Kiersten runs a website that helps people learn programming. Every month, Kiersten receives a subscription fee of \$10$10dollar sign, 10 from each of the subscribers to her website. Last month, the website had 100100100 subscribers. This month, 303030 new members joined the website, and 101010 members cancelled their subscription. How much money will Kiersten’s online business earn this month?

Answer:

17

Step-by-step explanation:

Eliminating parentheses, we have ...

... 1 - 5q +2·2.5q +2·8

... = 1 -5q +5q +16 . . . . . carry out the multiplication

... = q(-5+5) +(1 +16) . . . . group like terms

... = q·0 +17 . . . . . . . . . . . combine like terms

... = 17

Answer:

The correct answers are

(-2,1)

(-4,2)

(1,-1)

Step-by-step explanation:

hope dis help u guys, habe a good rest of ya day!

Julianna will earn $22,579.20 if she works for 46 weeks, which is closest to the option D) $22,500.

To calculate the amount of money Julianna will earn if she works for 46 weeks, we need to multiply the number of work hours per week (32) by the hourly wage ($14.15), and then multiply that result by the number of weeks worked (46). This can be calculated as follows:

Total earnings = (32 hours/week) × ($14.15/hour) × (46 weeks)

Total earnings = $22,579.20

Therefore, the closest amount of money Julianna will earn if she works for 46 weeks is $22,579.20, which is closest to option D) $22,500.

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Answer:

6 stickers.

Step-by-step explanation:

We have been given that Scott makes 6 rows with 5 stickers in each row.

The total number of stickers in 6 rows would be:

[tex]\text{The total number of stickers}=6\times 5[/tex]

[tex]\text{The total number of stickers}=30[/tex]

Therefore, there are 30 stickers in 6 rows.

We need to find the number of stickers in 5 rows having the same number of stickers.

Since there are 30 stickers in total, so the number of stickers in each row would be 30 stickers by 5 rows.

[tex]\text{The number of stickers in each row}=\frac{30}{5}[/tex]

[tex]\text{The number of stickers in each row}=6[/tex]

Therefore, there would be 6 stickers in each row.

Answer:

(2.2, 0.5, 1.1)

Step-by-step explanation:

The parametric equation of the line normal to the plane and through point (2, 0, 1) can be written ...

... L = (2, 0, 1) + t(2, 5, 1)

We want to find the value of t (and the corresponding point) that makes L satisfy the equation of the plane.

... 2(2+2t) +5(0 +5t) +1(1+t) = 8 . . . . . put values from L in for x, y, z in plane

... 5 + 30t = 8 . . . . . simplify

... t = (8 -5)/30 = 0.1 . . . . solve for t (subtract 5, divide by 30)

For this value of t, L is ...

... (2, 0, 1) + 0.1(2, 5, 1) = (2.2, 0.5, 1.1)

compare the 5 and 10 dimension then the 7 and 14 dimensions

10/5 =2

14/7 =2

 they both are the same ratio/factor

 the factor is 2

Answer:

2:1

Step-by-step explanation:

it asks for the scale since 5 x 2 = 10 and 7 x 2 = 14

the scale is 2 : 1

The likelihood that the flight will be no more than 5 minutes late would be equal to a flight that arrives between 2 hours and 2 hours and 10 minutes.

So the answer would be:

P (5 late or less) = 10 * 1/20 = 0.5

The answer is 0.5

But if we are looking for the probability of the flight that would be 10 minutes late, the likelihood of that event would be 0.25.

Final answer:

The probability that a Delta airlines flight from Cincinnati to Tampa will be no more than 5 minutes late, given that the actual flight times are uniformly distributed between 2 hours and 2 hours, 20 minutes, is 0.50.

Explanation:

The question deals with the uniform distribution of flight times and calculating the probability that a flight will be no more than 5 minutes late. Assuming flight times are uniformly distributed between 2 hours (120 minutes) and 2 hours, 20 minutes (140 minutes), the total range of flight times is 20 minutes (140 minutes - 120 minutes).

Delta airlines quotes a flight time of 2 hours, 5 minutes, which is 125 minutes. Therefore, a flight being no more than 5 minutes late means the flight time would be at most 130 minutes. To find the probability that the flight time is 130 minutes or less, we calculate the proportion of the distribution that falls within this range.

The probability that the flight will be no more than 5 minutes late is the length of the interval from 120 minutes to 130 minutes divided by the total range (20 minutes), which is (130 - 120) / 20 = 10 / 20 = 0.5. Thus, the probability is 0.50 to two decimal places.

Final Answer:

Word done = 31 units

Explanation:

To find the work done by a force field [tex]$\mathbf{F}(x, y, z)$[/tex] on a particle moving along a path, we can use the line integral of the force field along that path.

First, we need to define the force field [tex]$\mathbf{F}(x, y, z)$[/tex] and the parametric equations for the line segment.

Given the force field

[tex]$$\mathbf{F}(x, y, z) = \langle y + z, x + z, x + y \rangle,$$[/tex]
we have the following components:

[tex]F_x &= y + z, \\\\F_y &= x + z, \\\\F_z &= x + y.[/tex]

Now, let's find the parametrization of the line segment from the point (1,0,0) to the point (5,3,2). We can define a vector [tex]$\mathbf{r}(t)$[/tex] that describes this line segment by

[tex]$$\mathbf{r}(t) = \mathbf{r_0} + t(\mathbf{r_1} - \mathbf{r_0}),$$[/tex]

where [tex]$\mathbf{r_0} = \langle 1, 0, 0 \rangle$[/tex] is the starting point, [tex]$\mathbf{r_1} = \langle 5, 3, 2 \rangle$[/tex] is the ending point, and t is a parameter that goes from 0 to 1.

So we have

[tex]\mathbf{r}(t) &= \langle 1, 0, 0 \rangle + t(\langle 5, 3, 2 \rangle - \langle 1, 0, 0 \rangle) \\\\ &= \langle 1, 0, 0 \rangle + t\langle 4, 3, 2 \rangle \\\\ &= \langle 1 + 4t, 3t, 2t \rangle.[/tex]

From the parametric equation, we have the components:

[tex]x(t) &= 1 + 4t, \\\\y(t) &= 3t, \\\\z(t) &= 2t.[/tex]

The line integral of the force field along the path can be computed by

[tex]$$W = \int_C \mathbf{F} \cdot d\mathbf{r},$$[/tex]

where [tex]$d\mathbf{r}$[/tex] is the differential element along the path C, and the dot product represents the scalar product of the force field and the differential path element.

We can express [tex]$d\mathbf{r}$[/tex] in terms of t as

[tex]$$d\mathbf{r} = \mathbf{r}'(t) dt = \langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \rangle dt = \langle 4, 3, 2 \rangle dt.$$[/tex]

Then the work done is

[tex]W &= \int_{t=0}^{t=1} \mathbf{F}(x(t), y(t), z(t)) \cdot \mathbf{r}'(t) dt \\\\ &= \int_{0}^{1} \langle 3t + 2t, 1 + 4t + 2t, 1 + 4t + 3t \rangle \cdot \langle 4, 3, 2 \rangle dt \\\\ &= \int_{0}^{1} [(5t)(4) + (6t + 1)(3) + (7t + 1)(2)] dt \\\\ &= \int_{0}^{1} [20t + 18t + 3 + 14t + 2] dt \\\\ &= \int_{0}^{1} [52t + 5] dt \\\\ &= [26t^2 + 5t]_{0}^{1} \\\\ &= 26(1)^2 + 5(1) - (26(0)^2 + 5(0)) \\\\ &= 26 + 5 \\\\ &= 31.[/tex]


So, the work done by the force field [tex]$\mathbf{F}(x, y, z)$[/tex] on the particle as it moves along the line segment from (1, 0, 0) to (5, 3, 2) is 31 units of work.

Answer:

The correct answers are options B and C.                            

Step-by-step explanation:                      

B It has four right angles.

C It has two pairs of parallel opposite sides.

All angles are right angles by definition of a rectangle. The rectangle has 2 pairs of parallel opposite sides.              

Option A is wrong as its a square whose all 4 sides have equal length.

Answer:   [tex]x-2y+7=0[/tex]

Step-by-step explanation:

Given : The  line[tex]L_1[/tex] has equation [tex]2x + y = 8[/tex]

In intercept form : [tex]y =-2x+8[/tex]

The slope of [tex]L_1[/tex] =-2    [in intercept form equation of line [tex]y =mx+c[/tex], m is slope ]

Let p be slope of  [tex]L_2[/tex]

We know that when two lines are perpendicular then the product of their slopes is -1.

[tex]\Rightarrow\ -2\times p=-1\\\\\Rightarrow\ p=\dfrac{1}{2}[/tex]

The equation of line having slope 'm' and passing through (a,b) is given by :-

[tex](y-b)=m(x-a)[/tex]

Then , equation of line  [tex]L_2[/tex] will be :-

[tex](y-7)=\dfrac{1}{2}(x-7)\\\\\Rightarrow\ 2y-14=x-7\\\\\Rightarrow\ x-2y+7[/tex]

The sum of three consecutive integers where the last integer is z is 3z - 3.

The sum of three consecutive integers can be derived using the concept of arithmetic series. For instance, if z represents the last of the three consecutive integers, we know that the numerals preceding it are z-1 and z-2. Hence, the sum of these three integers would be (z-2) + (z-1) + z.

Simplifying this, we get 3z - 3 which corresponds to option c.)3z-3 in the question's list.

Remember, an important aspect of sequences and series is that the numbers are arranged in a specific order.

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Answer:

Stu estimated correctly by rounding up.

He ordered enough brick.

The exact area is [tex]228.75[/tex] square feet

Step-by-step explanation:

we know that

the area of the patio is equal to

[tex]A=bh[/tex]

In this problem we have

[tex]b=12.5\ ft, h=18.3\ ft[/tex]

substitute in the formula

[tex]A=12.5*18.3=228.75\ ft^{2}[/tex]

[tex]228.75\ ft^{2}< 247\ ft^{2}[/tex]

Answer:

Third option:

[tex]A(x)\approx 2,000(1.065)^x[/tex]

Step-by-step explanation:

Notice if we invest 2,000 then the first month then after the first 0.4% interest is earned we will have in the account:

2,000(1+0.004) = 2,000(1.004)

I just applied the simple interest formula.

Then after the second month we will have:

[tex]2,000(1.004)(1.004) = 2,000(1.004)^2[/tex]

And so on each month.

So at the end of the first year (12 months) we have:

[tex]2,000(1.004)^{12}[/tex]

But we also earn an additional 1.5% in the year, so we will have:

[tex]2,000(1.004)^{12}(1.015)[/tex]

At the end of the second year we will have:

[tex]2,000[(1.004)^{12}(1.015)]^2[/tex]

At the end of the third year:

[tex]2,000[(1.004)^{12}(1.015)]^2[/tex]

And so on,

So if the number of years is denoted with x, at the end of x years we will have the following amount in the savings:

[tex]2,000[(1.004)^{12}(1.015)]^x[/tex]

We use our calculator to simplify that inside the brackets:

[tex](1.004)^{12}(1.015)=1.06481[/tex]

Rounding  to the third decimal place we get: 1.065

So notice the amount in the savings after x years will be:

[tex]2,000(1.065)^x[/tex]

Since they want it in function notation we just write the A(x) that denotes the amount in the savings:

[tex]A(x)\approx 2,000(1.065)^x[/tex]