At the beginning of the month, Arnold’s bank account balance was $217.25. During the month he made deposits of $49.58 and $105.75 and a withdrawal of $25.45. What was Arnold’s bank balance at the end of the month?

Answer:

Step-by-step explanation:

Answer:

4 x 1/4

Step-by-step explanation:

The box is split into 4 representing the 1/4 and it goes to the 4 on the number line representing the.

Answer:

Umm this is like 2 grade stuff? Ummm well difference means to subtract soo just use a calutour the answer is 1.69

Step-by-step explanation:

Answer:

21

Step-by-step explanation:

63 divided by 3 is 21

210 divided by 10 is 21

The position of the center of mass of the weighted wheel is 5.11 cm out from the center. The moment of inertia about an axis through its CM, perpendicular to its face is roughly 26357 kg cm².

(a) We must take into account the masses and their various distances from the center in order to determine the location of the weighted wheel's center of mass (CM).

Let's use the symbol x to represent how far the CM is from the wheel's centre. Without a weight, the centre of the wheel, or CM, will be at a distance of 0 cm from the centre.

We can formulate the equation using the center of mass principle:

(m₁ × x₁ + m₂ × x₂)/(m₁ + m₂) = x

where m₁ is the weight's mass (2.0 kg), x₂ is the weight's distance from the center (23 cm), and m₁ is the mass of the wheel (7.0 kg), with x₁ being the distance from the wheel's center (0 cm).

Inserting the values:

(7.0 kg × 0 cm + 2.0 kg × 23 cm) / (7.0 kg + 2.0 kg) = x

(46.0 kg cm)/9.0 kg = x

x ≈ 5.11 cm

As a result, the weighted wheel's centre of mass is located 5.11 cm out from the centre.

(b) The parallel-axis theorem can be used to determine the weighted wheel's moment of inertia around an axis that passes through its CM and is perpendicular to its face. The following equation gives the wheel's moment of inertia about its own axis (via the center):

I(wheel) = (1/2) × m(wheel) × r²

where r is the wheel's radius (38 cm) and m(wheel) is the wheel's mass (7.0 kg).

Calculations for the weight's moment of inertia around the same axis are as follows:

I(weight) = m(weight) × d²

where d is the weight's distance from the axis (which is equal to the weight's distance from the center, 23 cm), and m(weight) is the weight's mass (2.0 kg).

By summing the inertial moments of the wheel and the weight, one may get the moment of inertia of the weighted wheel about the axis through its CM:

I(total) = I(wheel) + I(weight)

I(total) = (1/2) × m(wheel) × r² + m(weight) × d²

I(total) = (1/2) × 7.0 kg × (38 cm)² + 2.0 kg × (23 cm)²

I(total) ≈ 26357 kg cm²

Consequently, the weighted wheel's moment of inertia along an axis through its CM that is perpendicular to its face is roughly 26357 kg cm².

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Answer:y=3x-2

Step-by-step explanation:

Answer:

3.75k + 2.25p = 14

Step-by-step explanation:

The cost from the weights of kumquats can be represented by 3.75k.

The cost from the weights of Asian pears can be represented by 2.25p.

Using the standard form equation, Ax + By = C, replace Ax and By with these terms.

Then, plug in 14 as C:

Ax + By = C

3.75k + 2.25p = 14

So, the equation is 3.75k + 2.25p = 14

Answer:

$171

Step-by-step explanation:

=PRN

950×0.06×3

hope this helps and makes sense :))

Answer:

$171

Step-by-step explanation:

SI=950×6×3/100

=$171

HOPE IT HELPS

Answer:

$0.27

However answered below me is correct. I divided backwards.

My apologies. It is 5am here locally and I've been up doing math for too many hours.

Unit price = Total cost divided by quantity

Unit price = 1.91 ÷ 7= 0.2728571429

rounded to the nearest cent $0.27

Answer:

27 cents

Step-by-step explanation:

When multiplying the 2 and the -12 it would be -24, not 24. Which if my mental math is correct, would make the answer 6

Edit:

Note that subtracting a negative is the same as adding the absolute value.

so:

-24-(-30)

changes to

-24+30

and it equals

6

The parametric description of the surface is ⟨4sin(u)cos(v), 4sin(u)sin(v), 4cos(u)⟩.

To parametrically describe the given surface, we can use spherical coordinates since the equation \(x^2\) + \(y^2\) + \(z^2\) = 16 represents a sphere centered at the origin with a radius of 4.

In spherical coordinates, the surface can be described as:

x = 4sin(u)cos(v)

y = 4sin(u)sin(v)

z = 4cos(u)

where u represents the azimuthal angle in the range 0 ≤ u ≤ 2π, and v represents the polar angle in the range 23/​45 ≤ v ≤ 4.

Therefore, the parametric description of the surface is:

r(u, v) = ⟨4sin(u)cos(v), 4sin(u)sin(v), 4cos(u)⟩

where u ∈ [0, 2π] and v ∈ [23/45, 4].

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The millennium development goals (MDGs) were a set of eight global targets established by the United Nations in 2000 to address key global development challenges by 2015.

One conclusion that can be drawn about the millennium development goals is that they served as a catalyst for mobilizing global efforts and generating significant progress in areas such as poverty reduction, education, gender equality, and access to healthcare. Through the MDGs, there was increased international cooperation, financial support, and policy focus on these critical areas, leading to notable advancements.

Additionally, the MDGs brought global attention to the importance of setting specific targets and monitoring progress towards them. The goals provided a framework for governments, organizations, and stakeholders to align their efforts and work towards common objectives. This focus on accountability and monitoring helped in tracking progress, identifying gaps, and implementing targeted interventions.

It is important to note that while significant progress was made, there were also challenges and uneven progress across different regions and goals. Not all targets were achieved by the 2015 deadline, and disparities persisted in areas such as poverty, gender equality, and environmental sustainability.

Overall, the millennium development goals played a crucial role in galvanizing global action, raising awareness, and driving progress towards key development priorities.

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The correct ordering, assuming x is positive, is III: 2x < x² < 2 < 1/x²< 1.

Let's evaluate each option one by one:

I. x² < 2x < 1/x² < 2 < 1

If x is positive, x² will always be greater than 1/x². Therefore, this ordering is not possible.

II. x² < 1/x² < 2x < 1 < 2

Similarly, x² will always be greater than 1/x². Therefore, this ordering is also not possible.

III. 2x < x² < 2 < 1/x² < 1

For this ordering to be true, we need to confirm that 2x is indeed less than x². Since x is positive, we can divide both sides of the inequality by x to preserve the inequality direction. This gives us 2 < x. As long as x is greater than 2, this ordering holds true. Therefore, the correct ordering, assuming x is positive, is III: 2x < x² < 2 < 1/x²< 1.

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Answer: F

Step-by-step explanation:

Answer:

Answer in photo

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given points

Midpoint M is

Answer:

Step-by-step explanation:

The measure of angle A is 19.47°.

Trigonometric functions are the functions which are described as the ratio of sides of a right angled triangle.

Given a right angled triangle ABC.

∠C = 90° and AB = 6, is the hypotenuse.

We have to find the measure of angle A.

Given BC = 2, which is the opposite side with respect to ∠A.

We can use the definition of sine here.

Sine function of an angle is defined as the ratio of opposite side to the hypotenuse of the right angled triangle.

sin A = Opposite side / Hypotenuse

sin A = BC / AB

sin A = 2/6

sin A = 0.333

A = sin⁻¹ (0.333)

A = 19.47°

Hence the angle A has a measure of 19.47°.

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Your question is incomplete. The complete question is as given in the image below.

Answer:

The length of AB is \(6\sqrt{5}\) units

Step-by-step explanation:

The rule of the distance between two points (x1, y1) and (x2, y2) is

∵ The endpoints of AB are (-4, 5) and (2, -7)

→ Let point (-4, 5) be (x1, y1) and point (2, -7) be (x2, y2)

∴ x1 = -4 and y1 = 5

∴ x2 = 2 and y2 = -7

→ Substitute them in the rule above to find the length of AB

∵ \(AB=\sqrt{(2--4)^{2}+(-7-5)^{2} }=\sqrt{(2+4)^{2}+(-12)^{2}}\)

∴ \(AB=\sqrt{(6)^{2}+144}=\sqrt{36+144}\)

∴ \(AB=\sqrt{180}=6\sqrt{5}\)

∴ The length of AB is \(6\sqrt{5}\) units

Answer:

Te conozco y sé qué

Como Nuevo de fabrica el otro

Answer:

see below

We need to prove that KL ≅ JK .

Given :-

To prove:-

Proof:-

In ∆JKM and ∆LKM ,

Therefore by AAS congruence condition , ∆JKM ≅ ∆LKM .

Therefore,

Hence proved !

Answer:

The answer is "35,868,500 people".  

Step-by-step explanation:

In this question, the chance of people that have traveled above 50 miles is calculated. The chance of success is the probability.

\(n=50000\\\\ p=15595\)

\(\to P(x\geq 50)=\frac{15595}{50000}=0.3119\)

The population of the United States is at 1150000.

Let's determine that portion of the current population has gone more than 50 miles:

\(\to E(X)=np,N=115000000,p=0.3119\\\\=0.3119\times 115000000\\\\=35868500\)

Answer:

\(n=\{0,\: 4\}\)

Step-by-step explanation:

\(|3n-6| = 6\\\therefore 3n - 6 = \pm 6\\\therefore 3n = 6 \pm 6\\\therefore 3n =6+6\:\: or\:\: 3n = 6-6\\\therefore 3n =12\:\: or \:\:3n = 0\\\\\therefore n =\frac{12}{3}\:\: or\:\: n =\frac{0}{3}\\\\\therefore n = 4 \:\: or\:\: n = 0\\\\\therefore n=\{0,\: 4\}\\\)

Answer:

A,B and D

A: y= 2x -3/2

B: y =2x-4

C:y=2x-5

same gradient y=mx+c

m=2

Answer:

Hey there!

The answer is: \(\frac{9+n}{3}\)

Let me know if this helps :)

The procedure for finding the area between two curves Find the intersection points, set up the integral using the difference between the curves, integrate, take the absolute value, and evaluate the result and the area between the two curve in excercise 1 is 40/3

The procedure for finding the area between two curves involves the following steps:

Identify the two curves: Determine the equations of the two curves that enclose the desired area.

Find the points of intersection: Set the two equations equal to each other and solve for the x-values where the curves intersect. These points will define the boundaries of the region.

Determine the limits of integration: Identify the x-values of the intersection points found in step 2. These values will be used as the limits of integration when setting up the definite integral.

Set up the integral: Depending on whether the curves intersect vertically or horizontally, choose the appropriate integration method (vertical slices or horizontal slices). The integral will involve the difference between the equations of the curves.

Integrate and evaluate: Evaluate the integral by integrating the difference between the two equations with respect to the appropriate variable (x or y), using the limits of integration determined in step 3.

Calculate the absolute value: Take the absolute value of the result obtained from the integration to ensure a positive area.

Round or approximate if necessary: Round the final result to the desired level of precision or use numerical methods if an exact solution is not required.

In summary, to find the area between two curves, determine the intersection points, set up the integral using the difference between the curves, integrate, take the absolute value, and evaluate the result.

Here's the procedure explained using the exercises:

Exercise 1:

Consider the functions F(x) = \(x^2 + 2x + 1\)and f(x) = 2x + 5. To find the area between these curves, follow these steps:

Set the two functions equal to each other and solve for x to find the points of intersection:

\(x^2 + 2x + 1 = 2x + 5\)

\(x^2 - 4 = 0\)

(x - 2)(x + 2) = 0

x = -2 and x = 2

The points of intersection, x = -2 and x = 2, give us the bounds for integration.

Now, determine which curve is above the other between these bounds. In this case, f(x) = 2x + 5 is above F(x) =\(x^2 + 2x + 1.\)

Set up the integral to find the area:

Area = ∫[a, b] (f(x) - F(x)) dx

Area = ∫\([-2, 2] ((2x + 5) - (x^2 + 2x + 1)) dx\)

Integrate the expression:

Area = ∫\([-2, 2] (-x^2 + x + 4) dx\)

Evaluate the definite integral to find the area:

Area = \([-x^3/3 + x^2/2 + 4x] [-2, 2]\)

Area = [(8/3 + 4) - (-8/3 + 4)]

Area = (20/3) + (20/3)

Area = 40/3

Therefore, the area between the curves F(x) = \(x^2 + 2x + 1\)and f(x) = 2x + 5 is 40/3 square units.

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To obtain the mixed number answer to 19 ÷ 6 from the whole-number-with-remainder answer, divide the numerator (19) by the denominator (6).

To find the mixed number answer to 19 ÷ 6, we divide 19 by 6. The whole-number quotient is obtained by dividing the numerator (19) by the denominator (6), which in this case is 3. This represents the whole number part of the mixed number answer, indicating how many complete groups of 6 are in 19. Next, we consider the remainder. The remainder is the difference between the dividend (19) and the product of the whole number quotient (3) and the divisor (6), which is 1. The remainder, 1, becomes the numerator of the fractional part of the mixed number.

This method makes sense because it aligns with the division process and provides a clear representation of the result. It shows the whole number part as the number of complete groups and the fractional part as the remaining portion. This representation is helpful in various real-world scenarios, such as dividing objects or quantities into equal groups or sharing items among a certain number of people.

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Cases (iii) and (iv) are left as exercises, meaning the proof for those cases is not provided in the given information. To fully establish the Characterization Theorem, the proof for these remaining cases needs to be completed.

Theorem 2.5.1 (Characterization Theorem):

If S is a subset of R that contains at least two points and has the property that if x, y ES and x < y, then (x, y)C S, then S is an interval.Proof.

There are four cases to consider:

(i) S is bounded,

(ii) S is bounded above but not below,

(iii) S is bounded below but not above, and

(iv) S is neither bounded above nor below.

Case (i): Let a = inf S and b = sup S.

Then SC[a, b] and we will show that (a, b)C S. If a < z < b, then there exist x, y

ES such that x < z < y. Since x < y and S has property (1), we have (x, y)C S.

Since zEP(x, y), it follows that zES.

Thus (a, b)C S.

Now if a S and b S, then S =[a, b].

If a S and b S, then S=(a, b).

The other possibilities lead to either S = (a, b) or S = [a, b].

Case (ii): Let b = sup S.

Then SC (-[infinity]o, b] and we will show that (-oo, b)C S. For, if z < b, then there exists y

ES such that z < y < b.

Since b is the least upper bound of S and yES, it follows that y 6S. But then (z, y)C (-oo, b) and (z, y)C S.

Thus (-oo, b)C S. Now if S contains its smallest element a, then S = [a, b]. Otherwise, S=(a, b).

Cases (iii) and (iv) are left as exercises.

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\(\\ \sf\longmapsto V=5in^3\)

\(\\ \sf\longmapsto side^3=5\)

\(\\ \sf\longmapsto side=\sqrt[3]{5}\)

\(\\ \sf\longmapsto side=2.2in\)

Side gets tripled

\(\\ \sf\longmapsto Volume=6.6^3\)

\(\\ \sf\longmapsto Volume=287.4in^3\)

Answer:

Step-by-step explanation:

If the side of the cube is s and volume is 5 and you triple the side then:

The enlarged cube has the volume: