can someone please help me(20 points will give brainliest!!!)

The investment problem is modeled by an initial value problem (IVP) where the rate of change of the investment is determined by the initial investment, monthly deposits, and the interest rate.

a) The investment problem can be modeled by an initial value problem where the rate of change of the investment, y(t), is given by the initial investment, monthly deposits, and the interest rate. The IVP can be written as:

dy/dt = 0.09y + 200, y(0) = 500.

b) To solve the IVP, we can use an integrating factor to rewrite the equation in the form dy/dt + P(t)y = Q(t), where P(t) = 0.09 and Q(t) = 200. Solving this linear first-order differential equation, we obtain the solution for y(t).

c) To find the value of the investment after 2 years, we substitute t = 2 into the obtained solution for y(t) and calculate the corresponding value.

d) After 2 years, the monthly deposit increases to $300. To find the value of the investment a year later, we substitute t = 3 into the solution and calculate the value accordingly.

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Answer:

Step-by-step explanation:

Find the number that makes the ratio equivalent to 3:5

6:10

9:15

12:20

Answer: 15 total fruits

Step by Step:

1. set up equation = 10+5=_

2. then solve = 15

Answer:

I don't think you uploaded the entire question, could you possibility update the question so I could try helping you with it?

Given the price the stickers can be sold for, and the cost in total of the stickers, you can make profit.

The total revenue that can be made on the sale of the 600 stickers that you got, would be:

= ( Number of stickers in total / Number of stockers per pack ) x Price per sticker pack

= 600 / 10 x 2

= 60 x 2

= $ 120

Your profit would be:

= Total revenue from stickers - total cost

= 120 - 26. 97

= $ 93. 03

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The side lengths of Triangle representing a right angled Triangle is :

because , Pythagoras theorem is true for given Side lengths,

that is :

Answer:

7,24,25

Step-by-step explanation:

Use Pythagorean theorem

\(\\ \sf\longmapsto H^2=P^2+B^2\)

\(\\ \sf\longmapsto 25^2=24^2+7^2\)

\(\\ \sf\longmapsto 625=576+49\)

\(\\ \sf\longmapsto 625=625\)

\(\\ \sf\longmapsto verified\)

The solution to the equation \((\sum\limits^{\infty}_{i=1} \frac{1}{2^i}) + (\sum\limits^{\infty}_{i=1} \frac{9}{10^i}) + \sin^2(\theta) + \cos^2(\theta) + e^2\) is 10.3891

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

\((\sum\limits^{\infty}_{i=1} \frac{1}{2^i}) + (\sum\limits^{\infty}_{i=1} \frac{9}{10^i}) + \sin^2(\theta) + \cos^2(\theta) + e^2\)

Using the following trigonometry ratio

sin²(x) + cos²(x) = 1

We have

\((\sum\limits^{\infty}_{i=1} \frac{1}{2^i}) + (\sum\limits^{\infty}_{i=1} \frac{9}{10^i}) + \sin^2(\theta) + \cos^2(\theta) + e^2 = (\sum\limits^{\infty}_{i=1} \frac{1}{2^i}) + (\sum\limits^{\infty}_{i=1} \frac{9}{10^i}) + 1 + e^2\)

The sum to infinity of a geometric series is

S = a/(1 - r)

So, we have

\((\sum\limits^{\infty}_{i=1} \frac{1}{2^i}) + (\sum\limits^{\infty}_{i=1} \frac{9}{10^i}) + \sin^2(\theta) + \cos^2(\theta) + e^2 = \frac{1/2}{1 - 1/2} + \frac{9/10}{1 - 1/10} + 1 + e^2\)

So, we have

\((\sum\limits^{\infty}_{i=1} \frac{1}{2^i}) + (\sum\limits^{\infty}_{i=1} \frac{9}{10^i}) + \sin^2(\theta) + \cos^2(\theta) + e^2 = 1 + 1 + 1 + e^2\)

Evaluate the sum

\((\sum\limits^{\infty}_{i=1} \frac{1}{2^i}) + (\sum\limits^{\infty}_{i=1} \frac{9}{10^i}) + \sin^2(\theta) + \cos^2(\theta) + e^2 = 3 + e^2\)

This gives

\((\sum\limits^{\infty}_{i=1} \frac{1}{2^i}) + (\sum\limits^{\infty}_{i=1} \frac{9}{10^i}) + \sin^2(\theta) + \cos^2(\theta) + e^2 = 10.3891\)

Hence, the solution to the equation is 10.3891

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The table has a greater rate of change.

The rates of change are equal.

In the given problem, we have a table showing the relationship between x and y values. By comparing the change in y with the change in x, we can determine the rate of change. Looking at the table, we observe that for every increase of 1 in x, there is a corresponding increase of 1 in y. Therefore, the rate of change for this table is 1.

The slopes are equal.

The equation has a greater slope.

In problem 2, we are given a linear equation in the form y = mx + b, where m represents the slope. The given equation is y = 2x + 7, which means the slope is 2. To compare the rates of change, we compare the slopes. If the slopes are equal, the rates of change are equal. In this case, the slopes are equal to 2, so the rates of change are the same.

The function rule has a greater slope.

The slopes are equal.

In problem 3, we are told that as x increases by 1, y increases by 3. This information gives us the rate of change between x and y. The slope of a function represents the rate of change, and in this case, the slope is 3. Comparing the slopes, we find that they are equal, as both have a value of 3. Therefore, the rates of change are the same.

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Answer:

2714.34 in³ (nearest hundredth)

Step-by-step explanation:

A) Cone and hemisphere

B) volume of a cone = \(\dfrac13\pi r^2h\)

volume of a hemisphere = \(\dfrac23\pi r^3\)

(where r is the radius and h is the height)

C)

\(\implies \textsf{volume of cone}=\dfrac13\pi \times 9^2\times 14=378 \pi\)

\(\implies \textsf{volume of hemisphere}=\dfrac23\pi \times 9^3=486 \pi\)

\(\implies \textsf{total volume}=378 \pi+486 \pi=864\pi =2714.34 \textsf{ in}^3 \textsf{ (nearest hundredth)}\)

The dotplot of the simulation results and the sample of 10 buses provides convincing evidence that the true probability of a bus running off schedule is more than the claimed 10%. The data suggests that it is most likely for exactly one bus out of 10 to be running off schedule, and there is a low probability of 5 or more buses being off schedule.

Based on the dotplot of the simulation results and the sample of 10 buses, we can draw the following conclusions:

The true probability that a city bus is running off schedule is more than 10%:

The simulation results show that the frequency of buses running off schedule is concentrated towards the higher end of the spectrum. The dotplot indicates that 5 buses running off schedule is a relatively unusual occurrence, which suggests that the true probability is higher than the claimed 10%.

There is about a 3% chance that 5 or more buses are running off schedule:

Looking at the dotplot, we can see that the frequency of 5 buses running off schedule is quite low, with only 2 occurrences out of the 100 trials. This indicates that it is unlikely for 5 or more buses to be running off schedule under the assumption that only 10% of buses run off schedule.

It is most likely that exactly one out of 10 buses is running off schedule:

The dotplot shows a peak in the frequency at 1 bus running off schedule. This suggests that one bus running off schedule is the most probable scenario among the 10-bus samples.

If we continued to take more samples of 10 buses, the center of the distribution would shift to 1:

Based on the dotplot, the distribution appears to be centered around 1 bus running off schedule. This indicates that as more samples of 10 buses are taken, the average number of buses running off schedule would tend to be closer to 1.

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False. There is no strong evidence to support the claim that the temporal pattern of super-eruptions (M>8 eruptions) is random.

The statement claims that the temporal pattern of super-eruptions is random, implying that there is no specific pattern or correlation between the occurrences of these large volcanic eruptions. However, scientific studies and research suggest otherwise. While it is challenging to study and predict rare events like super-eruptions, researchers have analyzed geological records and evidence to understand the temporal patterns associated with these events.

Studies have shown that super-eruptions do not occur randomly but tend to follow certain patterns and cycles. For example, researchers have identified clusters of super-eruptions that occurred in specific geological time periods, such as the Yellowstone hotspot eruptions in the United States. These eruptions are believed to have occurred in cycles with intervals of several hundred thousand years.

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Answer:

Each student will have a maximum of 2 erasers. The other 9 would belong to the teacher.

Step-by-step explanation:

55 (erasers) /  23 (students) = 2 9/23

2 (erasers) * 23 (Students) = 46

The remaining, leftover 9 belong to the teacher bec its what you stated in your question ^^

Hope I helped! ^^

Answer:

C

Step-by-step explanation:

First, let us clarify what (11, 0) represent

(x, y) ---> (11, 0)

The values are as followed:

x = 11

y = 0

If the problem states that x is the number of days after the promotion started, then it will be 11 days.

When we assert that it has been 11 days, we rule answers A and D.

If the problem states that y is the number of free coffee samples that remain after 11 days (because x is included in this statement), then it will be 0 samples.

Therefore, the answer will be C. After 11 days of the promotion, there will be 0 samples that are left.

Answer:

the measure of angle 8 is 83 degrees

Step-by-step explanation:

Answer:

X= 23 degrees

Step-by-step explanation:

117 degrees minus 94 equals x

x=23

Answer:

where are the answer choices

Step-by-step explanation:

Answer:

2x + 3 = 7x + 4

3 + 2x = 7x + 4

3 + 2x = 4 + 7x

3 + 2x = 4 + 7x

3 + 2x + -7x = 4 + 7x + -7x

2x + -7x = -5x

3 + -5x = 4 + 7x + -7x

7x + -7x = 0

3 + -5x = 4 + 0

3 + -5x = 4

3 + -3 + -5x = 4 + -3

 3 + -3 = 0

0 + -5x = 4 + -3

-5x = 4 + -3

4 + -3 = 1

-5x = 1

 x = -0.2

Answer:

y = \(\frac{3}{2}\)x + \(\frac{7}{2}\)

Step-by-step explanation:

The base form of the slope-intercept form is y = mx + b. We need to think of this as a system of linear equations problem, where we are solving for m and x using 2 equations. Substituting the x and y coordinates of the 2 points, we have 5 = m + b and -1 = -3m + b.

Let's now solve the system by elimination. We first subtract to eliminate the variable b and we get 6 = 4m, so m = \(\frac{3}{2}\). Then we substitute \(\frac{3}{2}\) for m in any of the equations and we get b = \(\frac{7}{2}\). Thus, the equation of the line that goes through (1, 5) and (-3, -1) is y = \(\frac{3}{2}\)x + \(\frac{7}{2}\).

Equation y (1/8 x3 + 7) is rewritten by using a common factor. A literal equation is regarded as having at least two variables.

Given that,

Use a common factor to rewrite one eighth times x cubed times y plus seven eighths times x times y squared.

This is how the equation can be rewritten:

The first equation is written as 1/8 x3y + 7/8 xy2.

It is possible to rewrite the equation 1/8 xy(x2 + y) using the common factor.

Equation two:

1/4x.y.4x2 + 28y

1/4 .4 x³ .y + 28y

x3.y + 28y equation to be solved

Using the formula

y = (x3 + 28).

Third claim:

1/4. x. y. x2 + 7y

Fix the problem.

1/4 . x³ .y + 7y

Write y = (1/4x3 + 7) in a new way.

Fourth assertion:

1/8 x3y2.y + 7 x 1/8 x3y3 + 7 x

After removing the common element x(1/8 x2y3+ 7), rewrite the equation.

Fifth Declaration

1/8 x. y. x² + 7y

1/8 x³. y + 7y

Equation y (1/8 x3 + 7) is rewritten.

Thus, a literal equation is one that contains at least two variables. Solve for one variable in terms of the other variable to rephrase a literal equation.

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The rectangular prism has the greater volume because the cylinder fits within the rectangular prism with extra space between the two figures.

A prism is a three-dimensional object. There are triangular prism and rectangular prism.

We have,

We can see this by comparing the formulas for the volumes of the two shapes.

The volume V of a rectangular prism with length L, width W, and height H is given by:

\(\text{V} = \text{L} \times \text{W} \times \text{H}\)

The volume V of a cylinder with radius r and height H is given by:

\(\text{V} = \pi \text{r}^2\text{H}\)

Now,

We are told that the length of each side of the prism base is equal to the diameter of the cylinder.

Since the diameter is twice the radius, this means that the width and length of the prism base are both equal to twice the radius of the cylinder.

So we can write:

\(\text{L} = 2\text{r}\)

\(\text{W} = 2\text{r}\)

Substituting these values into the formula for the volume of the rectangular prism, we get:

\(\bold{V \ prism} = \text{L} \times \text{W} \times \text{H}\)

\(\text{V prism} = 2\text{r} \times 2\text{r} \times \text{H}\)

\(\text{V prism} = 4\text{r}^2 \text{H}\)

Substituting the radius and height of the cylinder into the formula for its volume, we get:

\(\bold{V \ cylinder} = \pi \text{r}^2\text{H}\)

To compare the volumes,

We can divide the volume of the cylinder by the volume of the prism:

\(\dfrac{\text{V cylinder}}{\text{V prism}} = \dfrac{(\pi \text{r}^2\text{H})}{(4\text{r}^2\text{H})}\)

\(\dfrac{\text{V cylinder}}{\text{V prism}} =\dfrac{\pi }{4}\)

1/1 is greater than π/4,

Thus,

The rectangular prism has a greater volume.

The cylinder fits within the rectangular prism with extra space between the two figures because the cylinder is inscribed within the prism, meaning that it is enclosed within the prism but does not fill it completely.

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To find:-

Answer:-

The given system of equation is ,

\(\begin{cases}y = 3x + 1 \dots (\rm 1 ) \\ y = -x + 5 \dots (\rm 2) \end{cases}\)

This given set of equations are linear equations in two variables , which can be solved by various methods like elimination method and substitution method .

We would use here substitution.

We cam here substitute the value of \( y \) from equation 1 into equation 2 as ,

\(\implies y = - x + 5\)

\(\implies 3x + 1 = - x + 5\)

Now solve for \(x\) ,

\(\implies 3x + x = 5 - 1 \)

\(\implies 4x = 4 \)

\(\implies x =\dfrac{4}{4} \)

\(\implies x = 1 \)

Now substitute this value of \( x\) into equation 2 to find out \(y\) as ,

\(\implies y = -1 + 5\)

\(\implies y = 4 \)

Hence we got our answer as , x = 1 and y = 4 , which is the solution to the given set of equations.

Hence we conclude that (1,4) is a solution to the system of equations.

Yes , (1,4) is a solution .

and we are done!

Answer:

the answer:  b= -1

Step-by-step explanation:

Answers:

1) 56.33

2) 44.55

3) 22.75

4) 84.57

5) 71.92

6) 39.2

7) 73.6

8) 29.67

9) 26.91

Hope this was helpful!

-Kai <3

Answer: 0.75

Step-by-step explanation:

The 3 × 3 matrix that translates a point in r 2 by (3, 1) then rotates the result 45 degrees about the origin is

M = | 1/√2    -1/√2    3/√2- 1/√2 |

      | 1/√2    1/√2     1/√2 + 1/√2 |

       | 0           0                1 |

To find the 3x3 matrix that translates a point in R^2 by (3, 1) and then rotates the result 45 degrees about the origin using homogeneous coordinates, we can follow these steps:

Translation Matrix:

The translation matrix for a 2D translation by (3, 1) is:

T = | 1 0 3 |

| 0 1 1 |

| 0 0 1 |

This matrix translates a point (x, y) by adding the translation vector (3, 1) to it.

Rotation Matrix:

The rotation matrix for a 2D rotation by 45 degrees counterclockwise is:

R = | cos(theta) -sin(theta) 0 |

| sin(theta) cos(theta) 0 |

| 0 0 1 |

Since we want to rotate by 45 degrees, theta = 45 degrees = π/4 radians. Therefore, the rotation matrix becomes:

R = | cos(π/4) -sin(π/4) 0 |

| sin(π/4) cos(π/4) 0 |

| 0 0 1 |

Simplifying, we have:

R = | 1/√2 -1/√2 0 |

| 1/√2 1/√2 0 |

| 0 0 1 |

Combine Translation and Rotation:

To obtain the combined transformation matrix, we multiply the translation matrix T by the rotation matrix R:

M = T * R

Performing the matrix multiplication:

M = | 1 0 3 | | 1/√2 -1/√2 0 |

| 0 1 1 | * | 1/√2 1/√2 0 |

| 0 0 1 | | 0 0 1 |

Simplifying, we have:

M = | 1/√2 -1/√2 3/√2 - 1/√2 |

| 1/√2 1/√2 1/√2 + 1/√2 |

| 0 0 1 |

This is the desired 3x3 matrix that translates a point in R^2 by (3, 1) and then rotates the result 45 degrees about the origin using homogeneous coordinates.

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If you flip a coin one time, the probability to get a Head is

p = 1/2

The probability of not getting a head in a single toss is :

q = 1 - 1/2 = 1/2

Thus there is only one unique situation to get the same number of heads and tails : in 10 toss you need to get exactly five heads, it will means that the rest is the tails.

Now using Binomial theorem of probability, the probability of getting exactly x = 5 heads in a total of n = 10 tosses is :

P(X = 5) = \(\frac{10!}{5!*5!} *(\frac{1}{2})^{5} *(\frac{1}{2})^{5}\)≈ 0.246

So the probability of that is 24,6 %

Good Luck

There are several different sampling methods that could be employed for this study, including:

Simple random sampling: This method involves randomly selecting a certain number of students from each campus and measuring the distance between their hometowns and their campuses.

Stratified random sampling: This method involves dividing the student population at each campus into different strata (e.g. by major or year in school) and then randomly selecting a certain number of students from each stratum.

Cluster sampling: This method involves randomly selecting a certain number of campuses and then measuring the distance between the hometowns of all students at those campuses and their respective campuses.

Multi-stage sampling: This method involves using a combination of the above methods, such as first using cluster sampling to select a certain number of campuses and then using stratified random sampling to select a certain number of students from each campus.

This study would be an enumerative study because it seeks to measure the characteristics of a population, in this case, the average distance between the hometowns of students and their campuses, by counting and measuring them.

It is not an analytic study because it doesn't involve testing a hypothesis or making causal inferences.

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The kite is flying 415.84 feet from the ground

The given parameters are:

Let the height be h.

So, we have the following sine function

sin(x) = h/L

Make h the subject

h = L * sin(x)

This gives

h = 725 * sin(35)

Evaluate

h = 415.84

Hence, the kite is flying 415.84 feet from the ground

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Yes , there can be a Right triangle with two side lengths as simple fraction and other side length as an integer .

In the question ,

we have to find , whether a Right Triangle can be made by two fractions and an integer length .

yes, we can make a Right Triangle with two lengths as fractions and other length as integer .

let us prove it with the help of an example .

Consider the triplet 7 , 24 , 25 .

let the two fractions side be 7/25 and 24/25 , and the hypotnuse of the right triangle be  1 .

We know that for the Right Triangle

(side1)² + (side2)² = hypotnuse²

(7/25)² + (24/25)² = 1²

49/625 + 576/625 = 1

(49+576) / 625 =1

625/625 = 1

1 = 1

hence ,it is a Right Triangle .

Therefore , Yes , there can be a Right triangle with two side lengths as simple fraction s and other side length as an integer .

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