Car A began a journey from a point at 9 am, traveling at 30 mph. At 10 am car B started traveling from the same point at 40 mph in the same direction as car A. At what time will car B pass car A?

The equation that represents the profit for the month is: y = 33.09x - (2,137.00 + 15.15x).

The ongoing expenses incurred from the normal day-to-day operations of a business are referred to as operating costs.

The equation that represents the profit for the month is:

y = 33.09x - (2,137.00 + 15.15x)

This equation takes into account the revenue generated by selling x units, which is 33.09x, and subtracts the total cost of producing those x units, which is (2,137.00 + 15.15x). The result is the total profit generated for the month.

We can simplify the equation by combining like terms:

y = 17.94x - 2,137.00

This equation can be used to determine the total profit for any given number of units produced.

The coefficient of x (17.94) represents the profit generated per unit sold, and the constant term (-2,137.00) represents the fixed costs that must be subtracted from the revenue to calculate the profit.

Therefore, the correct equation is y = 33.09x - (2,137.00 + 15.15x), because it takes into account both the revenue and the total costs associated with producing x units.

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we have

[tex](3x^{2} -2y^{2}+5x)+(4 x^{2}+12y^{2}-7x)[/tex]

Combine like terms

[tex](3x^{2} -2y^{2}+5x)+(4 x^{2}+12y^{2}-7x)\\=(3x^{2}+4x^{2})+(-2y^{2}+12y^{2})+(5x-7x)\\=7x^{2} +10y^{2}-2x[/tex]

therefore

the answer is the option C

He combined the terms [tex]-2y^{2}[/tex] and [tex]12y^{2}[/tex] incorrectly

The average daily balance for the given billing cycle is calculated by summing the product of each daily balance and the length of time it was held, and then dividing by the number of days in the billing cycle. After doing this, the average daily balance comes out to be $755.17.

The subject of this question is finance and it involves calculating the average daily balance for a billing cycle. In this method, the daily outstanding balances are added together and then divided by the total number of days in the billing period.

Here, the billing cycle lasts for 29 days, from 4/17 to 5/17.

Like this, you calculate the balance for each portion of the billing cycle, then multiply each by the number of days that balance was held, add those up, and finally, divide by the number of days in the billing cycle:

((1100*10) + (400*2) + (700*8) + (640*11)) / 29 = $755.17

Therefore, the average daily balance is $755.17.

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Random sample - choosing a sample so that every member has an equal chance of being chosen (this is the absolute definition of random sampling)

Sample - a portion of a large group is termed as a sample.

Biased sample -  choosing a sample from people who are under 15 years old (biased sampling states that all the members of a group are not included in survey)  

Estimating from a sample - using the responses of a small group to make conclusions about the large group.(estimation means roughly judging something based on some sample )

The equation in standard form passing through (-4, 2) with a slope of 9/2 is 9x - 2y = -10.

Since the equation is in standard form, rearrange it to slope-intercept form y = mx + b.

Given slope m = 9/2 and point (-4, 2), substitute these values to find the correct equation.

After substitution, the equation is 9x - 2y = -10.

Answer:

16 years

Step-by-step explanation:

Given,

Number of students in group A = X,

Sum of their ages = 221,

So, the average age of students in group A = [tex]\frac{\text{Sum of ages}}{\text{Total students}}[/tex]

[tex]=\frac{221}{X}[/tex]

According to the question,

[tex]\frac{221}{X}=\text{Integer}[/tex]

∴ X must be a factor of 221,

∵ 221 = 13 × 17,

Now, If X = 13,

Then the number of students in group B = 26 ( NOT POSSIBLE )

If X = 17,

Then the number of students in group B = 34

Which is between 30 and 40,

Now, Let S be the sum of ages of students in group B,

So, the average age in group B = [tex]\frac{S}{34}[/tex]

Again according to the question,

[tex]\frac{S+221}{17+34}=\frac{S}{34}-1[/tex]

[tex]\frac{S+221}{51}=\frac{S-34}{34}[/tex]

[tex]34S+7514 = 51S - 1734[/tex]

[tex]34S - 51S = -1734 - 7514[/tex]

[tex]17S = 9248[/tex]

[tex]\implies S = 544[/tex]

Therefore, the average age of B = [tex]\frac{544}{34}[/tex] = 16 years.

Answer:

c

Step-by-step explanation:

Answer: At the end of May, there will be $372.

Explanation:

Since we have given that

On the first day of May, each one save =$0.25

Number of friends = 3

Total save on the first day of May is given by

[tex]0.25\times 3=\$0.75[/tex]

Similarly,

On the second day of May, each one save =$0.25

Total save on the second day of May is given by

[tex]0.50\times 3=\$1.5[/tex]

Similarly,

On the third day of May, each one save =$0.25

Total save on the third day of May is given by

[tex]0.75\times 3=\$2.25[/tex]

This series becomes arithmetic progression,

[tex]0.75,1.5,2.25......[/tex]

Here, a = 0.75

d=common difference is given by

[tex]d=a_2-a_1=1.5-0.75=0.75\\d=a_3-a_2=2.25-1.5=0.75[/tex]

Since there are 31 days in the month of May,

So, n=31

We need to calculate the sum of 31 terms to get our answer,

[tex]S_n=\frac{n}{2}(2a+(n-1)d)\\\\S_{31}=\frac{31}{2}(2\times 0.75+(31-1)\times 0.75)\\\\=\frac{31}{2}(1.5+30\times 0.75)\\\\=\frac{31}{2}(24)\\\\=31\times 12\\\\=\$372[/tex]

Hence, at the end of May, there will be $372.

The total amount in the fund at the end of May is 239.25.

To find the total amount of money in the fund at the end of May, we need to calculate the sum of the money put into the fund on each day. The students decide to put 0.25 on the first day, 0.50 on the second day, 0.75 on the third day, and so on. This is an arithmetic sequence, where the first term (a) is 0.25 and the common difference (d) is 0.25. The formula to find the sum of an arithmetic sequence is:

S = (n/2)(2a + (n-1)d)

In this case, n = 31 (since there are 31 days in May), a = 0.25, and d = 0.25. Plugging these values into the formula, we get:

S = (31/2)(2 * 0.25 + (31-1) * 0.25) = 239.25

Therefore, at the end of May, there will be 239.25 in their fund.

Answer:

y = (x + 1)^2

Step-by-step explanation:

I used desmos and put each answer choice into the graphing calculator and answer choice D was the only one that matched with the image in the question.

In a scatterplot that depicts an exponential model, the data points will form a curve that resembles the shape of an exponential function.

In a scatterplot that depicts an exponential model, the data points will form a curve that resembles the shape of an exponential function. The x-axis represents the independent variable, while the y-axis represents the dependent variable. As the x-values increase, the corresponding y-values will increase at an exponential rate. For example, if we are graphing the growth of a population over time, the scatterplot will show the population size increasing rapidly as time progresses.

On a scatterplot depicting an exponential model, the data points are represented as points on a curvy line. The correct answer is option D) points on a curvy line.

In a scatterplot depicting an exponential model, the data points typically form a curvy line. Exponential relationships show rapid growth or decay, which is reflected in the scatterplot as a smooth, non-linear curve.

As the independent variable increases, the dependent variable changes exponentially, resulting in data points that do not lie along a straight line but instead form a distinctive curve. This curve can either rise sharply or decrease rapidly, depending on whether the exponential function involves growth or decay.

For instance, in exponential growth, the data points start small and then rapidly increase, forming a curve that steepens over time. Conversely, in exponential decay, the data points start large and decrease rapidly, forming a curve that flattens out as it approaches zero.

Therefore, on a scatterplot representing an exponential model, the data points are typically observed to form a curvy line, as seen in option **D)**.

Complete question : How is the data represented on a scatterplot that depicts an exponential model?

A) points on a line

B) random points

C) points on a horizontal line

D) points on a curvy line

Answer:

The correct option is 1.

Step-by-step explanation:

It is given that in ΔRST , segment SU is an altitude. It means the angle SUR and angle SUT are right angles.

It triangle RST and SUT,

[tex]\angle RST=\angle SUT=90^{\circ}[/tex]         (Definition of altitude)

[tex]\angle STR=\angle UTS[/tex]         (Common angle)

Two corresponding angles are equal. So by AA property of similarity,

[tex]\triangle RST\sim \triangle SUT[/tex]              .... (1)

It triangle RST and RUS,

[tex]\angle RST=\angle SUR=90^{\circ}[/tex]         (Definition of altitude)

[tex]\angle SRT=\angle URS[/tex]         (Common angle)

Two corresponding angles are equal. So by AA property of similarity,

[tex]\triangle RST\sim \triangle RUS[/tex]              .... (2)

According to transitive property of equality,

if a=b and b=c, then a=c.

From (1) and (2), we get

[tex]\triangle RUS\sim \triangle SUT[/tex]             ( Transitive property of equality)

Therefore the correct option is 1.