The amount of money in a savings account increases by 0.2% every month.

Part A

Write a function for the amount of money in the account, B, after t months with an initial deposit of $100.

B(t) =
(
)t

Part B

By what factor does the amount in the account increase every month? Every year? Every 5 years? Round answers to the nearest thousandth.

each month:


each year:


every five years:

Answer:  D.  -1

Step-by-step explanation:

Your equation:

[tex]\sqrt[4]{(\sqrt[3]{64}) ^{2} } =(\frac{1}{2} )^{x}[/tex]       >We are going to work from the inside first then out

                                   The cube root of 64 is 4 because 4*4*4=64

[tex]\sqrt[4]{(4) ^{2} } =(\frac{1}{2} )^{x}[/tex]            > 4² = 4*4=16

[tex]\sqrt[4]{(16) } =(\frac{1}{2} )^{x}[/tex]           > the 4th root of 16 is 2 because 2*2*2*2=16

[tex]2 =(\frac{1}{2} )^{x}[/tex]                   > if you have the same bases you can set the

                                   exponents equal.  They are not the same but we

                                    are going to make them the same.  

[tex]2^{1} =(\frac{1}{2} )^{x}[/tex]                  > 2 is the same as 2^1,  i can make the bases the

                                   same if I can make the 2 a reciprocal.  That

                                   happens when I take the negative exponent of the

                                   number

[tex](\frac{1}{2} )^{-1} =(\frac{1}{2} )^{x}[/tex]             >Now that my bases are the same, I can make the

                                exponents =

-1 = x

Answer:

$\boxed{52}$ dog owners would be expected to brush their dog's teeth.

Step-by-step explanation:

The transformation of f(x) to g(x) is (a) f(x) is shifted up 1 unit to g(x).

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

The functions f(x) and g(x)

In the graph, we can see that

So, we have

Difference = 1 - 0

Evaluate

Difference = 1

This means that the transformation of f(x) to g(x) is (a) f(x) is shifted up 1 unit to g(x).

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Answer:

We can write 4x^2 + y^2 - 4xy - 16 as the product:

(2x - y + 4)(2x - y - 4)

Step-by-step explanation:

To write 4x^2 + y^2 - 4xy - 16 as a product, we can use the technique of completing the square. First, we can rearrange the terms to group the x terms and the y terms:

4x^2 - 4xy + y^2 - 16

Next, we can complete the square for the x terms and the y terms separately. For the x terms, we can add and subtract (2x)^2:

4x^2 - 4xy + (2x)^2 - (2x)^2 + y^2 - 16

Simplifying the first three terms gives:

(2x - y)^2 - (2x)^2 + y^2 - 16

For the y terms, we can add and subtract 16:

(2x - y)^2 - (2x)^2 + (y - 4)^2 - 16^2

Simplifying the second term gives:

-(4x^2) + 16x - (y^2) + 16y - 16^2

Therefore, we can write 4x^2 + y^2 - 4xy - 16 as the product:

(2x - y + 4)(2x - y - 4)

Answer:

the answer to the question provided is y = x

"The line has a positive slope and passes through the x-axis at -2. Additionally, it crosses through point (2, 1).

The formula in this case is 2x-y = -4. When we set the value of y to 0, we can use this equation to calculate the x-intercept: 2x0=42x=4. When we multiply both sides by 2, we obtain 2x2=42x=2. The x-intercept is therefore -2.

We may use the slope-intercept version of the equation, y = mx + b, where m is the slope and b is the y-intercept, to graph the line y = 4x - 2.

We can observe that the slope is m = 4 and the y-intercept is b = -2 by comparing y = 4x - 2 to y = mx + b.

Starting with the y-intercept of -2 on the y-axis, we can graph line by finding other points on it using the slope of 4.

To get to the point, if we move two units to the right, we must move up eight units. (2, 6). To get to the point, if we move two units to the left, we must move down eight units. (-2, -10).

According to the description and choices given, "The line has a positive slope and passes through the x-axis at -2" is the option that most closely matches our graph. Additionally, it crosses through point (2, 1).

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Tasheena's grade percentage in the class so far is approximately 71.00935%.

The percentage can be calculated by dividing the value by the total value, and then multiplying the result by 100.

To compute Tasheena's grade percentage in the class so far, we can calculate the weighted average of her scores based on the weightage of each component of the final grade.

Given:

Quizzes: 15% weightage

Exams: 55% weightage

Projects: 25% weightage

Attendance: 5% weightage

Tasheena's scores:

Quizzes: 117 out of 150 points

Exams: 74, 86, and 91 on the first three exams (each worth 100 points)

Projects: 29 out of 25 possible points

Attendance: Perfect attendance

Let's calculate Tasheena's grade percentage:

Quizzes:

Tasheena's quiz score percentage = (117 / 150) * 100 = 78%

Exams:

Tasheena's exam average = (74 + 86 + 91) / 3 = 83.67

Tasheena's exam score percentage = (83.67 / 100) * 55 = 46.017%

Projects:

Tasheena's project score percentage = (29 / 25) * 100 = 116%

Attendance:

Tasheena's attendance score percentage = 100% (since she had perfect attendance)

Now, let's calculate the weighted average of Tasheena's scores:

Weighted average = (Quizzes weightage * Quiz score percentage) + (Exams weightage * Exam score percentage) + (Projects weightage * Project score percentage) + (Attendance weightage * Attendance score percentage)

= (15% * 78%) + (55% * 46.017%) + (25% * 116%) + (5% * 100%)

= 11.7% + 25.30935% + 29% + 5%

= 71.00935%

Hence, Tasheena's grade percentage in the class so far is approximately 71.00935%.

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The value of B is 6 when the value of sin x degrees equals to 4/5.

In the given triangle diagram, the opposite side of x = 3b

The hypotenuse of the given triangle = 22.5

Given triangle is a right-angled triangle, in trigonometry, we know that in a right-angled triangle the sin x = opposite side of x / hypotenuse side

So, sin x = 3b/22.5

But, the given value is 4/5. So,

3b/22.5 = 4/5

3b = 90/5

3b = 18

b = 18/3

b = 6

From the above explanation, we can conclude that the value of b is 6.

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The solution set for each rational inequality:

Case 1: - 9 ≤ x < - 5

Case 3: Every real number except x = 1.

In this problem we find two cases of rational inequality, whose solution sets can be found by using algebra properties and sign laws. Now we proceed to solve on each case:

Case 1

(3 · x + 7) / (x + 5) ≥ 5

(3 · x + 7) / (x + 5) - 5 ≥ 0

[(3 · x + 7) - 5 · (x + 5)] / (x + 5) ≥ 0

(- 2 · x + 18) / (x + 5) ≥ 0

- 2 · (x - 9) / (x + 5) ≥ 0

The inequality is positive for - 9 ≤ x < - 5.

Case 3

(- x² - 1) / (- x² + 2 · x - 1) > 0

[(- 1) · (x² + 1)] / [(- 1) · (x² - 2 · x + 1)] > 0

(x² + 1) / (x² - 2 · x + 1) > 0

(x² + 1) / (x - 1)² > 0

The inequality is positive for all real number except x = 1.

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The scenario that can be represented by the inequalities 1.25 < x < 1.5 is: A. A container of milk costs at least $1.25 but less than $1.50

The inequality 1.25 < x < 1.5 represents a range of values between 1.25 and 1.5, where x falls within that range.

Option A, which states that a container of milk costs at least $1.25 but less than $1.50, fits this range.

Option B, which states that a student spends at least 1 hour 15 minutes, but no more than 1 hour 30 minutes on homework, does not fit this range, as it represents a range of time values and not a range of numerical values.

Option C, which states that a tip added to a restaurant bill is less than or equal to 25% or less than or equal to 50%, does not fit this range either, as it represents two separate ranges of values.

Option D, which states that the point value of a test item is more than 1.25 points and less than 1.5 points, fits this range as well.

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The temperature at 1pm is -3°C, at 2pm it's -6°C, at 3pm it's -9°C, and at 6pm it's -18°C. The temperature t hours after noon is T = 0 - 3t

If the temperature is dropping at a rate of 3 degrees per hour, then the temperature t hours after noon can be represented by the equation T = 0 - 3t, where T is the temperature in degrees Celsius.

To find the temperature at a specific time after noon, we need to substitute the corresponding value of t into the equation and solve for T. For example:

At 1pm (t = 1), T = 0 - 3(1) = -3°C

At 2pm (t = 2), T = 0 - 3(2) = -6°C

At 3pm (t = 3), T = 0 - 3(3) = -9°C

At 6pm (t = 6), T = 0 - 3(6) = -18°C

To find the temperature t hours after noon, we can use the same equation: T = 0 - 3t. For example, if we want to find the temperature 4 hours after noon, we can substitute t = 4 into the equation: T = 0 - 3(4) = -12°C. So the temperature 4 hours after noon is -12°C.

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Answer:

54.6 square yards

Step-by-step explanation:

For any regular polygon, the area of the polygon is given by:

[tex]A_{regular~polygon}=\frac{1}{2}nsa[/tex], where

Heptagons are 7-sided polygons, so n=7.

The length of a side and the apothem are given, so substitute and calculate:

[tex]A_{heptagon}=\frac{1}{2}(7)sa[/tex]

[tex]A_{heptagon}=\frac{1}{2}(7)(3.9~yd)(4.0~yd)[/tex]

[tex]A_{heptagon}=54.6 ~yd^2[/tex]

Answer:

77.53

Step-by-step explanation:

You just have to sum them up

Answer:

i think the answer is 77.53

Step-by-step explanation:

Therefore , the solution of the given problem of parallel lines comes out to be parallel lines stay parallel option C is correct.

A parallelogram in Euclidean geometry is essentially a straightforward hexagon with two different groups & equal distances. When both sets of sides evenly share a horizontal path, a particular type of quadrilateral known as a parallelogram is created. There are four distinct types of parallelograms, three of them being incompatible. The four distinct forms are slightly parallelograms, rectangular shapes, squares, and parallelograms.

Here,

C. It is untrue that a line may be translated into two parallel lines.

A line can be translated into another line that runs parallel to the first line, but not into two parallel lines.

Every point on the line is rigidly transformed by a translation, which moves all of the points along the line uniformly and in one direction.

After a translation, parallel lines stay parallel.

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You are on the right track! To find the equation of the parabola given by the equation x^2 + 6x + 8 = 0, you can use the standard form of a quadratic equation, which is:

where (h, k) is the vertex of the parabola and a is a coefficient that determines the shape of the parabola.

To get the equation of your parabola, you first need to complete the square on the x terms of the given equation:

From this equation, you can see that the vertex of the parabola is at (-3, -1) and the value of a is positive. This means that the parabola opens upwards.

To find the value of a, you can compare the equation with the standard form of the quadratic equation:

where h = -3, k = -1, and a is the coefficient you need to find. Substituting these values into the equation gives:

So the equation of the parabola is:

To plot the parabola, you can use the vertex (-3, -1) as a starting point and then use the coefficient a to determine the shape of the parabola. Since a is positive, the parabola opens upwards.

The percentage of percentage of the apples have diameters that are between 6.34cm and 8.5cm is given as follows:

99.7%.

The Empirical Rule states that, for a normally distributed random variable, the symmetric distribution of scores is presented as follows:

The measures of 6.34 cm and 8.50 cm are the bounds exactly within three standard deviations of the mean, hence the percentage is given as follows:

99.7%.

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The volume of each cup shown above are as follows;

Volume of small cup = 50.24 in³.

Volume of medium cup = 127.56 in³.

Volume of large cup = 226.08 in³.

In Mathematics and Geometry, the volume of a cylinder can be calculated by using this formula:

Volume of a cylinder, V = πr²h

Where:

By substituting the parameters, we have:

Volume of small cylinder, V = 3.14 × 2² × 4

Volume of small cylinder, V = 50.24 in³.

Volume of medium cylinder, V = 3.14 × 2.5² × 6.5

Volume of medium cylinder, V = 127.56 in³.

Volume of large cylinder, V = 3.14 × (6/2)² × 8

Volume of large cylinder, V = 226.08 in³.

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The probability of randomly selecting a date in March and spinning the spinner once, resulting in an odd number and then blue, is 1/6.

To find the probability of an odd number and then blue, we need to consider the number of favorable outcomes for each event and the total number of possible outcomes.

Probability of an odd number:

The spinner has 6 equally likely outcomes (numbers 1 to 6), and out of these, 3 are odd numbers (1, 3, and 5).

Therefore, the probability of getting an odd number is 3/6, which simplifies to 1/2.

Probability of blue:

The spinner has 6 equally likely outcomes, and out of these, 2 are blue. Therefore, the probability of getting blue is 2/6, which simplifies to 1/3.

To find the probability of both events occurring, we multiply the probabilities of each event:

Probability of an odd number and then blue [tex]= Probability $ of an odd number \times Probability $ of blue[/tex]

[tex]= (1/2) \times (1/3)[/tex]

= 1/6.

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Question: A date in March is chosen at random, then the spinner below is spun once. Find the probability of an odd number, and then blue. Use the counting principle to find the probability.

I'm taking linear algebra right now so this one hits home :)

Elementary Row Operations (EROs) are very important and not too difficult, so let's dive into the problem!

You're given the matrix below and asked to perform a single ERO to produce a matrix with a 1 at the position (1,1):

[tex]\begin{bmatrix}3 & 10 & 5\\2 & -1 & 1\end{bmatrix}[/tex]

Think of the two rows as separate entities in the matrix. Ultimately we want to have the index (1,1) currently holding the number 3 to become the number 1. To do this, logically you just need to subtract 2. Now, looking at the rows we have, a simple row operation is quite apparent.

Simply subtract row 2 from row 1, shown below:

[tex]\begin{bmatrix}3-2 & 10-(-1) & 5-1\\2 & -1 & 1\end{bmatrix}[/tex]

Now, simplify and you will have the answer:

[tex]\begin{bmatrix}1 & 11 & 4\\2 & -1 & 1\end{bmatrix}[/tex]

Notice that our matrix now has the required number 1 in row 1 and column 1, therefore, the matrix above is our answer! Let me know if you have any questions!

The part of the cake with vanilla or lemon frosting is 1/5 by using arithmetic and algebraic expression.

Let's start by defining the variable "x" as the part of the cake with vanilla or lemon frosting.

According to the problem, Jonah uses vanilla frosting on 1/5 of the cake, which can also be written as x = 1/5.

Similarly, he uses lemon frosting on 2/5 of the cake, which can be written as x = 2/5.

We know that the sum of the parts of the cake with vanilla, lemon, and chocolate frosting should equal the whole cake. Since there are only three types of frosting, we can write:

x + x + chocolate frosting = 1

2x + chocolate frosting = 1

We also know that chocolate frosting covers the remaining part of the cake, which is 3/5. Therefore, we can write:

chocolate frosting = 3/5

Substituting this value into the previous algebraic expression, we get:

2x + 3/5 = 1

Subtracting 3/5 from both sides of the equation, we get:

2x = 2/5

x = 1/5

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The total cost of the fence needed for the Perimeter of the work site is $8,800. So the answer is option B.

Because the question involves many units of measurement, we must convert them all to the same unit in order to determine the answer.

1/4 mile is equivalent to 1320 feet (1 mile = 5280 feet).

The length of the fence required to surround the work site equals the perimeter of the work site. To calculate the length of the fence in yards, divide 1320 feet by 3 (since a yard is 3 feet long).

1320 feet ÷ 3 = 440 yards

So the length of the fence needed is 440 yards.

The fence costs $20.00 per yard, thus to get the total cost, multiply the length of the fence by the cost per yard:

440 yards x $20.00/yard = $8,800.00

Therefore, the total cost of the fence needed for the perimeter of the work site is $8,800.00.

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The length of OP in the given circle figure is 2.

We know that the area of a sector which obtain 'A' degree at center in a circle with radius 'r' is given by,

A = (A/360)*πr²

Here length of OP represents the radius of the circle with center O.

Let OP = R.

The area of the whole circle will be = πR²

Given that the sector which intends an angle of 72 degrees in center has area 4π/5.

According to the condition,

(72/360)* πR² = 4π/5

R²/5 = 4/5

R² = 4

R = 2 [since length cannot be negative so the negative value of square root is ignored.]

Hence the length of OP is 2.

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Answer:

The expression $11.75x + $20 represents the total price for the order, including delivery to the school.

In this expression, 'x' represents the number of boxes of paints being purchased, and $11.75 represents the cost of each box. The term $11.75x represents the total cost of the boxes of paints based on the number of boxes being purchased.

The term $20 represents the delivery cost to the school, which is added to the total cost of the boxes of paints.

By multiplying the cost per box ($11.75) by the number of boxes (x) and adding the delivery cost ($20), you can calculate the total price for the order, including delivery.

Log22/Log9 express log_9 22 in terms of common logarithms

Common logarithms, is commonly describd as base-10 logarithms.

It is a type of logarithm that involve taking the logarithm of a number with respect to the base of 10.

This means that common logarithms show how many powers of ten must be increased to get a particular number. The usual logarithm of 100, for example, is 2, because 10 raised to the power of 2 equals 100.

The answer provided is based on the full question below;

-------------express log_9 22 in terms of common logarithms

a. Log 22/9

b. Log 198

c. Log22/Log9

d. Log9/Log22

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Answer:

12.8

Step-by-step explanation:

Quotient means the answer of two things divided.

62.72/4.9=12.8

Answer: 64

Step-by-step explanation:

multiple all of them

x = -6.

[tex]\sf \dfrac{3x+4}{3} -\dfrac{2x}{x-3} =x[/tex]

Applying the distributive property of multiplication on the left hand side:

[tex]\sf (3)(\dfrac{3x+4}{3} -\dfrac{2x}{x-3}) =x(3)\\ \\ \\{3x+4} -\dfrac{(3)2x}{x-3}=3x\\ \\ \\{3x+4} -\dfrac{6x}{x-3}=3x[/tex]

Applying the distributive property of multiplication:

[tex]\sf (x-3)({3x+4} -\dfrac{6x}{x-3})=3x(x-3)\\ \\ \\(x-3)({3x+4}) -6x=3x(x-3)\\ \\ \\(x)(3x)+(x)(4)+(-3)(3x)+(-3)(4) -[6x]=3x(x-3)\\ \\ \\[/tex]

Check the image below to see an illustration of this process.

[tex]\sf 3x^{2} +4x-9x-12 -[6x]=3x(x-3)\\ \\ \\3x^{2} +4x-9x-12 -6x=3x(x-3)\\ \\ \\3x^{2} -11x-12 =3x(x-3)[/tex]

Now simplifying on the right hand side (applying the same logic as last step).

[tex]\sf 3x^{2} -11x-12 =3x(x-3)\\ \\ \\3x^{2} -11x-12 =(3x)(x)+(3x)(-3)\\ \\ \\3x^{2} -11x-12 =3x^{2}-9x[/tex]

[tex]\sf 3x^{2} -11x-12+9x =3x^{2}-9x+9x\\ \\ \\3x^{2} -2x-12 =3x^{2}[/tex]

[tex]\sf 3x^{2} -2x-12-3x^{2} =3x^{2}-3x^{2}\\ \\ \\-2x-12 =0[/tex]

[tex]\sf -2x-12+12=0+12\\ \\ \\-2x=12[/tex]

[tex]\sf \dfrac{-2x}{-2} =\dfrac{12}{-2} \\ \\ \\x =-6[/tex]

If "x= -6" is the correct answer, substituting "x" by "-6" on the original equation should return the same value on both sides of the equal (=) symbol. Let's test!

[tex]\sf \dfrac{3(-6)+4}{3} -\dfrac{2(-6)}{(-6)-3} =(-6)\\ \\-6=-6[/tex]

That's correct!

x = -6 is the corect answer.

1. normally distributed if the sample size is 30 or larger.

2. Not always normally distributed.

3. Skewed to the right is still normally distributed

4. normally distributed.

1. normally distributed if the sample size is 30 or larger.

2. If the population from which samples are drawn is not normally distributed, then the sampling distribution of the sample mean is not always normally distributed. It depends on the sample size and the shape of the population distribution.

3. The sampling distribution of the sample mean for a sample of 10 elements taken from a population with a bell-shaped distribution that is skewed to the right is still normally distributed, by the central limit theorem, as long as the sample size is sufficiently large (typically at least 30) or the population distribution is approximately normal. Therefore, the answer is normally distributed.

4. The sampling distribution of the sample mean for a sample of 36 elements taken from a population with a bell-shaped distribution is normally distributed regardless of the population's skewness. Therefore, the answer is "normally distributed".

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