4. What function represents how much more water can be added to the bucket before it
overflows? Explain how you solved this problem.

On solving the provided question we got to know that - Thus, the temperature will be 14°C below zero at midnight.

Since equations are essentially questions, efforts to systematically find answers to those questions have been the driving force behind the development of mathematics. From straightforward algebraic equations that just require addition or multiplication to differential equations, exponential equations that use exponential expressions, and integral equations, there are many different types of equations that range in complexity.

The answer to the query is provided as,

Temperature at the start, or at midday, is 10°C.

Temperature change at a rate of 2°C per hour

Then,

[tex]Temperature at 1 PM = 10 + (-2) = 10 - 2 = 8[/tex] °C

[tex]Temperature at 2 PM = 8 + (-2) = 8 -2 = 6\\[/tex]°C

[tex]Temperature at 3 PM = 6 + (-2) = 6 - 2 = 4\\[/tex]°C

[tex]Temperature at 4 PM = 4 + (-2) = 4 - 2 = 2\\[/tex]°C

[tex]Temperature at 5 PM = 2 + (-2) = 2 - 2 = 0\\[/tex]°C

[tex]Temperature at 6 PM = 0 + (-2) = 0 -2 = -2\\[/tex]°C

[tex]Temperature at 7 PM = -2 + (-2) = -2 -2 = -4\\[/tex]°C

[tex]Temperature at 8 PM = -4 + (-2) = -4 – 2 = -6\\[/tex]°C

[tex]Temperature at 9 PM = -6 + (-2) = -6 – 2 = -8[/tex]°C

At nine o'clock at night, it will be eight degrees below zero.

Then,

the temperature at 12 AM, or midnight

Temperature change over 12 hours = -2°C divided by 12 equals - 24°C

Therefore, the temperature at midnight will be 10 + (-24)

= - 14°C

Thus, the temperature will be 14°C below zero at midnight.

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The values are 5, 6, 8, 9 for the given function y=∛x+7.

So, option D is correct.

In mathematics, a function from a set X to a set Y assigns each element of X exactly one element of Y.

The earliest known concept of a function was created by two Persian mathematicians, Al-Biruni and Sharaf al-Din al-Tusi. [4] At first, functions were envisioned as the perfect representation of the relationship between two variables. In mathematics, a function from a set X to a set Y assigns each element of X exactly one element of Y. Simply described, a function is an association of inputs where each input is coupled to a single, distinct output. Each function has a codomain or range.

Given,

y=∛x+7

Let x=-8,

y=∛8+7

y=-2+7

y=5

x=-1, y=∛(-1)+7

y=-1+7

y=6

x=1, y=∛1+7

y=1+7

y=8

x=8, y=∛8+7

y=2+7

y=9

Therefore, the values are 5, 6, 8, 9

So, option D is correct

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The complete question is:

"Complete the table for the function y = ∛x+7

OA) -5, -6, -8, -9

OB) -9, -8, -6, -5

Oc9,8. 6. 5

OD) 5,6, 8, 9".

The difference between living with family and on campus is 27299 and she will be able to save an average of $4549.9 every month.

To determine how much she will save, we have to find the total cost of living with parents or relative and the cost of living on/off campus and then divide it by the total number of months which is six and the find their difference.

The total cost of living with parents or relative for 6 months is = 900 + 4158 + 2259 + 675 + 11205 + 19197 + 36587 + 44579 = 119560

The cost of living on or off campus for 6 months = 900 + 12798 + 2259 + 468 + 11205 + 27630 + 36587 + 55012 = 146859

The average cost of living with relative or family for 6 months = 119560 / 6 = 19926.6

The average cost of living on or off campus for 6 months = 146859 / 6 = 24476.5

The difference is = 24476.5 - 19926.6 = 4549.9

Amy will save approximately $4549.9 monthly

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1. Using division operations, Amy's monthly budget for textbooks and course supplies is $100.

2. Amy's monthly budget for rent, utilities, and food is $1,422.

3. Amy's monthly budget for personal and miscellaneous costs is $251.

4. Amy's total monthly budget is $1,825

The total monthly budget is the total budget divided by the period (nine months).

We can use division operations to determine the individual monthly budgets for each expense class.

Textbooks and Course Supplies =         $900      $100

Rent, Utilities, and Food =                   $12,798   $1,422

Personal & Miscellaneous Expense = $2,259     $251

Transportation =                                      $468       $52

Total non-tuition expenses =             $16,425  $1,825 ($16,425/9)

The number of months = 9 months

Monthly expenses = $1,825

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The missing height of parallelogram is 6 units

Define Area of Parallelogram

Given,

base = 9 units

Area = 54 unit²

Now, the formula for area of parallelogram is

Area = b * h

let, h = height of parallelogram which we need to find

so, now put the values

54 = 9 * h

h = 54 / 9

h = 6 units

Hence, the missing height of parallelogram is 6 units.

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9482.7 must Joe invest today to receive $22,000 each year for the next 22 years.

percentage, a relative value indicating hundredth parts of any quantity.

Given,

Joe Jan wants to receive $22,000 each year for the next 22 years.

6% interest rate compounded annually.

Convert 6% to decimal value

6/100=0.06

A=P(1+rt)

22000=P(1+0.06(22))

22000=P(1+1.32)

22000=P(2.32)

Divide both sides by 2.32

P=9482.7

Hence, 9482.7 must Joe invest today to receive $22,000 each year for the next 22 years.

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The angle between the two sides of lengths a and b that maximizes the area of the triangle is θ = π/2.

The area of a triangle whose sides of lengths a and b with an angle θ between them is

A = 1/2ab sinθ

The given triangle has two sides of lengths a and b.

Consider θ is the angle between the a and b sides. Then the area of the triangle is

A = 1/2ab sinθ

For maximizing the area of the triangle, derivating the area we have and equate to zero i.e.,

dA/dθ = 0

⇒ dA/dθ = 1/2 ab cosθ = 0

⇒ cosθ = 0

∴ θ = π/2

And the condition for the maximum area is d²A/dθ² < 0.

I.e., d²A/dθ² = -1/2ab sinθ

⇒ d²A/dθ² = -1/2ab sin(π/2) = -1/2ab < 0

So, the maximum area is  

Δ(max) = 1/2 ab sin(π/2) = 1/2ab

Thus, the angle between the two sides of lengths a and b of a triangle is π/2 when its area is maximum.

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Therefore , the best choice is you to join jmu and after that you will able to earn the same amount you spend on the graduation in 2 year.

Equation: A statement that two variable or integer expressions are equal. In essence, equations are questions, and the motivation for the development of mathematics has been the systematic search for the answers to these questions.

Here,

Given:

You would earn $37,870 per year at walmart

Tuition at jmu is $20,000 per year, textbook expenses are $2,699 per year, and room and board expenses are $15,000 per year.

and you will get a $75,000 per year after graduation.

So the total expense is = 20,000 + 15000 + 2699 =37,699

Therefore , the total expense at jmu = $37,699

and total 4 year graduation course fees = 37,699 * 4=1,50,796

Thus, total fees you have to pay = $1,50,796

and after this you will get a job  of $75,000 per year.

you will able to earn the same amount you spend on the graduation in 2 year.

Therefore , the best choice is you to join jmu and after that you will able to earn the same amount you spend on the graduation in 2 year.

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The congruency of the right triangles by the Hypotenuse-Leg Congruency Theorem can be made true by the following sides.

1. [tex]\overline{RP}[/tex] ≅ [tex]\overline{MF}[/tex] and   [tex]\overline{DP}[/tex] ≅ [tex]\overline{QF}[/tex]

2. [tex]\overline{CI}[/tex] ≅ [tex]\overline{CE}{}[/tex] and    [tex]\overline{AI}[/tex] ≅ [tex]\overline{GE}[/tex]

3. [tex]\overline{TR}[/tex] ≅  [tex]\overline{RT}[/tex]  and  ;[tex]\overline{TQ}[/tex] ≅ [tex]\overline{RS}[/tex]

4. [tex]\overline{DG}[/tex] ≅  [tex]\overline{KN}[/tex] and  [tex]\overline{AD}[/tex] ≅ [tex]\overline{HK}[/tex]

Two right triangles are congruent if  the hypotenuse side and one leg  of one triangle is congruent to  the hypotenuse side and one leg of the other triangle.

The hypotenuse leg congruency theorem states that two right triangles are congruent if the hypotenuse and one leg of one triangle are congruent to the hypotenuse and one leg of the other triangle

The hypotenuse side of ΔDPR = [tex]\overline{RP}[/tex]
The hypotenuse side of ΔQMF = [tex]\overline{MF}[/tex]

The corresponding legs of both triangles are;

[tex]\overline{DP}[/tex] corresponds to [tex]\overline{QF}[/tex]

[tex]\overline{DR}[/tex] corresponds to [tex]\overline{MQ}[/tex]

Therefore, ΔDPR is congruent to ΔQFM if  [tex]\overline{RP}[/tex] ≅ [tex]\overline{MF}[/tex]

And the leg  [tex]\overline{DP}[/tex] ≅ [tex]\overline{QF}[/tex] or  [tex]\overline{DR}[/tex] ≅ [tex]\overline{MQ}[/tex]

2. The sides that makes the congruency statement, ΔACI ≅ ΔGCE are found as follows;

The hypotenuse side of  ΔACI = [tex]\overline{CI}[/tex]

The hypotenuse side of triangle ΔGCE = [tex]\overline{CE}{}[/tex]

Therefore,

ΔACI ≅ ΔGCE ican be proved when    [tex]\overline{CI}[/tex]  is congruent to [tex]\overline{CE}{}[/tex], and

 [tex]\overline{AI}[/tex]  is congruent to [tex]\overline{GE}[/tex] or    [tex]\overline{AC}[/tex]  is congruent to [tex]\overline{CG}[/tex]

3. In the figure the specified information are;

[tex]\overline{TR}[/tex] is congruent to [tex]\overline{RT}[/tex] by reflexive property

The triangles are congruent if either;[tex]\overline{TQ}[/tex] iis congruent to [tex]\overline{RS}[/tex] or;[tex]\overline{TS}[/tex] iis congruent to [tex]\overline{RQ}[/tex]

4. The appropriate sides that can be used to prove that  ΔADG ≅ ΔHKN  are the hypotenuse side of both triangles;[tex]\overline{DG}[/tex] and  [tex]\overline{KN}[/tex]

and either the legs; [tex]\overline{AD}[/tex] and [tex]\overline{HK}[/tex] or  [tex]\overline{AG}[/tex] and  [tex]\overline{HN}[/tex]

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,

The appropriate sales journal entry for the building plans prepared for Spangler Construction Company is as follows:

Debit Accounts Receivable (Spangler Construction Company) $900

Credit Drafting Fees Earned $900

(To record the revenue from drafting fees.)

Sales or service journals are used to record sales or service revenue transactions.

When the services are performed on credit, the Accounts Receivable account is debited while the Service Revenue account is credited.

Accounts Receivable (Spangler Construction Company) $900 Drafting Fees Earned $900

Thus, the above demonstrates the appropriate recording of the service revenue transaction in the sales/service revenue journal.

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Prepare the appropriate journal entry.

the inverted coefficient matrix is

[tex]\left[\begin{array}{ccc}2&-1&-5\\3&-2&5\\-3&5&2\end{array}\right][/tex]

1. The inverted coefficient matrix is found by taking the inverse of the coefficient matrix.

3x1 + x2 = 4

5x1 + 2x2 = 7

3. Therefore, the inverted coefficient matrix is

2 -1 -5

3 -2 5

-3 5 2

It is used to represent and manipulate complex data, such as linear equations, vectors, and polynomials. A matrix can also be used to represent a system of linear equations or to graph a function. Matrices are key elements in linear algebra, a branch of mathematics that studies vector spaces, linear transformations, and systems of linear equations.

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The right tail of the chi-square distribution curve's value for 19 degrees of freedom and an area of 0.010 is 3.162.

To find the value of chi-square for 19 degrees of freedom and an area of 0.010 in the right tail of the chi-square distribution curve, we need to consult a chi-square distribution table. The table will list the chi-square values corresponding to different degrees of freedom and areas of the right tail of the distribution. In this case, the table should list the chi-square value corresponding to 19 degrees of freedom and an area of 0.010 in the right tail of the distribution. The value of this chi-square is 3.162, which is the exact answer from the chi-square distribution table.

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if a polygon layer and a line layer are intersected, the geometry of the output feature class may be chosen as either a polygon or a line.

If the line layer is used to clip the polygon layer, the output feature class will be a polygon feature class that contains the part of the polygon layer that intersects with the line layer. The output will be a polygon feature class that contains the same attributes as the input polygon layer, plus a new attribute field indicating the line layer's object ID (OID) that was used to clip the feature. If the line layer is used to split the polygon layer, the output feature class will be a line feature class that contains the line features that were used to split the input polygon layer. The output will be a line feature class that contains the same attributes as the input line layer, plus a new attribute field indicating the polygon layer's OID that was used to split the features.

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The equation of the line perpendicular to 7x+5y = 4 is y = 5/7(x) -15/7. The equation of the line parallel to 7x+5y = 4 is  y = -7/5(x) -53/5

Given that: the line is perpendicular/parallel to the line 7x+5y = 4 and it passes through (−4,-5)

The slope-intercept form of a straight line is:

y = mx + c

where m is the slope and c is the y-intercept

For perpendicular case:

When the two lines are perpendicular, the product of their slope is -1 i.e.

m₁ x m₂ = -1

where m₁ and m₂ represent the slope of the lines

Let the slope of the given line be m₁ and the slope of the unknown line be m₂

Line1(given):

7x+5y = 4  => y = (-7/5)x + 4/5

Thus,  m₁ = -7/5

m₁ x m₂ = -1

-7/5 x m₂ = -1

m₂= 5/7

Line 2(unknown) with the point (−4,-5):

y = mx + c

-5 = 5/7 (-4) + c

-5 = -20/7 + c

c = -15/7

Thus, the equation of the line is y = 5/7(x) -15/7

For parallel case:

When the two lines are parallel, they have equal slope i.e. m₁ = m₂

Line1(given):

7x+5y = 4  => y = (-7/5)x + 4/5

Thus,  m₁ = -7/5

m₂= -7/5

Line 2(unknown) with the point (−4,-5):

y = mx + c

-5 = -7/5(-4) + c

-5 = 28/5 + c

c = -53/5

Thus, equation of the line is y = -7/5(x) -53/5

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The magnitude of u + v is 6.08 and the direction of the vector u + v is 4.7°.

What is rounded to the nearest tenth?

To write a decimal number up to one decimal place, round it to the nearest tenth. This is done in a way that leaves one digit after the decimal point after rounding.

Given that vector u is 7 (cos 330°, sin 330°)

Vector v equal to is 4 (cos 90°, sin 90°)

Calculating the value of cos 330° and  sin 330°:

cos 330°

= cos (3×90° + 60°)

= sin 60°

= √3/2

sin 330°

= sin (3×90° + 60°)

= -cos 60°

= -1/2

The vector u can be written as 7 (cos 330°, sin 330°) = 7(√3/2, -1/2) =(7√3/2, -7/2) = 7√3/2 i - 7/2j

The vector v can be written as 4 (cos 90°, sin 90°) = 4(0,1) = (0,4) = 0i + 4j

Add vector u and v:

u + v

= 7√3/2 i - 7/2j +  0i + 4j

= 7√3/2 i + 1/2 j

= 6.06 i + 0.5 j

The magnitude of the vector  u+v is √( 6.06² + .5²) = 6.08

The angle is

tan θ = y/x

tan θ =0.5/6.06

θ = 4.7°

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The population in the year 2000 is 395.808

Exponential Function

Exponential functions have the form f(x) = bx, where b > 0 and b ≠ 1. Just as in any exponential expression, b is called the base and x is called the exponent. An example of an exponential function is the growth of bacteria. Some bacteria double every hour

Where t is the number of months since the population was first recorded. And the population after 36 months so we need to replace t=36 months into the function

Probability

Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur, or how likely it is that a proposition is true. The probability of an event is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty.

So then can conclude that after 38 months the population of mouse is between 1963 and 2000.

A(t)=336e^(0.031t)

The population can be represented with this formula:

A(38)=336e^(0.031*38)

Where t is the number of months since the population was first recorded. And  the population after 38 months so we need to replace t=38 months into the function and we got:

A(38)=336e^(0.031*38)

A(38)=336e^(1.178)

       =395.808

So then can conclude that after 38 months the population of mouse is between 1963 and 2000.

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There are no crictical numbers of the given function 12-5e^(-x).

Let's Consider

[tex]f(x) = 12 - 5*e^-x[/tex].

What is critical number of a function.

If a number causes the derivative of an expression to equal 0, it is critical.

As a result, we must set the derivative of the expression to 0.

Local Maxima and Local Minima points are also consider as critical number of a function.

f`(x) = 0 or f`(x) =1/0 by solving both equation we can find all the critical points of a function f(x).

Derivative of given funciton f(x) is f`(x).

f`(x) = [tex]0 - 5*(e^-x)*(-1).[/tex]

f`(x) = [tex]5*(e^-x)[/tex].

f`(x) = 0

[tex]5*e^-x[/tex] = 0

[tex]e^-x[/tex] = 0

[tex]e^-x[/tex] never going to zero because e^-x is an exponential decreasing function.

So there are no crictical numbers of the given function f(x).

So We enter none in this question.

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The value for  given trigonometric functions are csc 30° = 2, cot 60°  = √3/3, cos 30 ° = √3/2, and cot 30° = √3.

A branch of mathematics called trigonometry examines correlations between side lengths and angles of the triangle. Trigonometric Identities are equality conditions that apply to all values of the variables in the equation and which require trigonometry functions.

Given, sin 30° = 1/2 and tan 30° = √3/3.

The formula for csc or cosec x = 1/sinx.

Then,

[tex]\begin{aligned}\csc 30^{\circ} &=\frac{1}{\frac{1}{2}}\\&=2\end{aligned}[/tex]

The formula for cot θ = tan(90°- θ)

Then,

[tex]\begin{aligned}\cot 60^{\circ} & = \tan(90^{\circ}-60^{\circ}) \\&= \tan 30^{\circ}\\&=\frac{\sqrt{3}}{3}\end{aligned}[/tex]

The formula for cos θ = sin θ/tan θ.

Then,

[tex]\begin{aligned}\cos 30^{\circ} &= \frac{\sin 30^{\circ}}{\tan 30^{\circ}}\\&= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{3}}\\&=\frac{3}{2\sqrt3}\times \frac{\sqrt3}{\sqrt3}\\&=\frac{\sqrt3}{2} \end{aligned}[/tex]

The formula for cot θ = 1/tan θ.

Then,

[tex]\begin{aligned}\cot 30^{\circ}&=\frac{1}{\tan30^{\circ}}\\&=\frac{1}{\frac{\sqrt3}{3}}\\&=\frac{3}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}\\&=\sqrt{3}\end{aligned}[/tex]

Therefore, the answers are 2, √3/3, √3/2, and √3.

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Answer:

x= 40 degree

y = 80 degree

Step-by-step explanation:

where is z?