3(4x +2) +3
I need help!!
Thx!

The man still owes R87,726 in taxes.

To calculate the tax amount the man still owes, we need to determine his taxable income first.

Gross income: R500,000

Less: PAYE deductions (17% of gross income): R500,000 * 0.17 = R85,000

Taxable income: R500,000 - R85,000 = R415,000

Next, we can use the tax table information provided to calculate the tax owed based on the taxable income.

Tax owed: R93,135 + 36% of (taxable income - R393,200)

Tax owed: R93,135 + 0.36 * (R415,000 - R393,200)

Tax owed: R93,135 + 0.36 * R21,800

Tax owed: R93,135 + R7,848

Tax owed: R100,983

Finally, we subtract the rebate amount of R13,257 from the tax owed to find the amount the man still owes.

Tax owed - Rebate: R100,983 - R13,257 = R87,726

Therefore, the man still owes R87,726 in taxes.

The correct option is not provided among the given answer choices.

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Answer:

If we denote the unknown number as "x", then the expression representing "9 less than the unknown number" would be:

x - 9

The exponential function Q=Q0 (1+r) t can be used to describe exponential growth or decay depending on the value of r. If r is less than 0, the function can be used to describe exponential growth, whereas if r is greater than 0, the function can be used to describe exponential decay.

The exponential function Q=Q0 (1+r) t can be used to describe exponential growth or decay. The function is determined by three variables: Q0, which is the starting value of the function; r, which is the growth or decay rate of the function; and t, which is the time when the function occurs. If r is less than 0, then the function can be used to describe exponential growth, meaning that the output of the function will increase over time. On the other hand, if r is greater than 0, then the function can be used to describe exponential decay, meaning that the output of the function will decrease over time. The exponential function can be used to describe a variety of different processes, such as population growth, compound interest, and radioactive decay. It can also be used to predict future values based on past values by using the growth or decay rate. Thus, the exponential function is a powerful tool for describing and predicting exponential growth and decay.

Q=Q0 (1+r) t

For example, if Q0 = 10, r = 0.5, and t = 5, then

Q=10 (1+0.5)5

Q= 10 x 1.53

Q= 15.3

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Based on the histograms, it appears that Los Angeles has a higher typical temperature than New York City during the summer month.

In statistics, a distribution is a way of showing how data is spread out or arranged. It describes the pattern of values in a dataset, and provides information on the frequency, range, and possible values of the data.

A distribution can take many forms, depending on the type of data being analyzed. For example, a distribution can be:

Symmetrical: where the data is evenly distributed around a central value, such as a normal distribution.

Skewed: where the data is not evenly distributed and is clustered more towards one end of the range than the other end, such as a positively skewed distribution.

Bimodal: where there are two distinct peaks in the data, such as in a distribution of test scores for a class with both high-performing and low-performing students.

There are many ways to represent a distribution, but some common methods include histograms, box plots, and probability density functions.

Understanding the distribution of data is important in statistics because it can help identify patterns, trends, and outliers in the data. It can also help inform decisions about which statistical methods to use when analyzing the data.

Here,

In the Los Angeles histogram, the peak of the distribution is around 80 degrees Fahrenheit, whereas in the New York City histogram, the peak is around 75 degrees Fahrenheit. However, the New York City histogram appears to have more variation in temperature than the Los Angeles histogram. This can be seen in the wider spread of the distribution, with a range of temperatures from around 65 to 95 degrees Fahrenheit, compared to the narrower range of temperatures in the Los Angeles distribution, which ranges from around 70 to 90 degrees Fahrenheit.

In other words, while the typical temperature in Los Angeles during the summer month is higher than in New York City, the temperature in New York City varies more widely from day to day.

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\(\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta r^2}{2} ~~ \begin{cases} r=radius\\ \theta =\stackrel{radians}{angle}\\[-0.5em] \hrulefill\\ \theta =\frac{\pi }{2}\\[1em] r=\frac{8}{3} \end{cases}\implies A=\cfrac{1}{2}\cdot \cfrac{\pi }{2}\cdot\left( \cfrac{8}{3} \right)^2 \\\\\\ A=\cfrac{1}{2}\cdot \cfrac{\pi }{2}\cdot \cfrac{64}{9}\implies A=\cfrac{16\pi }{9}\implies A\approx 5.59\)

Answer:#1: 1225

#2: 1149

Step-by-step explanation:

Hope this helps! :) ❤❤❤

Answer:

$1470 coupon of their product I have

Answer:

B- 20

D- no solutions

Step-by-step explanation:

Cuz its light work lol

Answer:

For the first question, the answer is x=20 so the answer is B

For the second I don't know, sorry :(

Step-by-step explanation:

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above

\(3x+y=5\implies y=\stackrel{\stackrel{m}{\downarrow }}{-3}x+5\qquad \impliedby \qquad \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\)

so we're really looking for the equation of a line that has a slope oif -3 and it passes through (2 , -4)

\((\stackrel{x_1}{2}~,~\stackrel{y_1}{-4})\hspace{10em} \stackrel{slope}{m} ~=~ - 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-4)}=\stackrel{m}{- 3}(x-\stackrel{x_1}{2}) \implies y +4 = - 3 ( x -2) \\\\\\ y+4=-3x+6\implies {\Large \begin{array}{llll} y=-3x+2 \end{array}}\)

now, keeping in mind that perpendicular lines have negative reciprocal slopes

\(\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{ -3 \implies \cfrac{-3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-3} \implies \cfrac{1}{ 3 }}}\)

so for this one, we're looking for the equation of a line whose slope is 1/3 and it passes through (2 , -4)

\((\stackrel{x_1}{2}~,~\stackrel{y_1}{-4})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-4)}=\stackrel{m}{ \cfrac{1}{3}}(x-\stackrel{x_1}{2}) \implies y +4 = \cfrac{1}{3} ( x -2) \\\\\\ y+4=\cfrac{1}{3}x-\cfrac{2}{3}\implies y=\cfrac{1}{3}x-\cfrac{2}{3}-4\implies {\Large \begin{array}{llll} y=\cfrac{1}{3}x-\cfrac{14}{3} \end{array}}\)

Answer:

The weight that does not round to 2.46 pounds is C 2.454 pounds.

Step-by-step explanation:

Based on the given information, the weights of the four similar packs of tomatoes are as follows:

Malcolm rounds the weights to the nearest hundredth pound. To round to the nearest hundredth pound, we look at the digit in the hundredths place. If it is 5 or greater, we round up the digit in the tenths place. If it is less than 5, we leave the digit in the tenths place as it is. Therefore, we can obtain the rounded weights as follows:

From the above rounded weights, we see that Pack C rounds to 2.45 pounds and does not round to 2.46 pounds. Therefore, the weight that does not round to 2.46 pounds is C 2.454 pounds.

Answer:

3x^2-3x-6=3(x^2-x-2)=3(x-2)(x+1)

x=2/x=-1

Step-by-step explanation:

The number of 4-fluidounce bottles that should be dispensed to meet the dosing requirements for the treatment of oropharyngeal candidiasis is 37.5 bottles.

To determine the number of 4-fluidounce bottles, we apply mathematical operations of addition, subtraction, division, and multiplication.

First, the total dosage is determined as 1,500 mg.

Since 40 mg/mL is available in 4-fluidounce bottles, the total dosage is divided by 40 mg/mL to determine the number of bottles required.

Dose of posaconazole for the first day = 100 mg twice

= 200 mg (100 mg x 2)

The total number of days for the dosage = 14

Dose for next 13 days = 1,300 mg (100 mg x 13)

Total dosage = 1,500 mg (200 mg + 1,300 mg)

40 mg/mL = 4-fluidounce bottle

1,500 mg = 37.5 bottles (1,500/40)

Thus, the number of 4-fluidounce bottles is 37.5 bottles.

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Answer:100 chocolates

Step-by-step explanation:

20 chocolate in each box so 20 time 5 is 100 so 20 time 11 is 220

Answer:

3

Step-by-step explanation:

Answer:

This is very cute.

Step-by-step explanation:

Answer:

Cute

Step-by-step explanation:

Answer:

7 schools have fewer than 50 classrooms.

For this, let's go through each problem carefully and step-by-step.

According to the question, the rate of change of the temperature of any object that is defined by T, is directly proportional to the difference of T and the temperature of the environment around it, which we'll denote as X.

\(\frac{dT}{dt}\)\(= k (T-X)\)

K is a constant of proportionality here. And the temperature of the surrounding environment is said to be (-18°C). Thus,

\(\frac{dT}{dt} = k(T+18)\).

For part A, in order to find the differential equation for T, we need to solve for k. So we separate the variables and then integrate to solve the equation.

\(\int\limits{\frac{dT}{T+18} } = \int\limits {k} \, dt\)

\(ln(T+18) = kt+c\)

Now thw inital temperature of a pot of chili is 23°C, so at \(t = 0, T_0 = 23*C\).

Substituting 23 for T and 0 for t, we have the following:

\(ln(23+18) = k(0)+c\)

\(ln(41) = c\)

We know the temperature of chili after 2 hours is 7°C, so we know that when \(t = 2, T_1 = 7\)

Substituting t for 2, and T for 7, we get:

\(ln(7+18) = 2k+ln(41)\)

\(ln(25) = 2k + ln(41)\)

Solving for 2k

\(2k = ln(25) -ln(41)\)

\(2k = ln(\frac{25}{41})\)

\(k = \frac{1}{2}ln(\frac{25}{41})\).

Substituting the value of \(\frac{dT}{dt} = k (T+18)\), the differential equation obtained is \(\frac{dT}{dt} = \frac{1}{2}ln(\frac{25}{41})(T+18)\).

For part B, to find the temperature of the chili after four hours, we first need to solve the above differential equation.

The solution of the differential equation is given by the equation \(ln(T+18) = kt+c\). Substituting the values of k and c, we have:

\(ln(T+18) = \frac{1}{2}ln(\frac{25}{41})t+ln(41)\).

Using the above relation, at any time (t), the temperature (T) can be found out in the following.

At \(t = 4, T_2 = \phi\)

\(ln(T_2+18)=\frac{1}{2}ln(\frac{25}{41})*4+ln(41)\)

\(ln(T_2+18)=2ln(\frac{25}{41})+ln(41)\)

\(ln(T_2+18)=-0.989 + 3.714\)

\(ln(T_2+18)\) ≅ \(2.725\)

Solving the natural logarithm,

\(T_2+18 = e^{2.725} = 15.256\)

\(T_2 =15.256 - 18\)

\(T_2 = -2.744\).

So the temperature of the chili after four hours would be -2.744°C approximately.

To find part C in what time the chili would be 10°C, we need to substitute again.

\(t = \phi\)\(, T = -10\)

\(ln(-10 + 18) = \frac{1}{2}ln(\frac{25}{41})t + ln(41)\)

\(ln(8) = \frac{1}{2}ln(\frac{25}{41})t + ln(41)\)

Solving for \(\frac{1}{2}ln(\frac{25}{41})t\),

\(\frac{1}{2}ln(\frac{25}{41})t = ln(8) - ln(41)\)

\(\frac{1}{2}ln(\frac{25}{41})t = ln(\frac{8}{41})\)

\(\frac{1}{2}ln(\frac{25}{41})t = -1.634\)

\(ln(\frac{25}{41})t\)\(= -1.634 * 2\)

\((-0.494)t=-3.268\)

\(t = \frac{-3.268}{-0.494}\)

\(t=6.615\) hours, approximately.

Thus, the chili would reach -10°C at around 6.615 hours.

Hope this helped. This took me a long time.

Answer:

b,c,e,f is perfect square and other are not a perfect square

Answer: i'll show you how to do it.

Step-by-step explanation:

Is there any number that multiplied by itself gives you one of the options obove?

Example:

81 = 9×9... So it is a perfect square

36 = 6x6... It's a perfect square too

27 = ther isn't a number that multipied by itself is equal to 27... Then, it is not a perfect square

The solution set of the inequality for R is given by \(-\frac{1172}{85} \leq x\leq \frac{1228}{55}\) .

An inequality in mathematics is a relation that compares two numbers or other mathematical expressions in an unequal way.  It is frequently used to compare two numbers on the number line based on their sizes. Different types of inequalities are represented by a variety of notations such as < ,> ,≤ and ≥.

The modulus function takes the absolute value of a function.

To solve the given inequality we have simplify the inequality first.

\(x-\frac{7}{9} |(\frac{12}{5} -6x)|+50\ge -30\)

\(or, x-\frac{7}{9}\times \frac{6}{5} |(2-5x)|+50\ge -30\\or,x-\frac{14}{15}|2-5x|\ge-80\\\)

Now we separate the inequalities:

\(x-\frac{14}{15}(2-5x)\ge-80\)  for 2x-5 ≥ 0

\(x-\frac{14}{15}(-(2-5x))\ge-80\)  for 2x-5 ≤ 0

Now we will solve the two inequalities separate and find the intersection.

\(x-\frac{14}{15}(2-5x)\ge-80\\or,15x-28+70x\ge-1200\\or,x\ge -\frac{1172}{85}\)

Again

2x-5 ≥ 0

or, \(x\le\frac{2}{5}\)

hence \(-\frac{1172}{85}\le x\le\frac{2}{5}\)

Similarly solving the other inequality we get

\(\frac{2}{5}\le x\le\frac{1228}{55}\)

Now we find the intersection of the two inequalities:

\(-\frac{1172}{85}\le x\le\frac{1228}{55}\)

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The value of sin A is, 7/25.

The circle is a closed two dimensional figure , in which the set of all points is equidistance from the center.

We have to given that;

The terminal side of angle A goes through the point (−24/25,7/25) on the unit circle.

Now, By definition we get;

⇒ cos A = - 24/25

⇒ Sin A = 7/25

Thus, The value of sin A is, 7/25.

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The value of the equations are a = 10 , b = -5.1 and c = 1

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as A

Now , the value of A is

Substituting the values in the equation , we get

a + ( -7 ) = 3   be equation (1)

On simplifying the equation , we get

a - 7 = 3

Adding 7 on both sides of the equation , we get

a = 3 + 7

a = 10

Substituting the values in the equation , we get

1.9 + b = -3.2   be equation (2)

Subtracting 1.9 on both sides of the equation , we get

b = -3.2 - 1.9

b = -5.1

Substituting the values in the equation , we get

-1/6 + ( 7/6 ) = c   be equation (3)

c = ( 7 - 1 ) / 6

c = 6/6

c = 1

Hence , the equations are solved

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The probability distribution for x is:

x P(x)

0 0.214

1 0.571

2 0.214

Since there are 8 cameras in the box, the total number of ways to choose 2 cameras from the box is given by the combination formula:

C(8,2) = 8!/(2!×(8-2)!) = 28

So there are 28 possible ways to choose a sample of 2 cameras.

Now, let's calculate the probability of each possible value of x:

When x = 0, both cameras in the sample are non-defective. The number of ways to choose 2 non-defective cameras from the 4 non-defective cameras in the box is given by the combination formula:

C(4,2) = 4!/(2!×(4-2)!) = 6

So the probability of x = 0 is:

P(x=0) = (number of ways to choose 2 non-defective cameras)/(total number of ways to choose 2 cameras)

= 6/28

= 0.214

When x = 1, one camera in the sample is defective and the other is non-defective. The number of ways to choose 1 defective camera from the 4 defective cameras in the box and 1 non-defective camera from the 4 non-defective cameras in the box is given by the product of the corresponding combinations:

C(4,1) × C(4,1) = 4×4 = 16

So the probability of x = 1 is:

P(x=1) = (number of ways to choose 1 defective camera and 1 non-defective camera)/(total number of ways to choose 2 cameras)

= 16/28

= 0.571

When x = 2, both cameras in the sample are defective. The number of ways to choose 2 defective cameras from the 4 defective cameras in the box is given by the combination formula:

C(4,2) = 4!/(2!×(4-2)!) = 6

So the probability of x = 2 is:

P(x=2) = (number of ways to choose 2 defective cameras)/(total number of ways to choose 2 cameras)

= 6/28

= 0.214

Therefore, the probability distribution for x is:

x P(x)

0 0.214

1 0.571

2 0.214

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The equation X^2/2 - 8 = 2X - 2 has two solutions, X = 6 and X = -2. These are the values of X that satisfy the equation and make both equations Y = X^2/2 - 8 and Y = 2X - 2 intersect on the graph.

To find the solutions to the equation X^2/2 - 8 = 2X - 2, we need to set the two equations equal to each other and solve for X.

The equation is:

X^2/2 - 8 = 2X - 2

To simplify the equation, let's multiply both sides by 2 to eliminate the fraction:

X^2 - 16 = 4X - 4

Next, we rearrange the equation to have all terms on one side:

X^2 - 4X - 12 = 0

Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use factoring in this case:

(X - 6)(X + 2) = 0

Setting each factor equal to zero gives us two possible solutions:

X - 6 = 0 --> X = 6

X + 2 = 0 --> X = -2

So the solutions to the equation X^2/2 - 8 = 2X - 2 are X = 6 and X = -2.

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Answer:

The maximum possible volume = 332.685 cubic inches

Step-by-step explanation:

From the information given:

Let the height be = x

The length be = 24 - 2x

The width be = 12 - 2x

Then V = x (24 -2x) ( 12 - 2x)

V = x ( 288 -48x - 24x +4x²)

V = x(288 - 72x + 4x²)

V = 288x - 72x² + 4x³

\(\dfrac{dV}{dx}= (3x^2 - 36x+ 72)\)

\(\implies x^2 - 12x + 24 = 0\)

\(=\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(=\dfrac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(24)}}{2(1)}\)

\(=\dfrac{12 \pm \sqrt{144 - 96}}{2}\)

\(=\dfrac{12 \pm \sqrt{48}}{2}\)

where;

\(x \ne \dfrac{12 + \sqrt{48}}{2}\)

∴ \(x= \dfrac{12 - \sqrt{48}}{2}\)

x = 2.536 ( since the length cannot be negative)

So, the length x = 24 - 2(2.535) = 18.93

The width = 12 - 2(2.535) = 6.93

heigth = 2.536

∴

V = 18.93 ×  6.93 ×  2.536

V = 332.685 cubic inches

    A box is made from a plastic sheet by removing square from each corner, maximum volume of this box will be 332.55 cubic inches.

           Volume = Length × Width × Height

Given in the question,

Let the measure of each side of a square removed = x inches

Therefore, length of the box = (12 - 2x) inches

Width of the box = (24 - 2x) inches

Height of the box = x inches

Volume of the box (V) = (12 - 2x) × (24 - 2x) × (x)

                                V  =  4x³ - 72x² + 288x

To find the maximum volume of the box differentiate the expression with respect to 'x' and equate it to zero.

V' = 12x² - 144x + 288

For V' = 0,

12x² - 144x + 288 = 0

x² - 12x + 24 = 0

\(x=\frac{12\pm\sqrt{(12)^2-4(1)(24)} }{2(1)}\)

\(x=(6\pm2\sqrt{3})\)

\(x=9.46,2.54\)

For x = 9.46,

Volume of the box = 4(9.46)³- 72(9.46)² + 288

                               = -2769.03 cubic inches

For x = 2.54,

Volume of the box = 4(2.54)³- 72(2.54)² + 288

                               =  332.55 cubic inches

       Therefore, maximum volume of the box will be 332.55 cubic inches.

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Answer:

The cash flow statement paints a picture as to how a company’s operations are running, where its money comes from, and how money is being spent. So i think D is the answer

Starting from the x-value (A column) for -7 and going down the y-values values (B column) are 5, 4, 3, 2, 1, 0, 1, 2

The function is given to us in the B column. We can use this to find the y-values without having to plug them in each time.

Finding the Vertex

In absolute value functions, there is a vertex. The vertex, in this case, is the minimum where the y-values change from decreasing to increasing. The absolute value formula is y = a|x - h| + k.

The vertex is equal to (h,k). The h-value in the function above is -2 and the k-value is 0. Thus, the vertex is at (-2,0).

Finding Slope

The slope of an absolute value function is written before the x. Since there isn't anything before the x in this function, the slope must be 1. So, the values either increase or decrease by 1.

Like all positive absolute value functions, the graph will decrease, reach 0, then begin to increase. The graph will reach 0 at the vertex.

Writing Out the Values

The table shows us that the y-values begin at 6. Then, we decrease by one until we get to 0.

Each of these x-values matches up with one of the y-values. Then, the graph begins to increase once again.

So together the y-values for the x-values -8 through 0 are 6, 5, 4, 3, 2, 1, 0, 1, 2.

The average rate of change of the function f(x) = √(x+4) over the interval 2 ≤ x ≤ 6 is approximately 0.29 to the nearest hundredth.

To find the average rate of change of the function f(x) = √(x+4) over the interval 2 ≤ x ≤ 6, we need to calculate the change in the function divided by the change in the input variable over that interval.

The change in the function between x = 2 and x = 6 is:

f(6) - f(2) = √(6+4) - √(2+4) = √10 - √6

The change in the input variable between x = 2 and x = 6 is:

6 - 2 = 4

So, the average rate of change of the function over the interval 2 ≤ x ≤ 6 is:

(√10 - √6) / 4

To approximate the answer to the nearest hundredth, we can use a calculator or perform long division to get:

(√10 - √6) / 4 ≈ 0.29

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Answer: A) max at (14, 6) = 64,   min at (0,0) = 0

Step-by-step explanation:

Graph the lines at look for the points of intersection.

Input those points into the Constraint function (2x + 6y) and look for the maximum value and minimum value.

Points of Intersection: (0, 0), (17, 0), (0, 10), (14, 6)

Point     Constraint 2x + 6y

(0, 0):                    2(0) + 6(0) = 0              Minimum

(17, 0):                    2(17) + 6(0) = 34

(0, 10):                    2(0) + 6(10) = 60

(14, 6):                    2(14) + 6(6) = 64           Maximum

Answer:

1)A) Minimize C = 40x + 100y.

2) d) 7x + 8y ≤ 190, 5x + 4y ≤ 22, x ≥ 0, y ≥ 0.

3) B) Min C = 295x + 416y; s.t 23x + 15y ≥ 154, 17x + 25y ≥ 140, x ≥ 0, y ≥ 0.

Step-by-step explanation:

1) Energize pills cost 40 cents each, while Excel pills cost 100 cents each. If x represent the energize pills and y represent the excel pills, the objective function would be:

minimize cost = cost of energize pill + cost of excel pill

Minimize cost = 40x + 100y

2) Let x = the number of mountain bikes they produce, and let y = the number of racing bikes they produce.

7 hours of assemble time to produce a mountain bike and 8 hours of assemble time to produce a racing bike. Since the company has at most 190 hours for assemble time, it can be represented as:

7x + 8y≤ 190

5 hours of mechanical tuning to produce a mountain bike and 4 hours of mechanical tuning to produce a racing bike. Since the company has at most 22 hours for mechanical tuning, it can be represented as:

5x + 4y ≤ 22

Also assemble time and mechanical time must be greater than 0, hence:

x ≥ 0, y ≥ 0

3) The fieldwork constraint can be represented as:

23x + 15y ≥ 154

The time to be spent at university research center can be represented as:

17x + 25y ≥ 140

Both fieldwork time and time spent at university research center mut be greater than 0, hence:

x ≥ 0, y≥ 0

The cost equation is:

Minimize cost = 295x + 416y

The meaning of the x-value on the line when y = 100 is the number of hours a student should spend on their homework to expect a score of 100 on the quiz.

1. The given information suggests that the math teacher plots student grades on the y-axis (vertical axis) and the number of hours they study on the x-axis (horizontal axis).

2. The teacher then draws a line of best fit, which represents the general trend or relationship between the quiz scores and the number of study hours.

3. We are asked to determine the meaning of the x-value on the line when y = 100, which implies finding the number of study hours required to expect a score of 100 on the quiz.

4. To find this value, we look for the point on the line where the y-coordinate is 100.

5. By observing the graph or equation of the line, we can determine the corresponding x-value or the number of study hours.

6. In this case, the x-value corresponding to y = 100 represents the number of hours a student should spend on their homework to expect a score of 100 on the quiz.

7. Therefore, the x-value on the line when y = 100 provides an estimate of the amount of time a student needs to allocate for studying in order to achieve a score of 100 on the quiz.

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