-3(x + 1) - 5
(-3)(x) + (-3)(1) -5
-3x - 3 -5
-3x - 8
-2(u + 3) + 4u
-2u - 6 + 4u
2u - 6
3. Trigonometric Function a. Describe two real-world situations that could be modelled by a trigonometric function. Cannot be Ferris Wheel ride, tides, hours of daylight. Cite any Internet source you may have used for reference. b. Clearly define all variables in the relationship. c. Clearly justify why this model fits the real applications with specific reference to key features of the function. d. Your justification should also include reference to the graphical and algebraic models. e. Accurately describe what changes to the base function y = sin x would be necessary to fit both real applications.
For this problem, we need to describe a real-life situation where trigonometric functions can be used to model the problem.
Let's assume that a certain vehicle's position is controlled by the speeds of the wheels on each side of the car. Whenever the speeds on the left wheels and right wheels are equal, then the car moves forward, if the speed on the left side is greater than the one on the right side the car goes right, and if the speed on the right side is greater, then the vehicle goes to the left side. This type of car is called a differential drive car, and it's very common on remote-controlled (RC) vehicles.
If we want to model the speed of the car in a two dimensional grid, such as below:
We need to assume that the vehicle will have two components of velocity, one that is parallel to the x-axis and one that is parallel to the y-axis. These will form the linear velocity for the vehicle. We also need an angular velocity, which is the rate at which the angle of the vehicle changes.
If we assume that the wheels of the vehicles are at a distance of "L" apart from each other, then we can model the angular velocity of the vehicle as:
[tex]\omega=\frac{v_r-v_l}{L}[/tex]Where "vr" is the speed on the right wheel, and "vl" is the speed on the left wheel. The movement will happen with the center of the car as the center of the movement, with this we can assume that the velocity of the vehicle on the two axes should be:
[tex]\begin{gathered} v_x=\frac{1}{2}(v_r+v_l)\cdot cos(\theta)\\ \\ v_y=\frac{1}{2}(v_r+v_l)\cdot sin(\theta) \end{gathered}[/tex]Therefore we can describe the vehicle speed with the following equations:
[tex]\begin{gathered} \omega=\frac{v_{r}-v_{l}}{L}\\ \\ v_x=\frac{1}{2}(v_r+v_l)cos(\theta)\\ \\ v_y=\frac{1}{2}(v_r+v_l)s\imaginaryI n(\theta) \end{gathered}[/tex]The input variables are "vr" and "vl" which are the speeds of each wheel and the angle of the vehicle "theta", the output is the speed at the x coordinate and the speed at the y coordinate, and the angular speed.
This works very well because if the vehicle is moving parallel to the x-axis, the angle will be 0, the cosine of 0 is 1, therefore the speed on the y axis will be 0 and the speed on the x-axis will be given by 0.5(vr+vl). The opposite happens when the vehicle is moving parallel to the y-axis.
400 meters to 350 meters increase or decrease
In this case we have a negative change, therefore decrease
answer: decrease
Determine the probability of the given opposite event.What is the probability of rolling a fair die and not getting an outcome less than 3?
The Opposite Event rule is the probability that event A happens is equal to one minus the probability that A does not happen.
If P(A) is the probability of A happening, and N(A) is the probability of A don't happen, we can write:
[tex]P(A)=1-N(A)[/tex]Now we can see:
[tex]N(A)=1-P(A)[/tex]This, we if we calculate the probability of getting less than 3, ve can calculate the probability of not getting less than 3.
Then, what are the results that are less than 3? Those are 1 and 2. Thus are the favorable outcomes, and since is a fair dice, there are 6 total possible outcomes.
The probability of A = getting less than 3, is:
[tex]\begin{gathered} P(A)=\frac{2}{6} \\ P(A)=\frac{1}{3} \end{gathered}[/tex]
Now we can calculate the probability of not getting less than 3:
[tex]\begin{gathered} N(A)=1-\frac{1}{3} \\ \end{gathered}[/tex][tex]N(A)=\frac{2}{3}[/tex]
The probability of not getting less than 3 is:
[tex]Probability=\frac{2}{3}\approx0.666[/tex]Or in percentage:
[tex]Probability=66.67\%[/tex]Tj earns a 20% commission on all sales plus a base salary of 40k. his total income last year was at least 70k. which inequality can be used to calculate the minimum of Tj sales.
Let x be the all sale for individual.
Determine the expression for total income of individual.
[tex]\frac{20}{100}x+40000=0.2x+40000[/tex]The total income was at least 70000. So last year income is 70000 or more than 70000.
Setermine the inequality for the sales.
[tex]\begin{gathered} 0.2x+40000-40000\ge70000-40000 \\ \frac{0.2x}{0.2}\ge\frac{30000}{0.2} \\ x\ge150000 \end{gathered}[/tex]A car is purchased for 19,00. Each year it loses 25% of its value. After how many years will the car be worth 5800. dollars or less? Write the smallest possible whole number answer
5 years
Explanation
Given
Cost price = $ 19,000
Depreciation yearly is % 25
What to find
Time to depreciate to $ 5, 800 or less
Step- by - Step Solution
After first year St
[tex]\begin{gathered} 25\%\text{ }of\text{ 19,000} \\ \\ \frac{25}{100}\times\text{ 19,000 = 4,750} \\ \\ 19,000\text{ - 4750 = 14, 250} \end{gathered}[/tex]After the year the second year
[tex]\begin{gathered} \frac{25}{100}\text{ }\times\text{ 14, 250 = 3,562.5} \\ \\ 14,250\text{ - 3,562.5 =10, 687.5} \end{gathered}[/tex]After Third year
[tex]\begin{gathered} 25\%\text{ of 10,687.5} \\ \\ \frac{25}{100\text{ }}\times\text{ 10, 687.5 = 2,671.875} \\ \\ 10,687.5\text{ - 2,671.875 = 8,015.625} \\ \end{gathered}[/tex]After Fourth year
[tex]\begin{gathered} 25\%\text{ of 8,015.625} \\ \\ \frac{25}{100}\times\text{ 8,015.625 = 20003.906} \\ \\ 8\text{,015.625 - 20,003.906 = 6011.719} \end{gathered}[/tex]After Fifth year
[tex]\begin{gathered} 25\%\text{ of 6011.719} \\ \\ \frac{25}{100}\times\text{ 6011.719 = 1502.930} \\ \\ 6011.719-1502.930\text{ = 4508.789} \end{gathered}[/tex]Therefore after 5 years the car be worth 5800. dollars or less Therefore after 5 years the car be worth 5800. dollars or less
Kaitlin got a prepaid debit card with $15 on it. For her first purchase with the card, she bought some bulk ribbon at a craft store. The price of the ribbon was 21
cents per yard. If after that purchase there was $4.92 left on the card, how many yards of ribbon did Kaitlin buy?
yards
Answer:
Kaitlin bought 48 yards of ribbon.
Step-by-step explanation:
Hey! Let's help you with your question here!
So, let's start by figuring out what we know and what we need to figure out. First of all, we started with $15 and ended with $4.92. We also know that the price of the ribbon is $0.21 per yard and we need to figure out how many yards of ribbon she purchased.
In order to figure this out, we first want to know the difference in the price between what we started and what we ended up with. So, we can subtract! It would look like this:
[tex]15-4.92=10.08[/tex]
So, we figured out that the difference in the price is $10.08, but how do we find out how many yards of ribbon Kaitlin bought? Well, since we know that it is $0.21 cents for a yard of ribbon, we can just take the difference in price and divide it by how much a ribbon cost for a yard of it. So it would look like this:
[tex]10.08/0.21=48[/tex]
We have a nice whole number and that's our answer! Therefore, Kaitlin bought 48 yards of ribbon.
Mathematical Way:
To do it in a more mathematical way, we can put it in the form of a formula. We know that the end total is $4.92 and the initial is $15. We also know that it's $0.21 cents per yard of ribbon but we don't know how many yards she bought. We can let the number of yards she bought represent x in the formula, so we have:
[tex]15=0.21x + 4.92[/tex]
This formula makes sense because we start with $15 at the beginning, so we want to add $4.92 from 0.21x because the end total is the remainder of how many yards Kaitlin bought. The process is essentially the same as the method above. If we were to solve the formula, it would give us the same answer:
[tex]15=0.21x+4.92[/tex]
[tex]15-4.92=0.21x[/tex] - Moving the 4.92 over to the left side, beginning to isolate x.
[tex]10.08=0.21x[/tex] - Subtracting $4.92 from $15.
[tex]\frac{10.08}{0.21} =\frac{0.21x}{0.21}[/tex] - We divide by $0.21 to solve for x.
[tex]48=x[/tex]
And here, we get the exact same answer, 48 yards of ribbon.
Given the following piecewise function, determine the value of g(4) - 3g(3).
Piecewise Function
We are given the piecewise function shown in the figure.
We are required to calculate g(4) - 3g(3).
First, we calculate g(4). Since 4 is greater than 3, we use the second function:
[tex]g(4)=4^4+4^2+4-3=273[/tex]Now we need to calculate g(3). We use the same function because 3 is greater or equal to 3:
[tex]g(3)=3^4+3^2+3-3=90[/tex]Now we calculate:
g(4) - 3g(3) = 273 - 3*90 = 273 - 270 = 3
Answer: 3
1 pur Una foto de 4 pulgadas por 6 pulgadas se coloca en un marco de imagen con un borde de ancho constante. Si el área del marco, incluida la imagen, es de 80 pulgadas cuadradas, ¿qué ecuación podría usarse para determinar el ancho del borde, x? *
Tenemos una foto de 4 x 6 pulgadas.
Tenemos un marco con borde de ancho constante, lo que significa que alrededor de la foto siempre tenemos un "espesor" constante, que llamaremos "e".
También sabemos que el area total del marco es 80 pulgadas cuadradas.
Podemos dibujar esto así:
El marco tiene un ancho total de 4+2e y un largo de 6+2e.
Entonces, podemos calcular el espesor "e" a partir de calcular el area del marco e igualarlo a 80 pulgadas cuadradas. El area del marco sera igual al ancho por el largo:
[tex]\begin{gathered} A=(4+2e)(6+2e)=80 \\ 4\cdot6+4\cdot2e+2e\cdot6+2e\cdot2e=80 \\ 24+8e+12e+4e^2=80 \\ 4e^2+20e-56=0 \\ 4(e^2+5e-14)=0 \\ e^2+5e-14=0 \end{gathered}[/tex]Debemos aplicar la ecuación cuadrática para calcular e:
[tex]\begin{gathered} e=-\frac{b}{2a}\pm\frac{\sqrt[]{b^2-4ac}}{2a} \\ e=-\frac{5}{2}\pm\frac{\sqrt[]{5^2-4\cdot1\cdot(-14)}}{2} \\ e=-\frac{5}{2}\pm\frac{\sqrt[]{25+56}}{2} \\ e=-\frac{5}{2}\pm\frac{\sqrt[]{81}}{2} \\ e=-\frac{5}{2}\pm\frac{9}{2} \\ e_1=-\frac{5}{2}-\frac{9}{2}=-\frac{14}{2}=-7 \\ e_2=-\frac{5}{2}+\frac{9}{2}=\frac{4}{2}=2 \end{gathered}[/tex]e=-7 no es una solución válida, ya que el espesor debe ser positivo.
Entonces, la solución es e=2.
Respuesta: el ancho del borde es 2 pulgadas.
Can you please help me out with a question
Given data:
The given radius is r=16 ft.
The expression for the surface area is,
[tex]\begin{gathered} SA=4\pi(r)^2 \\ =4\pi(16ft)^2 \\ =1024\pi ft^2 \end{gathered}[/tex]The expression for the volume of the sphere is,
[tex]\begin{gathered} V=\frac{4}{3}\pi(r)^3 \\ =\frac{4}{3}\pi(16ft)^3 \\ =\frac{16384\pi}{3}ft^3 \end{gathered}[/tex]Thus, the surface area is 1024π sq-ft, and volume is (16384π)/3 cube-ft.
i have questions on a math problem. i can send when the chats open
The random sample is determined as the simplest forms of collecting data from the total population.
Under random sampling, each member of the subset carries an equal opportunity of being chosen as a part of the sampling process.
So according to the question given
Assign each person of the population a number. Put all the numbers into bowl and choose ten numbers.
is the random sample because every person carries an equal opportunity of being chosen from the total population.
Hence the correct option is A.
x=-3
f(x)= -2
f’(x)=2
g(x)=3
g’(x)=-1
h(x) = g(x)/2f(x)
Find h'(-3)
Answer: [tex]-1[/tex]
Step-by-step explanation:
Using the quotient rule,
[tex]h'(x)=\frac{2f(x)g'(x)-2g(x)f'(x)}{(2f(x))^2}\\\\h'(3)=\frac{2f(3)g'(3)-2g(3)f'(3)}{(2f(3))^2}\\\\=\frac{2(-2)(-1)-2(3)(2)}{2(-2)^2}\\\\=-1[/tex]
The cargo of the truck weighs at most 2,800 pounds. Use w to represent the weight (in pounds) of the cargo.To get the 10% discount, a shopper must spend no less than $100. Use d to represent the spending (in dollars) of a shopper who gets the discount
We can write this inequalities as:
If the cargo W has to be 2,800 pounds at most, then:
[tex]W\le2,800[/tex]The shopper has to spend $100 or more to get a discount, so the spending d to get a discount can be written as:
[tex]d\ge100[/tex]How many solutions does the equation 5(m + 3) = 6-7m have? Explain how you found your answer.
Expand the left hand side using distributive property:
[tex]\begin{gathered} 5\cdot m+5\cdot3=6-7m \\ 5m+15=6-7m \\ \text{Add 7m to both sides:} \\ 5m+15+7m=6-7m+7m \\ 12m+15=6 \\ \text{subtract 15 from both sides:} \\ 12m+15-15=6-15 \\ 12m=-9 \\ \text{divide both sides by 12:} \\ \frac{12}{12}m=-\frac{9}{12} \\ m=-\frac{3}{4} \end{gathered}[/tex]Which systems of inequalities represents the number of apartments to be built
Given that:
- The office building contains 96,000 square feet of space.
- There will be at most 15 one-bedroom units with an area of 800 square feet. The rent of each unit will be $650.
- The remaining units have 1200 square feet of space.
- The remaining units will have two bedrooms. The rent for each unit will be $900.
Let be "x" the number of one-bedroom apartments and "y" the number of two-bedroom apartments.
• The words "at most 15 one-bedroom units" indicates that the number of these apartments will be less than or equal to 15 units:
[tex]x\leq15[/tex]• You know that the remaining units are two-bedroom apartments. And the number of them is greater than or equal to zero. Then, you can set up the second inequality to represent this:
[tex]y\ge0[/tex]• You know the area of each one-bedroom apartment, the area of each two-bedroom apartment, and the total area that the office building contains. The sum of the areas of the apartments must be less than or equal to the total area of the office building.
Then, the inequality that represents this is:
[tex]800x+1200y\leq96,000[/tex]• Therefore, you can set up this System of Inequalities to represent that situation:
[tex]\begin{gathered} \begin{cases}x\leq15 \\ \\ y\ge0 \\ \\ 800x+1200y\leq96,000 \\ \end{cases} \\ \end{gathered}[/tex]Hence, the answer is: Last option.
(b) The area of a rectangular window is 6205 cm .If the width of the window is 73 cm, what is its length?Length of the window: 0cm
We have that the area is 6205 cm^2 and the widht is 73 cm.
since it is a rectangle, we must use
[tex]A_{rect}=widht\cdot length[/tex]Now, we only replace values and find the value of the length
[tex]\begin{gathered} 6205cm^2=73\operatorname{cm}\cdot length \\ \text{length }=\frac{6205\operatorname{cm}}{73\operatorname{cm}} \\ \text{length }=85cm \end{gathered}[/tex]The length of the window is 85 cm.
Find the coordinates of the stationary points of the curve and use the secondderivative to determine the type of each.
Calculate the derivative of the function, as shown below
[tex]\begin{gathered} y=3x+\frac{108}{x}=3x+108x^{-1} \\ \Rightarrow y^{\prime}=3+108((-1)x^{-1-1})=3-108x^{-2} \\ \Rightarrow y^{\prime}=3-108x^{-2} \end{gathered}[/tex]Set y'=0 and solve for x, as shown below
[tex]\begin{gathered} y^{\prime}=0 \\ \Rightarrow3-108x^{-2}=0,x\ne0 \\ \Rightarrow3=\frac{108}{x^2} \\ \Rightarrow x^2=\frac{108}{3} \\ \Rightarrow x^2=36 \\ \Rightarrow x=\pm\sqrt[]{36} \\ \Rightarrow x=\pm6 \end{gathered}[/tex]Their corresponding y-coordinates are
[tex]\begin{gathered} x=\pm6 \\ \Rightarrow y=3(6)+\frac{108}{6}=18+18=36 \\ \Rightarrow(6,36) \\ \text{and} \\ 3(-6)+\frac{108}{-6}=-18-18=-36 \\ \Rightarrow(-6,36) \end{gathered}[/tex]Therefore, the two stationary points are (6,36) and (-6,-36).
Using the second derivative test,
[tex]\begin{gathered} y^{\prime}=3-108x^{-2} \\ \Rightarrow y^{\doubleprime}=-108(-2x^{-2-1})=216x^{-3} \end{gathered}[/tex]Then,
[tex]\begin{gathered} y^{\doubleprime}(6)=\frac{216}{(6)^3}=1>0\to\text{ local minimum at x=6} \\ \text{and} \\ y^{\doubleprime}(-6)=\frac{216}{(-6)^3}=-1<0\to\text{ local maximum at x=-6} \end{gathered}[/tex](6,36) is a local minimum and (-6,-36) is a local maximum.
Mr. and Mrs. Hill hope to send their son to college in fourteen years. How much money should they invest now at an interest rate of 9.5% per year, compounded continuously, in order to be able to contribute $8500 to his education?Round your answer to the nearest cent.
continuouslyUsing the formula for a compounded continously
[tex]P=P_0\cdot e^{r\cdot t}[/tex]where P is the amount on the account after t years compounded at an interest rate r when Po is invested in an account.
then,
[tex]\begin{gathered} 8500=P_0\cdot e^{0.095\cdot14} \\ 8500=P_{0^{}}\cdot e^{1.33} \\ P_0=\frac{8500}{e^{1.33}} \\ P_0=2248.056\approx2248.06 \end{gathered}[/tex]I need help in math can you please help me
We have the following:
[tex]\begin{gathered} \sin \theta=-\frac{8}{17} \\ \theta=\sin ^{-1}(-\frac{8}{17}) \\ \theta=-28.07 \end{gathered}[/tex]now, in Quadrant III (180° to 270°):
[tex]\theta=180+28.07=208.7[/tex]now, for cosine:
[tex]\cos 2\theta=\cos (2\cdot208.7)=0.538=\frac{539}{1000}[/tex]The answer is 539/1000
Help 50 points (show ur work)
1. The value of 34% of 850 is 289.
3. The amount that Kepley paid for the tool is $120.
How to calculate the value?From the information, we want to calculate 34% of 850. This will be calculated thus:
= 34% ×850
= 34/100 × 850
= 0.34 × 850
= 289
The amount paid for the tool will be:
= Price or tool - Discount
= $200 - (40% × $200)
= $200 - $80
= $120
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Abby scored 88, 91, 95, and 89 on her first four history quizzes. What score does Abby need to get on her fifth quiz to have an average of exactly 90 on her history quizzes? a.85b.86c.87a.88
Solution
For this case we can use the definition of average given by:
[tex]\text{Mean}=\frac{x_1+x_2+x_3+x_4+x_5}{5}[/tex]The final score needs to be 90 so we can do this:
[tex]90=\frac{88+91+95+89+x_5}{5}[/tex]And solving for x5 we got:
5*90 = 88+91+95+89+ x5
x5= 450 - 88- 91- 95 -89 = 87
Final answer:
c.87
a relationship between decimal, fraction, or 3 Three students wrote percentage. Maggie wrote 75% = Bieber wrote 0.05 = 50% Lee Yung wrote == 0.375 Whích students wrote a correct equation? A. All the above B. None of the above C. Beiber and Lee Yung only D. Lee Yung only 8
To change decimal or fraction to percent multiply them by 100
Example: 1/4 x 100% = 25%, 0.2 x 100% = 20%
Let us check the answer of the 3 students
Maggie wrote 75% = 3/5
Since
[tex]\frac{3}{5}\times100=\frac{300}{5}=60[/tex]Then 3/5 = 60%, not 75%
Maggie is wrong
Bieber wrote 0.05 = 50%
Let us check
0.05 x 100% = 5%, not 50%
Bieber is wrong
Yung wrote 3/8 = 0.375
Let us check
[tex]\begin{gathered} \frac{3}{8}\times100=\frac{300}{8}=\frac{\frac{300}{2}}{\frac{8}{2}}=\frac{150}{4} \\ \frac{150}{4}=\frac{\frac{150}{2}}{\frac{4}{2}}=\frac{75}{2}=37.5 \end{gathered}[/tex]Since 0.375 x 100% = 37.5%
Yung is right
The answer is Lee Yung only
The answer is D
A total of $5000 is invested: part at 5% and the remainder at 15%. How much is invested at each rate if the annual interest is $540?
Answer
The amount invested at
Step-by-step explanation:
The total amount invested is $5000
Let x be the investment at 5%
Let y be the investment at 15%
Mathematically, this can be expressed as
x + y = 5000 -- equation 1
Since the first part of the investment is invested at 5% and the second part is at 15%
0.05x + 0.15y = 540 --------- equation 2
The systems of equations can be solved simultaneously using the substitution method
x + y =5000 ----- equation 1
0.05x + 0.15y = 540 ------ equation 2
Isolate x in equation 1
x = 5000 - y
Substitute the value of x into equation 2
0.05(5000 - y) + 0.15y = 540
Open the parenthesis
250 - 0.05y + 0.15y = 540
Collect the like terms
-0.05y + 0.15y = 540 - 250
0.1y = 290
Divide both sides by 0.1
0.1y/0.1 =290/0.1
y = $2900
Recall that equation 1 is
x + y = 5000
y = $2900
x = 5000 - y
x = 5000 - 2900
x = $ 2100
Hence, the investment at 5% is $2100 and the investment at 15% is $2900
Which type of statically graphic uses bars to describe the data ?Dot plot Box plot Histogram
HISTOGRAM
Explanations:
Data are reported using visuals to make reporting easier and ease the understanding of the audience.
Some of the graphic used in statistics to report data and make inference include:
• Bar charts
,• line charts
,• Dot plot
,• Box plot
,• Histogram etc.
Bar charts and histograms make use of bars to report data. This charts are important to detect outliers that may be present in our data.
We can therefore conclude that the type of statically graphic that uses bars to describe data is the HISTOGRAM
Select the correct answer. Angela is driving across the state to her friend's house. She just filled her fuel tank to its maximum capacity of 26 gallons. If the amount of gas in her car decreases by 2 gallons every 48 miles, which of the following graphs best represents the number of gallons of fuel remaining?
Let L be the amount of gas Angela has at distance d. At d=0 she has 26, and we know that every 48 miles the gas decreases 2 gallons, so the rate of decrease of gas per mile is
[tex]\frac{2\text{ }}{48}=\frac{1}{24}[/tex]Then, the linear equation that models this problem is
[tex]L=-\frac{1}{24}d+26[/tex](I used the minus sign since the amount decreases).
The gas will run out of gas whe she has driven
[tex]\begin{gathered} 0=-\frac{1}{24}d+26 \\ \frac{1}{24}d=26 \\ d=624\text{ miles} \end{gathered}[/tex]Then the graph that best fits the model is number Z. And the answer is D.
During the spring, Mr. Salina's grass grows at a rate of 1.5 inches per week. During a rainy stretch in the summer, his grass grew a total of 8 inches in 4 weeks.
Based on the growth rate of Mr. Salina's grass per week in the summer, and in spring, the relationship is not proportional.
How are relationships proportional?When relationships are said to be proportional, it means that they increase or decrease by the same rate.
In the spring, Mr. Salina's grass grows at a rate of 1.5 inches per week.
In the rainy stretch of summer, this rate goes to:
= Total number of inches / Number of weeks
= 8 / 4
= 2 inches per week
This means that the relationship is not proportional and one rate is higher than the other.
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Brett colors 25% of the total shapes on his paper. He colors 14 shapes. How many total shapes are there on Brett’s paper?
Answer:
56 i think because if 25% = 1/4 and 14 is 25% you would need to multiply 14 by 4 to get 100% or 4/4
a circular cylinder with a diameter of 12 cm and a height of 27 cm is filled with water. An aquarium is in the shaoe of a rectangular prism with the dimensions 35 cm 40cm by 42cm. what isvthe maximum number of full cylinders that can be poured into the fish tank without overflowing it?
Given data:
The diameter of cylinder is d=12 cm.
The height of the cylinder is h= 27 cm.
The dimension of the aquarium is V=(35 cm)(40 cm)( 42 cm).
The volume of the cylinder is,
[tex]\begin{gathered} V^{\prime}=\frac{\pi}{4}(d)^2h \\ =\frac{\pi}{4}(12cm)^2(27\text{ cm)} \\ =3053.628cm^3 \end{gathered}[/tex]The volume of the aquarium is,
[tex]\begin{gathered} V=(35\text{ cm)(40 cm)(42 cm)} \\ =58800cm^3 \end{gathered}[/tex]The number of cylinders that can be pour into aquarium is,
[tex]\begin{gathered} n=\frac{V}{V^{\prime}} \\ =\frac{58800}{3053.628} \\ =19.25 \end{gathered}[/tex]Thus, the number of cylinders that can be pour into aquarium is 19.25.
Write the equation of the line that passes through the points (12, 4) and (22,9).
Given the following points that pass through the line:
Point A : 12,4
Point B : 22,9
Step 1: Let's determine the slope of the line (m).
[tex]\text{ m = }\frac{y_2-y_1}{x_2-x_1}[/tex][tex]\text{ = }\frac{9\text{ - 4}}{22\text{ - 12}}[/tex][tex]\text{ m = }\frac{5}{10}\text{ = }\frac{1}{2}[/tex]Step 2: Let's determine the y-intercept (b). Substitute m = 1/2 and x,y = 12,4 in y = mx + b.
[tex]\text{ y = mx + b}[/tex][tex]\text{ 4 = (}\frac{1}{2})(12)\text{ + b }\rightarrow\text{ 4 = }\frac{12}{2}\text{ + b}[/tex][tex]\text{ 4 = 6 + b}[/tex][tex]4\text{ - 6 = b}[/tex][tex]\text{ -2 = b}[/tex]Step 3: Let's complete the equation. Substitute m = 1/2 and b = -2 in y = mx + b.
[tex]\text{ y = mx + b}[/tex][tex]\text{ y = (}\frac{1}{2})x\text{ + (-2)}[/tex][tex]\text{ y = }\frac{1}{2}x\text{ - 2}[/tex]Therefore, the equation of the line is y = 1/2x - 2.
If y varies directly with x and y = 90 when 3 = 15, then what is y when = 4?y =+
Recall than a direct variation implies the following type of relationship between y and x:
y = k * x
where k is a constant value
Then you have (by dividing by x, the following:
y / x = k (the constant)
then, we are told that when y = 90 , x = 15, so we have:
90 / 15 = k
6 = k
so,now that we know what the constant k is (6), we can answer the question: What is y when x = 4?
so we write:
y = k * x
y = 6 * 4
y = 24
This is the value of y when x is 4 since the constant k is 6 as we found above.
Another example:
We need to find the variation relationship for a case that when y = 6, x = 12
We think the same way we did before, starting with the fact that a direct variation is of the form:
y = k * x
given the info that when x = 12, y = 6, we can find the constant k:
6 = k * 12
divide by 12 both sides:
6/12 = k
1/2 = k
So k is 1/2 (one half)
Then we can write the variation as:
y = (1/2) x
(the product of 1/2 times x)
Identify each pair of angles as corresponding, alternate interior, alternate exterior, consecutiveinterior, vertical, or adjacent.
SOLUTION
Given the image on the answer tab;
Explanation;
The two angles are said to be adjacent angles when they share the common vertex and side.
Considering our question;
