the base radius and height of a right circular cylinder are measured as 8 cm and 16 cm, respectively, with a possible error in measurement of as much as 0.2 cm each. use the total differential to estimate the maximum error in the calculated volume of the cylinder.

The number of orders placed in one quarter of the year and shipped in the next quarter will vary depending on the segment. To determine the exact number, you would need to analyze the data from each segment separately.

To determine the number of orders placed in one quarter of the year but shipped in the next quarter, you will need to analyze the data from each segment separately. Begin by extracting the data from the segment you are interested in and sorting it by order date and ship date. For example, if you are interested in orders placed in Q1 2009 and shipped in Q2 2009, you will need to identify all orders with an order date in Q1 2009 and a ship date in Q2 2009. Once you have identified these orders, you can count them up to determine the total number of orders in the segment that fit this criteria. You can then repeat this process for the other segments to determine the total number of orders placed in one quarter of the year but shipped in the next quarter across all segments.

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If Brenda plans to reduce her spending by $70 a month, the future value of this reduced spending over the next 12 years is $14710.5.

We are given that Brenda will be saving $70 every month for a period of 12 years.

So, the future value of the annuity formula will be used for the calculation of future value.

Future value = R [( (1 + i)^n - 1 ) / i], where R is the regular payment, i is the interest rate and n is the number of payments

Here R = monthly amount to be saved = $70

i = interest rate = 6/12 = 0.5%

n = Number of payments = 12 x 12 = 144 months

Future value=70 [( (1+0.5%)^144 - 1) / 0.5%] = $14710.5

Hence, the future value of this reduced spending over the next 12 years is $14710.5.

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The number of different groups of 4 people that a company could send to a job site, which is 27,720.

In the context of this question, a company has 30 people and wants to send 4 people to a job site.

To find out how many different groups of 4 people they could make, we need to use the concept of permutation.

To find the number of permutations, we use the formula nPk, where n is the total number of objects (30 in this case) and k is the number of objects we want to arrange (4 in this case).

So, the number of permutations will be

=> 30P4 = 30!/(30-4)!.

This simplifies to

=> 30!/(26!) = 27,720.

Therefore, the company has 27,720 different groups of 4 people that they could send to a job site. This means that they can send 27,720 different combinations of 4 people from their 30 employees to a job site.

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Salut/Hello!

Answer: w = 20 cm and l = 28 cm

Step-by-step explanation:
l - length
w - width
P - perimeter
/ - fraction line
=> - results

we know that P = 2 x (l + w)
we also know that our ratio is l/w = 7/5

l/w = 7/5 => l = 7/5 x w
so now we can go back to the perimeter and replace l

96 = 2 x (7/5 x w + w)
96 = 2 x (7/5 x w/1 + w)   Why w/1? - We can't multiply unless w is also in a fraction, so we put it as w/1
96 = 2 x (7w/5 + w)
96 = 2 x (7w/5 + w/1)  and we amplify w/1 with 5
96 = 2 x (7w/5 + 5w/5)
96 = 2 x 12w/5
48 = 12w/5
48/1 = 12w/5
48 x 5 = 12w
240 = 12w
w = 20 cm

with that we can find l
96 = 2 x (l + 20)
48 = l + 20
l = 48 - 20
l = 28 cm

I hope it was helpful! :]

Answer:

y = 8x+3

Step-by-step explanation:

hoping its correct for you

Answer:

see explanation

Step-by-step explanation:

for the given angles to form a triangle they must sum to 180°

given

31 [tex]\frac{3}{4}[/tex]°  , 53 [tex]\frac{1}{2}[/tex]° , 94 [tex]\frac{3}{4}[/tex] ° ( changing to decimal form and adding )

= 31.75° + 53.5° + 94.75°

= 180°

Thus the 3 given angle measures can form a triangle.

The system of equation has no solution

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

2y = 6

y = -15

Since these are two different equations with the same variable, y, we can use substitution to find a solution.

Starting with the second equation, y = -15, we can substitute this value into the first equation:

2y = 6

2(-15) = 6

-30 = 6

This equation is false, meaning there is no solution for y that satisfies both equations.

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The probability that X is between 0.5 and 1 is (C) 19/24.

The probability that X is between 0.5 and 1 can be found by calculating the definite integral of the PDF f(x) over the interval [0.5, 1]:

P(0.5 <= X <= 1) = ∫f(x)dx from 0.5 to 1

Given the PDF f(x) = ax + X^2 for 0<=x<=1, we can evaluate this definite integral as:

P(0.5 <= X <= 1) = ∫(ax + X^2)dx from 0.5 to 1

= [ax^2/2 + x^3/3] from 0.5 to 1

= (1.5a + 1/3) - (0.5a + 0.125)

= (1 + 1/6) - (0.5 + 1/8)

= (2/3 - 3/8)

= 5/24

Since 5/24 is equal to 0.2083, the closest answer choice to this value is 19/24.

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Answer:

It's already in order

[tex]\frac{3}{4}[/tex] × [tex]\frac{4}{9}[/tex], [tex]\frac{7}{7}[/tex] × [tex]\frac{4}{9}[/tex]  , [tex]1\frac{2}{3}[/tex] × [tex]\frac{4}{9}[/tex], [tex]2[/tex] × [tex]\frac{4}{9}[/tex]

Step-by-step explanation:

[tex]\frac{3}{4}[/tex] × [tex]\frac{4}{9}[/tex]

[tex]=\frac{1}{3}[/tex]

[tex]\frac{7}{7}[/tex] × [tex]\frac{4}{9}[/tex]

[tex]=\frac{4}{9}[/tex]

[tex]1\frac{2}{3}[/tex] × [tex]\frac{4}{9}[/tex]

[tex]=\frac{20}{27}[/tex]

[tex]2[/tex] × [tex]\frac{4}{9}[/tex]

[tex]=\frac{8}{9}[/tex]

Change them all to the same common denominator of 27

[tex]\frac{9}{27}, \frac{12}{27}, \frac{20}{27}, \frac{24}{27}[/tex]

A statement could be used to describe the functions include the following: D. the domain of f(x) is (−∞, 0] while the domain of g(x) is [0, ∞).

In Mathematics, a domain can be defined as the set of all real numbers for which a particular function is defined.

Additionally, the horizontal extent of any graph of a function represents all domain values and they are read and written from smaller to larger numerical values, and from the left of a graph to the right.

By using interval notation, the domain of this function shown in the graph above can be written as follows;

Domain of f(x) = {-∞, 0}

Domain of g(x) = {0, ∞}

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The numeric value of the function for each case is given as follows:

5. f(-11) = 1841.

6. g(1/3) = 14.78.

7. h(-1/2) = -5/8.

To find the numeric value of a function or of an expression, we replace each instance of the variable in the function or in the expression by the value at which we want to find the numeric value.

Hence item 5 is solved as follows:

f(-11) = -(-11)³ + 5(-11)² + 9(-11) + 4

f(-11) = 1841.

Item 6 is solved as follows:

g(1/3) = 3(1/3)³ + 6(1/3)² + 12(1/3) + 10

g(1/3) = 14.78.

Item 7 is solved as follows:

h(-1/2) = 9(-1/2)³ - 8(-1/2)² + 11(-1/2) + 8

h(-1/2) = -5/8.

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The volume of the sawdust pile is 3754.667 cubic feet

A cone is a three-dimensional shape that has a circular base with tapers from the flat base into the vertex.

The formula for searching the volume of the cone can be described below :

V = [tex]\frac{1}{3}[/tex] π [tex]r^{2}[/tex] h

which

V = volume of the cone

r = radius of cone circular base

h = height of cone ( tall of the cone )

From the question, we have following information :\

1. Base diameter = 32 feet

  Base radius = Base diameter / 2 = 32 feet / 2 = 16 feet

2. Tall / Height of the cone = 14 feet

Since the known parameter is enough to be put into the formula, we can simply subtitute the formula with known parameter

V = [tex]\frac{1}{3}[/tex] π [tex]r^{2}[/tex] h

  = [tex]\frac{1}{3}[/tex] x π x [tex]16^{2}[/tex] x 14

  = 3754.667 cubic feet

Hence the volume of the cone is 3754.667 cubic feet

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The number of students actually enrolled is 6000.

The mathematical expression combines numerical variables and operations denoted by addition, subtraction, multiplication, and division signs.

Mathematical symbols can be used to represent numbers (constants), variables, operations, functions, brackets, punctuation, and grouping. They can also denote the logical syntax's operation order and other properties.

Given that a large university accepts 50​% of the students who apply. Of the students the university​ accepts, 40​% actually enroll. If 30,000.

The number of students will be calculated as:-

Number = ( 30000 x 0.5 x 0. 4 )

Number = 6000

Therefore, the number of students will be 6000.

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Answer:

Each pound of sugar costs $0.64

Step-by-step explanation:

4x = 2.56     Set up the equation. Divide both sides by 4 to isolate the x.

 x = 0.64

Answer:

1. B) x = 6

2. No solutions

3. D) x = 4​

4. A) x = -4

Step-by-step explanation:

1. -4x + 8 = -3x + 2

8 = x + 2  ==> add 4x on both sides to move x to one side of the equation

x = 6  ==> subtract 2 on both sides

B. x = 6

2. 9x - 7 = 5x - 19

4x - 7 = -19 ==> subtract 5x on both sides to move x to one side of the

                        equation

4x = -12  ==> add 7 on both sides to isolate x

x = -3  ==> divide 4 on both sides

x = -3 isn't one of the options, so problem 2 has no solutions.

3. 2x - 7 = 5x - 19

-7 = 3x - 19  ==> subtract 3x on both sides to move x to one side of the

                        equation

12 = 3x  ==> isolate x by adding 19 on both sides

x = 4  ==> divide both sides by 3

D. x = 4​

4. -7x + 3 = -3x +19

3 = 4x + 19  ==> add 7x on both sides to move x to one side of the equation

-16 = 4x  ==> subtract 19 on both sides to isolate x

x = -4  ==> divide both sides by 4

A. x = -4

Answer:

1

Step-by-step explanation:

[tex]\cos \left(\arccos \left(\frac{3}{5} \right)+\arcsin(-\frac{4}{5} \right)=\cos(\arccos(3/5))\cos(\arcsin(-4/5))-\sin(\arccos(3/5))\sin(\arcsin(-4/5)) \\ \\ =(3/5)\cos(\arcsin(-4/5))-(-4/5)(\sin(\arccos(3/5)) \\ \\ =(3/5)(3/5)-(-4/5)(4/5) \\ \\ =1[/tex]

The two expressions that could represent the dimension of the rectangle are x + 2 and x - 7

The area of a rectangle can be represented by the product of its length and width. The formula for the area is given by:

A = l x w

where A is the area, l is the length and w is the width.

If the area is given by the expression x^2 - 5x - 14, then the two expressions for the dimensions can be determined by getting the factors of the expression.

x^2 - 5x - 14 = (x + 2)(x - 7)

Therefore, the dimensions of the rectangle can be expressed as x + 2 and x - 7.

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Answer: 9

Step-by-step explanation:

g + 9 = 18

g = 18-9

  =9

The rate at which the area of the circle is increasing when the radius is 7 m is 527.52 m^2.

We already know that the area of a circle can easily be calculated using the formula -

= A = π(r^2)

So, in order to find out the rate at which it increases, we have to -

= dA/dt = dπ(r^2)/dt

= dA/dt = dπ(r^2)/dr × dr/dt

= dA/dt = 2πr × dr/dt

It has been mentioned that the value of dr/dt is 12 m/s.

So, after using this value, we find that the rate of increase is -

= dA/dt = 2π × 7 × 12

= dA/dt = 527. 52 m^2

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The mean of the sample size of 25 would fall between 68 and 72 68% of the time.

Using the 68-95-99.7 rule, if a variable is normally distributed with mean μ and standard deviation σ, then:

Since the interval we are interested in (68 to 72) falls within the 68% of the data that falls within 1 standard deviation of the mean, it means that 100% - 32% = 68% of the data falls within this interval.

Thus, the mean of this sample size of 25 would fall between 68 and 72 68% of the time.

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The number of campers is y = 650(676/650)^x

Exponential equations are equations in which variables occur as exponents.

An exponential equation in the form y = a(b)^x can be used to model the number of campers, y, x years after the camp opened. Let's use x = 0 for the first year the camp opened (last year) and x = 1 for this year.

Then, we have:

y = a(b)^0 = 650 (when x = 0)

y = a(b)^1 = 676 (when x = 1)

We can use these two points to solve for the values of a and b.

First, we'll use the first equation to solve for a:

a = 650

Next, we'll use the second equation to solve for b:

676 = 650(b)^1

676/650 = (b)^1

b = 676/650

So, the exponential equation in the form y = a(b)^x that can model the number of campers is:

y = 650(676/650)^x

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Answer: the answer is A), normal is 28 + 4fee, but if she buys LOTS its 22 +8

contact me through discord Acathia#0103, its easier to help, I can help with multiple questions aswell

Step-by-step explanation:

Answer:

a) 144π cm²

b) 248π cm²

Step-by-step explanation:

Using the formula, curved surface area of a cone = πrl where r = radius and l = slant height,

Curved surface area of large cone = π (10) (12 + 3) = π (10)(15) = 150π

Curved surface area of small cone = π(2)(3) = 6π

a) Curved surface area of frustum = 150π - 6π = 144π cm²

b)
The frustum has a top surface which is a circle of radius 2 and a bottom surface which is also a circle of radius 10

The area of a circle is πr²

So Total area of both circles = π(2)² + π(10)² = 4π + 100π = 104π

Total surface area of frustrum = 144π + 104π = 248π cm²

Answer:

-5

Step-by-step explanation:

Δy =9-7=2

Δx = -12-(-2) =-10

-10/2= -5

Answer:

the number of revolutions that are required to level the playground is 194 revolutions

Step-by-step explanation:

We know that LSA, that is lateral surface area of a cylinder is the area of a rectangular sheet, which when spread onto the ground, covers the area equal to the LSA of the cylinder.

Hence, area covered by the roller in one revolution = area of LSA of cylinder  = 32 cm².

Now, to cover the are of 6200 cm²,

the number of revolutions required  = Total area of the playground/ LSA of the cylinder = 6200/32 = 193.75 = 194

Hence the number of revolutions that are required to level the playground is 194.

The integral using integration by parts with the indicated choices of u and dv is equal to,   [tex]\int\limits 5x^{2} lnx dx[/tex] = [tex]\frac{5x^{3} }{3} (ln(x)-\frac{1}{3} )[/tex] + c

Basic Power Rule:

f(x) = cxⁿ

f’(x) = c· n xⁿ⁻¹

Integration

Integrals

[Indefinite Integrals] Integration Constant C

Integration Property [Multiplied Constant]:              

[tex]\int\limits cf(x)dx[/tex] = c [tex]\int\limitsf(x) dx[/tex]

Integration Rule [Reverse Power Rule]:

[tex]\int\limits x^{n} dx[/tex] = [tex]\frac{x^{n+1} }{n+1}[/tex] + c

Integration by Parts:  

[tex]\int\limits u dv[/tex] = uv - [tex]\int\limits v du[/tex]

Given that,

= [tex]\int\limits 5x^{2} lnx dx[/tex]

Rewrite [Integration Property - Multiplied Constant]:

[tex]\int\limits 5x^{2} lnx dx[/tex] = 5 [tex]\int\limits x^{2} lnx dx[/tex]

Set u:  

u = [tex]lnx[/tex]

[u] Logarithmic Differentiation:          

du = [tex]\frac{1}{x}[/tex] dx

Set dv:  

dv = [tex]x^{2}[/tex]                                                                                                        

[dv] Integration Rule [Reverse Power Rule]:  

v = [tex]\frac{x^{3} }{3}[/tex]

Integration by Parts:  

[tex]\int\limits 5x^{2} lnx dx[/tex] = 5 [tex]( \frac{x^{3}ln(x) }{3} - \int\limits \frac{x^{2} }{3} dx )[/tex]

Rewrite [Integration Property - Multiplied Constant]:  

[tex]\int\limits 5x^{2} lnx dx[/tex] = 5 [tex](\frac{x^{3}ln(x) }{3} -\frac{1}{3} \int\limits x^{2} dx )[/tex]

Factor:  

[tex]\int\limits 5x^{2} lnx dx[/tex] = [tex]\frac{5}{3} (x^{3} ln(x) - \int\limits x^{2} dx )[/tex]

Integration Rule [Reverse Power Rule]:  

[tex]\int\limits 5x^{2} lnx dx[/tex] = [tex]\frac{5}{3} (x^{3} ln(x) - \frac{x^{3} }{3} )[/tex] + c

Factor:

[tex]\int\limits 5x^{2} lnx dx[/tex] = [tex]\frac{5x^{3} }{3} (ln(x)-\frac{1}{3} )[/tex] + c

Therefore,

The integral using integration by parts with the indicated choices of u and dv is equal to,   [tex]\int\limits 5x^{2} lnx dx[/tex] = [tex]\frac{5x^{3} }{3} (ln(x)-\frac{1}{3} )[/tex] + c

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The required solution of the quadratic equation is given as x = -2 and -5. Option D is correct.

The process in mathematics to operate and interpret the function to make the function or expression simple or more understandable is called simplifying and the process is called simplification.

Here,
x² + 7x + 10 = 0

Factorize the above expression,

(x + 5)(x + 2) = 0

Now,
x + 5 = 0  ; x + 2 = 0
x = -5        : x = -2

Thus, the required solution of the quadratic equation is given as x = -2 and -5. Option D is correct.

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The shape of Banana is taking the shape of a parabola.

a plane curve created when a point moves so that the distance between it and a fixed point or line is equal.

y2=4ax, where is the distance from the parabola's vertex to focus, is the standard equation for a normal parabola.

Several examples of parabolas in real life

The wheel stance in yoga resembles a parabola.

Banana's shape is beginning to resemble a parabola.

A parabola is emerging from the Rainbow.

Antennas with a parabolic dish are resembling a parabola.

Mirror's concave surface.

Any point on a parabola is at an equal distance from both the focus, a fixed point, and the directrix, a fixed straight line. A parabola is a U-shaped plane curve. The topic of conic sections includes parabola, and all of its principles are discussed here.

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