we get:

Y(s) = [1/(s^2 + 4s + 5)] * e^(-2πs)

To find the inverse Laplace transform, we can complete the square in the denominator and use the following formula:

L{e^(-as)sin(bs)} = (b/(s^2 + a^2 - b^2))

So, we have:

Y(s) = [1/((s+2)^2 + 1)] * e^(-2πs)

= [1/(2*(s+2))] * [2*e^(-2πs)/(s+2)^2 + 2/(s+2)]

= [1/(2*(s+2))] * L{e^(-2πs)sin(t-2π)} + [1/(2*(s+2))] * L{1}

Applying inverse Laplace transform to both terms, we get:

y(t) = (1/2)*e^(-2(t-2π))sin(t-2π) + (1/2)e^(-2t)

Therefore, the solution to the given initial value problem is:

y(t) = (1/2)*e^(-2(t-2π))sin(t-2π) + (1/2)e^(-2t) with y(0) = 0, y'(0) = 0.

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To solve the initial-value problem using Laplace transform, apply it to both sides, solve for Y(s), and take the inverse Laplace transform.

To solve the given initial-value problem using Laplace transform, we can apply the Laplace transform to both sides of the differential equation. The Laplace transform of the Dirac delta function is given by 1. Thus, we have:

s^2Y(s) + 4sY(s) + 5Y(s) = e^(-2πs)

Using the initial conditions, we can solve for Y(s). Taking the inverse Laplace transform of Y(s) will give us the solution in the time domain.

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Answer:

There're 6 lines.

Step-by-step explanation:

you can count the arrow heads and divide it by 2. cuz 1 arrow has 2 arrow heads.

The law of sines can be used to solve a triangle in the ASA and SAS cases. However, it cannot be used in the SSA case, and in the SSS case, we need additional information to solve the triangle.

The law of sines can be used to solve a triangle in the following cases: ASA, SAS, and SSA.

1. ASA (Angle-Side-Angle): In this case, we know two angles and the side between them. To solve the triangle using the law of sines, we can use the following steps:

  a. Use the given angles to find the third angle by subtracting the sum of the known angles from 180 degrees.

  b. Apply the law of sines to find the length of the missing side.

2. SAS (Side-Angle-Side):

In this case, we know two sides and the angle between them. To solve the triangle using the law of sines, we can use the following steps:
  a. Apply the law of sines to find one of the unknown angles.

  b. Use the known angle and the sum of angles in a triangle (180 degrees) to find the third angle.

  c. Apply the law of sines again to find the remaining side.

3. SSA (Side-Side-Angle): In this case, we know two sides and one angle that is not between the sides.

However, the law of sines cannot be used to solve the triangle in this case. This is because the given information is not sufficient to uniquely determine the triangle.

It can lead to two different possible triangles, known as the ambiguous case.

4. SSS (Side-Side-Side): The law of sines cannot be used to solve a triangle when we only have the lengths of all three sides.

In this case, we need additional information, such as angles or other side lengths, to solve the triangle.

To summarize, the law of sines can be used to solve a triangle in the ASA and SAS cases. However, it cannot be used in the SSA case, and in the SSS case, we need additional information to solve the triangle.

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The solutions are

a) The smallest possible value of x - y is A = -6

b) The largest possible value of 2x / y is B = 6

c) The largest possible value of x² + y² is C = 100

What is an Inequality Equation?

Inequalities are the mathematical expressions in which both sides are not equal. In inequality, unlike in equations, we compare two values. The equal sign in between is replaced by less than (or less than or equal to), greater than (or greater than or equal to), or not equal to sign.

In an inequality, the two expressions are not necessarily equal which is indicated by the symbols: >, <, ≤ or ≥.

Given data ,

Let the first inequality equation be represented as P

The value of P is

-4 ≤ x ≤ 6

Let the second inequality equation be represented as Q

The value of Q is

2 ≤ y ≤ 8

Now , the smallest value of inequality P is = -4

The largest value of inequality P is = 6

And , the smallest value of inequality Q is = 2

The largest value of inequality Q is = 8

a)

Let the smallest possible value of x - y be = A

The value of x - y is calculated as

The smallest value of P - smallest value of Q

Substituting the values in the equation , we get

A = -4 - 2

A = -6

Therefore ,  The smallest possible value of x - y is A = -6

b)

Let the largest possible value of 2x / y be = B

The value of 2x/y is calculated as

Largest value of x goes in numerator and smallest value of y goes in denominator to get the largest value of 2x/y

So , substituting the values in the equation , we get

B = ( 2 x 6 ) / 2

B = 12 / 2

B = 6

Therefore , The largest possible value of 2x / y is B = 6

c)

Let the largest possible values of x² + y² be = C

The value of x² + y²  is calculated as

The largest value of x squared + the largest value of y squared

Substituting the values in the equation , we get

C = 6² + 8²

C = 36 + 64

C = 100

Therefore , The largest possible value of x² + y² is C = 100

Hence , The solutions are

a) The smallest possible value of x - y is A = -6

b) The largest possible value of 2x / y is B = 6

c) The largest possible value of x² + y² is C = 100

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Answer:

A: C=45h+30

B: C=$322.50

Step-by-step explanation:

A: We'll start with the equation for the total charge. Juan is saying he will pay a one time fee of $30, meaning there is 30 dollars added to what ever the hourly wage is. This can be represented by the +30. Now if "h" represents the variable for which the hourly wage will be calculated, and he pays $45 dollars per hour this will be represented as 45h. As an example if I pay you 2 dollars per hour using the same variable "h", this would be represented as 2h. So if you worked for two hours you would get 4 dollars, this is proven by the fact the 2(2) (remeber im replacing the "h" with the hours you worked) obviously 2 times 2 is 4 proving my point. This information will give you the equation you see above.

B: Onto solving for how much Juan will pay. Now you say youre supposed to have 9 answer but this can't be true. All you need to do is plug the hours given into "h" and solve the equation. It should look something like:

C=45h+30

C=45(6.5)+30

Im using a decimal because it should be clear 1/2=0.5

C= 292.5+30

45x6.5 always go with PEMDAS, we are multiplying first. Then add 292.5 plus 30

This results in the answer of:

C=$322.50

The area enclosed by the given ellipse is A = πab.



We can start by noting that the given equations for the ellipse are in parametric form, with t representing the angle parameter. To find the area enclosed by the ellipse, we can use the formula for the area of a sector of an ellipse, which is given by:

A = ½ abθ

where a and b are the lengths of the major and minor axes of the ellipse, respectively, and θ is the central angle that the sector subtends. In our case, we want to find the area enclosed by the entire ellipse, which corresponds to a full 360-degree rotation. Thus, we have:

A = ½ ab(2π) = πab




To fully understand how we arrived at the formula for the area of a sector of an ellipse, we can look at the geometry of the ellipse itself. An ellipse is defined as the set of all points in a plane whose distances from two fixed points (called the foci) sum to a constant. Alternatively, we can think of an ellipse as a stretched circle, with one axis longer than the other. The lengths of the major and minor axes are denoted by a and b, respectively.

Now, consider a sector of the ellipse, defined by two rays emanating from one of the foci and intersecting the ellipse at two points. Let the central angle that the sector subtends be denoted by θ,

To find the area of this sector, we can first find the area of the corresponding sector of a circle, with radius a. This is given by:

A_circle = ½ a²θ

However, since our sector is part of an ellipse, we need to adjust this formula to take into account the fact that the radius varies along the ellipse. Specifically, the radius at any point on the ellipse is given by:

r = a√[1 - (sin t)²]

(where t is the angle that the point makes with the x-axis). To account for this, we need to multiply the area of the circle sector by a scaling factor that accounts for the variation in radius. This factor is simply the ratio of the length of the minor axis to the length of the major axis:

scaling factor = b/a

Thus, the area of the sector of the ellipse is given by:

A_ellipse = ½ a²θ (b/a)

= ½ abθ


In summary, to find the area enclosed by an ellipse given in parametric form, we can use the formula A = πab, which is derived from the formula for the area of a sector of an ellipse. This formula takes into account the varying radius of the ellipse and the lengths of the major and minor axes.

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Let x ϵ ker T.

That is Tx = 0.

So T* Tx = 0 for all x.

Hence x ϵ ran T*

Therefore ker T is a

subset

of (ran T*)+.

Now let x ϵ (ran T*)+.

Then there exists a

sequence

{y n} ⊂ H such that y n → x and T*y n → 0.

For any x ϵ H, we haveT* Tx = 0, which implies x ϵ ker T*.

Let x ϵ (ker T)⊥.

That is, (x, y) = 0 for all y ϵ ker T.

Then (Tx, y) = (x, T*y) = 0 for all y ϵ H.

Hence x ϵ ran T*.

Thus (ker T)⊥ ⊂ ran T* and by taking orthogonal

complements

, we get (ker T) = ran T*.

Let T be one-to-one.

Then ker T = {0} and we have the equality ran T* = (ker T)⊥ = H.

Thus ran T* is dense in H.

Conversely, let ran T* be dense in H.

Suppose there exist x 1, x 2 ϵ H such that Tx 1 = Tx 2. Then T(x 1 - x 2) = 0,

so x 1 - x 2 ϵ ker T = (ran T*)+.

Hence there exists a sequence {y n} ϵ H such that y n → x 1 - x 2 and T*y n → 0. So we have Ty n → Tx 1 - Tx 2 = 0. Then(Ty n, z) = (y n , T*z) → 0 for all z ϵ H. Hence y n → 0 and hence x 1 = x 2.

Therefore T is one-to-one.

Hence, we have proved that T is one-to-one if and only if ran T* is

dense

in H.

Hence, it has been proven that, let T € B(H), if (a) ker T = (ran T*)+, (b) (ker T) = ran T* and (c) T is one-to-one if and only if ran T* is dense in H.

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Answer:

48.

Step-by-step explanation:

To find area, you need to multiply. If you multiply 12x4 you get 48, if that makes sense.

Answer:

t = 1 s to reach its maximum height.

Step-by-step explanation:

The function that represents the height of the rocket after s seconds is given by :

\(h(s)=-3s^2+6s+45\) ...(1)

For maximum height,

\(h'(s)=0\\\\-6s+6=0\\\\s=1\)

It is the time the rocket will take. So, the ball will take 1 second to reach its maximum height.

Kareem's gas tank has a capacity of 27 gallons, meaning it can hold a maximum of 27 gallons of gas.

Let's solve this problem step by step. We'll use algebraic equations to find the number of gallons Kareem's gas tank can hold.

Let's assume that Kareem's gas tank can hold 'x' gallons.

According to the problem, initially, the gas tank is 1/9 full. We can represent this as (1/9)x gallons.

After Kareem buys 6 gallons of gas, the tank becomes 1/3 full. This can be represented as (1/3)x gallons.

We can set up an equation based on the information given:

(1/9)x + 6 = (1/3)x

Now, let's solve this equation to find the value of 'x' (the total capacity of the gas tank).

Multiply both sides of the equation by 9 to eliminate the fractions:

x + 54 = 3x

Subtract 'x' from both sides:

54 = 2x

Divide both sides by 2:

x = 54/2

x = 27

Therefore, Kareem's gas tank can hold 27 gallons.

To summarize, Kareem's gas tank has a total capacity of 27 gallons. Initially, it was 1/9 full, but after buying 6 gallons, it became 1/3 full. By solving the equation (1/9)x + 6 = (1/3)x, we found that 'x' equals 27. Hence, Kareem's gas tank can hold 27 gallons.

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Note thecomplete question is

Kareem's gas tank is 1/9  full. After he buys 6 gallons of gas, it is 1/3 full. How many gallons can Kareem's tank hold?

Answer:

The equation is ( 32/2 ) × 3 = 48.

Step-by-step explanation:

In order to find the answer, we must find one third of last years trip. To do this, we can simply divide the cost of this year's trip by two, 32/2=16.

Now if we multiply this years trip by 3, 16×3=48, we can see that the total cost of last years trip was $48.

If we'd like to turn this into a equation, it looks like this:

( 32/2 ) × 3 = 48.

The circumference of the figure given is 43.96 mi   and the area is 153.86 mi²  .

In the question ,

it is given that ,

the radius of the given figure (circle) is = 7 mi

We know that formula for circumference of  circle is = 2πr

Substituting the radius as 7 mi , in the formula ,

we get ,

Circumference = 2*π*7

= 2*3.14*7              ,.......by using the value π = 3.14

= 43.96 mi

The formula for the Area of the Circle is given by

Area = π*r²

Substituting the radius = 7 mi in the area formula ,

we get ,

Area = π*7²

= π*49

= 3.14*49           ,.......by using the value π = 3.14

= 153.86 mi²

Therefore , The circumference of the figure given is 43.96 mi   and the area is 153.86 mi²  .

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Step 1

Given; Which toolkit function is this? I am always continuous and have a point of inflection at the origin. My range is from a negative one to one.

Step 2

The function that is always continuous and has a point of inflection at the origin with its range going from -1 to 1 is

From the graph, the range is from -1 to 1.

Answer;

The number of seniors that are in the school is 240.

The first step is to determine the number of seniors that prefer gym equipment.

Let:

number of students that prefer gym equipment = x

total number of seniors = number of students that prefer gym equipment + number of seniors that prefer computer

= 160 + x

probability that a senior prefers gym equipment = 1/3

[x / (160 + x)] = 1/3

In order to determine the value of x, take the following steps:

Multiply both sides of the equation by 3.

3x = 160 + x

Combine and add similar terms:

3x - x = 160

2x = 160

x = 160 / 2

x = 80

Total number of seniors = 160 + 80 = 240

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The function is not normalizable and cannot be a wave function.

A wave function must satisfy two important conditions. First, it must be normalizable, meaning its integral over all space is finite.

Second, it must be single-valued and continuous.

(i) The function Ψ1​(x) = \(e^{-x^{5} }\) is acceptable as a wave function, since it is continuous everywhere and decays to zero as x → ±∞.

However, we must check if it is normalizable. The normalization condition for a one-dimensional wave function Ψ(x) is:

∫(-∞) to ∞ |Ψ(x)|²dx = 1

In this case, we have:

∫(-∞) to ∞ |Ψ₁(x)|² dx

= ∫(-∞) to ∞ \(e^{-2x^{5} }\) dx

This integral can be quite challenging to evaluate, but we can see that the integrand goes to zero faster than any power of x as x → ±∞.

Therefore, the integral is convergent and the function is normalizable.

Therefore, Ψ₁(x) is an acceptable wave function.

(ii) The function Ψ₂​(x) = cos(x - 2π/3) is not acceptable as a wave function, since it is not normalizable.

The normalization condition for a one-dimensional wave function Ψ(x) is:

∫( -∞) to ∞ |Ψ(x)|² dx = 1

In this case, we have:

∫(-∞) to ∞ |Ψ₂​(x)|² dx = ∫(-∞)to∞ cos²(x - 2π/3) dx

This integral is not convergent, since the integrand oscillates between 0 and 1 over an infinitely repeating interval.

Therefore, the function is not normalizable and cannot be a wave function.

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Intellectually gifted people score in the top 1 - 2percent on a standard IQ test.

From the question, we have the following parameters that can be used in our computation:

IQ of intellectual gifted people

The general rule is that

People with intellectuals are usually ranked high and the range is usually small

Next, we test the options

A. 10

The top 10% has a high range

B. 5-10

The top 10% omits people in top 5%

C. 5

The top 5% has a high range

D. 1-2

This has a small range and it is the highest

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Answer:

10c + 5b + 420 + 4c

Step-by-step explanation: 10 dollars for each child, 5 dollars per a pair of rubber boots, party room is 420 and 4 is per each child's pizza

Dr. Hellsing is in the process of establishing a norm-referenced assessment.

Norm-referenced assessment is a method that evaluates an individual's aptitude in comparison with the outputs of a larger, representative sample of test takers from the corresponding population. This technique ranks persons on a scale relative to each other instead of objectively evaluating their actual abilities.

Therefore, by giving the exam to a significant group of high school graduates, Dr. Hellsing will be able to generate standards to interpret separate test scores and equate them to the outcome of others in the same demographic.

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Step-by-step explanation:

integral from 0 to 2pi of cos3x

(sin3x)/3 from 0 to 2 pi

sin 6pi /3 - sin 0 /3 = 0

Answer:

rational

Step-by-step explanation:

Answer: Integer and rational

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

Explanation:

The set of whole numbers is defined to be {0, 1, 2, 3, ....} so it's composed of numbers that don't have any decimal or fractional components. We start at 0 with this set. We don't include negative numbers here.

Integers are basically the same as whole numbers, but now we include negative values {..., -3, -2, -1, 0, 1, 2, 3, ...} so we see that -15 is in the set of integers.

The value -15 is also a rational number because -15 = -15/1, meaning we can write -15 as a fraction of integers. Any rational number is of the form a/b with a,b being integers and b is nonzero.

Based on the given angles:

From the given lines and angles, we can identify:

Vertical angles are pair of angles with no adjacent line but share one meeting point and have congruent angles. Four pairs of vertical angles are:

Alternate interior angles are pair of angles within the parallel line (inside) that have congruent angles. Each pair cross the transversal line "t". Two pairs of alternate interior angles are:

Cogruent angles might come from several types of relationships, such as:

All angles congruent to angle ∠1 are:

All angels congruent to angle ∠2 are:

Supplementary angles are a pair of angles whose its sum is 180°. Supplementary angles might formed by several types of angles relationship, such as:

Pairs of supplementary angles are:

Consecutive interior angles:

Adjacent angles:

Corresponding angles are a pair of unadjacent angles but have congruent angles because they face the same direction. Pairs of corresponding angles:

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Yes, you can show significant differences using superscripts in an ANOVA (Analysis of Variance) single-factor test.

In an ANOVA test, superscripts are commonly used to indicate significant differences between the means of different groups or treatments.

Typically, letters or symbols are assigned as superscripts to denote which groups have significantly different means. These superscripts are usually presented adjacent to the mean values in tables or figures.

The specific superscripts assigned to the means depend on the statistical analysis software or convention being used. Each group or treatment with a different superscript is considered significantly different from groups with different superscripts. On the other hand, groups with the same superscript are not significantly different from each other.

By including superscripts, you can visually highlight and communicate the significant differences between groups or treatments in an ANOVA single-factor test, making it easier to interpret the results and identify which groups have statistically distinct means.

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Answer:

C

Step-by-step explanation:

Answer:

Equals

Step-by-step explanation: