The average velocity of the particle in the time interval between 3s and 5s is 20 ms⁻¹ and its instantaneous velocity at 4s is 20 ms⁻¹.
How to determine average velocity and instantaneous velocity?
Average velocity is defined as the body's overall displacement divided by its time of motion. While instantaneous velocity is defined as a body's speed at a certain instant in time, or its displacement at that instant. When the velocity is constant, average and instantaneous velocities will equalize at just one condition.
The definition of instantaneous velocity is the rate of change of position over a relatively brief time period (almost zero). Simply said, the speed of an object at that precise moment. The definition of instantaneous velocity is "The velocity of an item in motion at a certain point in time." The instantaneous velocity of an object may be equal to its standard velocity if it has uniform velocity.
Mathematically, average velocity = [s(t₂) - s(t₁)]/[t₂ - t₁]
Instantaneous velocity at time, t is = (ds/dt) at time = t
Given, the displacement for the particle is given by s = 3t² - 4t + 5
Time interval, t₁ = 5s and t₂ = 3s;
Using formula in literature, average velocity of the particle in the time interval between 3s and 5s is:
Average velocity = (s(5) - s(3))/(5 - 3) = (60 - 20)/2 = 20 ms⁻¹
Instantaneous velocity at t = 4 is ds/dt at that time-frame:
Now, v = ds/dt = 6t -4
Now, v(4) = 6(4) - 4 = 20 ms⁻¹
The average velocity of the particle in the time interval between 3s and 5s is 20 ms⁻¹ and its instantaneous velocity at 4s is 20 ms⁻¹.
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Mimstoon started with at most 2 boondins (y). Every day (x), he
bought at most 1/2 more of them.
Write an inequality to model this relationship.
Step-by-step explanation:
y <= x/2 + 2
the maximum is to have 2 boondins and add 1/2 boondin every day.
but every time Mimstoon did not use the max. possible, the resulting total sum of boodins stays smaller that the maximum, and is therefore a valid data point.
(6, 7) is not in the inequality relationship.
because the max. boondins after 6 days is 2 (from the beginning) and 1/2 every day = 2 + 1/2 × 6 = 2 + 3 = 5.
but the data point shows 7 as y value (which is larger than the allowed max. of 5).
therefore the inequality is false for this data point, and therefore the data point is not in the inequality relationship.
The Massive Chemical Corporation starts an annuity to pay for the huge government penalties they expect in 11 years when a pending case finally gets litigated.Their lead attorney informs them that they can expect a $4,000,000 fine. An investment house offers 11.5% interest on annuities of that size, compoundedsemiannually. What will the semiannual payment be on this annuity? Round to the nearest cent.
To solve this problem we can use the formula for compound interes so the equation is:
[tex]Y=P_o(1+I)^t[/tex]The measures of 6 interior angles of a heptagon are: 111, 110, 121, 135, 139 and 92.Find the measure of the seventh interior angle. show work please:)
The sum of the interior angles of a heptagon 900º, if you know 6 of the 7 interior angles of the heptagon, you can determine the measure of the seventh angle as follows:
[tex]\begin{gathered} 900=111+110+121+135+139+92+x \\ 900=708+x \end{gathered}[/tex]Subtract 708 from both sides of the equal sign to reach the measure of the seventh angle:
[tex]\begin{gathered} 900-708=708-708+x \\ 192=x \end{gathered}[/tex]The seventh interior angle has a measure of 192º
Helena is watching a movie on television she notices that there are 9 1/4 min of commercial for every 1/2 hour of the movie
If Helena watches a movie that shows a commercial for 9 1/4 mins of every 30 minutes (half hour), then the unit rate of commercials per hour is derived as;
[tex]\begin{gathered} \text{Com /30 mins=9}\frac{1}{4} \\ \text{Com /60 mins=9}\frac{1}{4}\times2 \\ \text{Com /hour=}\frac{37}{4}\times2 \\ \text{Com /hour=}\frac{37}{2} \\ Com\text{ /hour=18}\frac{1}{2} \end{gathered}[/tex]The unit rate of commercials per hour is 18 1/2 minutes (eighteen and half minutes)
A 17 m guy wire attached to the top of a tower (the height of the tower is not yet known) is anchored on the ground, 8 m away from the base of the tower. A second guy wire needs to be attached to the centre of the tower and then anchored to the same ground-anchor as the first wire.Draw and label a diagram.How long does the second guy wire need to be?Determine the measure of the angle formed between the two wires.
Solution
Step 1:
Draw the diagram to illustrate the information.
Step 2:
Use the Pythagoras theorem to find the height of the pole.
[tex]\begin{gathered} 17^2\text{ = h}^2\text{ + 8}^2 \\ 289\text{ = h}^2\text{ + 64} \\ h^2\text{ = 289 - 64} \\ h^2\text{ = 225} \\ h\text{ = }\sqrt{225} \\ \text{h = 15m} \end{gathered}[/tex]Step 3
b) The height of the second guy wire = d
[tex]\begin{gathered} \text{Apply the pythagoras theorem} \\ d^2=\text{ \lparen}\frac{15}{2}\text{\rparen}^2+\text{ 8}^2 \\ d^2\text{ = 56.25 + 64} \\ d^2\text{ = 120.25} \\ \text{d = }\sqrt{120.25} \\ \text{d = 10.97m} \end{gathered}[/tex]c)
[tex]\begin{gathered} sin(\theta\text{ + }\alpha)\text{ = }\frac{Opposite}{Hypotenuse} \\ sin(\theta\text{ + }\alpha)\text{ = }\frac{15}{17} \\ \theta\text{ + }\alpha\text{ = sin}^{-1}(\frac{15}{17}) \\ \theta\text{ + }\alpha\text{ = 61.9}^o \end{gathered}[/tex][tex]\begin{gathered} sin\alpha\text{ = }\frac{7.5}{10.97} \\ \alpha=\text{ sin}^{-1}(\frac{7.5}{10.97}) \\ \alpha\text{ = 43.1} \end{gathered}[/tex]The angle between the two guys' wires = 61.9 - 43.1
Measure of the angle formed between the two wires = 18.8
if you can't read it, it says : Add -7d-3 and 10d-6. Show all steps
-7 d - 3 + 10d - 6
First step
Group in parenthesis and
Add both letter terms
(-7d + 10d)= -3d
Second step
Group and add numbers only
(-3 - 6) = -9
Now add both results
=-3d - 9
Answer is -3d - 9
Which relation is a function? 4 15 O -6- 7 C fy 0 -H O O O -2 -4
The most appropriate choice for functions will be given by -
Third option is correct
Third relation is a function.
What is a function?
A function from A to B is a rule that assigns to each element of A a unique element of B. A is called the domain of the function and B is called the codomain of the function.
There are different operations on functions like addition, subtraction, multiplication, division and composition of functions.
Here,
For a function a point in the domain has a unique image.
Here values of x axis represents domain and values of y axis represents the range
For the first option,
x = -1 has two images, y = -1 and y = 3
so x = -1 do not have a unique image
So the first relation is not a function
For the second option,
x = 0 has two images, y = -1 and y = 2
so x = 0 do not have a unique image
So the second relation is not a function
For the third option,
Every point of the domain has a unique image
So the third relation is a function
For the fourth option,
x = -2 has two images, y = 1 and y = -2
so x = -2 do not have a unique image
So the fourth relation is not a function
so, Third option is correct
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Which set of expressions are not equivalent?(4 1\2 a + 2) + 3 and 5 + 4 a18 a + 14 and 6 a + 12 + 2 + 1 2a0.5 a + 3.5 + 4 a and 3.5 + 4.5 a3 a + 7 +2 a and 7 a + 5
Solution:
Solution:
To find which of given expressions are equivalent
For the first expressions
[tex](4\frac{1}{2}a+2)+3\text{ and 5}+4\frac{1}{2}a[/tex]Solving the expression
[tex]\begin{gathered} 4\frac{1}{2}a+2+3=5+4\frac{1}{2}a \\ 5+4\frac{1}{2}a=5+4\frac{1}{2}a \end{gathered}[/tex]The expressions are equivalent
For the second expression
[tex]18a+14\text{ and 6a}+12+2+12a[/tex]Solving the expression
[tex]\begin{gathered} 18a+14=6a+12a+12+2 \\ 18a+14=18a+14 \end{gathered}[/tex]The expressions are equivalent
For the third expression
[tex]0.5a+3.5+4a\text{ and 3.5}+4.5a[/tex]Solving the expression
[tex]\begin{gathered} 3.5+0.5a+4a=3.5+4.5a \\ 3.5+4.5a=3.5+4.5a \end{gathered}[/tex]The expressions are equivalent
For the fourth expression
[tex]3a+7+2a\text{ and 7a}+5[/tex]Solving the expression
[tex]\begin{gathered} 3a+2a+7=7a+5 \\ 5a+7\ne7a+5 \end{gathered}[/tex]The expressions are not equivalent.
Hence, the set of expressions that are not equivalent is
[tex]3a+7+2a\text{ and 7a}+5[/tex]I don’t know how to do 4 - 6mod6
Modulus (mod) symbol is a symbol that is used to find the remainder of the quotient of two integers
6mod6 means we need to find the result of the remainder if 6 is divided by 6.
Take the ratio of the integers
6/6 = 1.0
Multiply the result by the denominator
1.0 * 6 = 6
Take the difference of the result and the denominator
Remainder = 6 - 6
Remainder = 0
This shows that 6mod6 = 0
Substitute into the original expression
[tex]4-6mod6=4-0=4[/tex]A professional football prospect runs a 40 yard dash in 5 seconds.What is the player's average speed over this distance?a.) 8 yards per secondb.) 0.625 yards per secondc.) 0.125 yards per secondd.) 5 yards per second
What is the volume of a cone with height of 80 km and diameter 21 km?
The volume of a cone with height of 80 km and diameter 21 km is 9240km³.
How to determine the volume of a cone?Volume of a cone=R^2H/3
Diameter=21
To get the radius is half of diameter,:
Radius(R)= Diameter/2= 21/2=10.5kM
Height=80km
Volume=22/7×10.5^2×80/3
The volume of the Cone= 9240km^3
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Sam takes 4 hours to peel 32 potatoes, while Josh takes 5 hours to peel 30 potatoes. How mnay potatoes will they peel together in 12 hours
To determine the amount of potato peeled in 12 hours:
Sam takes 4 hours to peel 32 potatoes
Sam take 1 hour to peel 32 / 4 potato
[tex]\begin{gathered} 4\text{ hour }\Rightarrow\text{ 32 potato} \\ 1\text{ hour }\Rightarrow\text{ }\frac{32}{4} \\ 1\text{ hour }\Rightarrow\text{ 8 potato} \\ \end{gathered}[/tex]Josh takes 5 hours to peel 30 potatoes
Josh takes 1 hour to peel 30 / 5 potato
[tex]\begin{gathered} 5\text{ hour }\Rightarrow\text{ 30 potato} \\ 1\text{ hour }\Rightarrow\text{ }\frac{30}{5} \\ 1\text{ hour }\Rightarrow\text{ 6 potato} \end{gathered}[/tex]Sam peels 8 potato
They will peel hepotatoeel together in 12 hours
Answer:
168
Step-by-step explanation:
Solve using the quadratic formula
(x*y)a=xa*ya illustrates which property of exponents
Consider the given equation,
[tex](x\cdot y)^a=x^a\cdot y^a[/tex]This equation represents the Product Rule of the exponents.
The properties of exponents are the general rules and laws, followed by the exponents. These are used very frequently, so it is better to memorize these standard properties.
Susan works as a Hostess where she makes $8.75 per hour she also babysitting earns $10 per hour last week she made $165 if she works 6 hours more as a Hostess than she did babysitting how many hours did she babysit last week?
6 hours as babysitter last week
1) Gathering the data
Hostess
8.75h
Babysitting
10h
2) Based on that, we can write out the following equation, considering that Susan worked 6 hours more:
8.75(h+6) +10h= 165 Distribute the factor 8.75
8.75h +52.5 +10h = 165 Combine like terms
18.75h = 165 -52.5 Subtract from both sides 52.5
18.75h = 112.5 Divide both sides by 18.75
h =6
3) Hence, we can conclude that Susan worked as a hostess for 12 hours and as a baby sitter 6 hours last week
A student has 30 dimes and nickels. Their total value is $2.70. If d = number of dimes and n = number of nickels, this system of equations represents the situation:
d + n = 30
0.10d + 0.05n = 2.70
How many of each type of coin are there? Solve the system to answer the question.
A.
24 dimes and 6 nickels
B.
20 dimes and 10 nickels
C.
10 dimes and 20 nickels
D.
6 dimes and 24 nickels
Taking into account the definition of a system of linear equations, a student has 24 dimes and 6 nickels.
System of linear equationsA system of equations is a set of equations in which there are two or more variables, the goal of which is to find all ordered pairs (x, y) that satisfy the equation, where x and y are real numbers.
In other words, a system of linear equations is a set of two or more first degree equations that relate two or more unknowns, in which it is desired to find the value of each unknown so that all the equations of the system are satisfied.
This caseIn this case, a system of linear equations must be proposed taking into account that:
d = number of dimes.n = number of nickels.You know that:
A student has 30 dimes and nickels. The total value is $2.70.The system of equations to be solved is
d + n = 30
0.10d + 0.05n = 2.70
It is decided to solve it using the substitution method. This method consists of isolate one of the variables in one of the equations of the system and substituting its value in the other equation.
In this case, isolating the variable "d" from the first equation:
d= 30 -n
Replacing the expression in the second equation:
0.10×(30 - n) + 0.05n = 2.70
Solving:
0.10×30 - 0.10n + 0.05n = 2.70
3- 0.10n + 0.05n = 2.70
- 0.10n + 0.05n = 2.70 -3
-0.05n= -0.3
n= (-0.3)÷ (-0.05)
n= 6
Remembering that d=30 -n you get:
d= 30 - 6
d= 24
There are 24 dimes and 6 nickels.
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Show where the expression in number 8 would be on a unit circle
Solution
Step 1:
Write the expression
[tex]cos270\text{ }[/tex]Step 2
[tex]\begin{gathered} cos270\text{ = 0} \\ sin270\text{ = -1} \\ (0,\text{ -1\rparen} \end{gathered}[/tex]Question 8 Second part
[tex]\begin{gathered} sin(\frac{16\pi}{3})\text{ = sin960 = sin\lparen960 - 2}\times360)\text{ = sin240} \\ sin(\frac{16\pi}{3})\text{ = }\frac{-\sqrt{3}}{2} \\ cos(\frac{16\pi}{3})\text{ = }\frac{-1}{2} \end{gathered}[/tex][tex](\frac{-1}{2},\text{ }\frac{-\sqrt{3}}{2})[/tex]the circumference of the circle is 76 cm. what is the area? using 3.14 for pi round to the nearest square cm
The circumference of a circle is gotten using the formula below;
[tex]\begin{gathered} C=2\pi r \\ \text{Where r is the radius of the circle} \end{gathered}[/tex]Step 1: Let's find the value of the radius;
[tex]\begin{gathered} C=2\pi r \\ r=\frac{C}{2\pi} \\ r=\frac{76}{2\times3.14} \\ r=\frac{76}{6.28} \\ r=12.10\operatorname{cm} \end{gathered}[/tex]Step 2: The area of the circle is given as;
[tex]\begin{gathered} A=\pi r^2 \\ \text{Where r is the radius of the circle} \\ A=3.14\times(12.10)^2 \\ A=459.73 \\ A=460\operatorname{cm}^2 \end{gathered}[/tex]Hence, the area of the circle is 460 square centimeters.
Suppose you choose four booksto read from a summer readinglist of 12 books. How manydifferent combinations of booksare possible?Note: nCrn!r!(n-r)!
Answer
495 different combinations of books are possible.
Explanation
The different combinations of books possible can be known using the given formula
[tex]_nC_r=\frac{n!}{r!(n-r)!}[/tex]Choosing four books to read from a list of 12 books means r = 4 and n = 12
Therefore
[tex]\begin{gathered} _nC_r=\frac{n!}{r!(n-r)!}=\frac{12!}{4!(12-4)!}=\frac{12!}{4!\times8!}=\frac{12\times11\times10\times9\times8!}{4\times3\times2\times8!}=495 \\ \\ \end{gathered}[/tex]Hence, there are 495 different combinations of books are possible
The prism below is made of cubes which measure 3 of a centimeter on one side. What is the volume? NORSKEREN OA. LICO cubic cm OB. 8 cubic cm oc. i cubic cm 8 OD cubic cm 27
Recall that for a cube of this form
where the side is of length a, then the volume is calculated by the expression
[tex]a^3[/tex]So, to define the volume of the given cube, we should calculate the side's length of the bigger cube, based on the small cube
If we take a look at the front side of the cube we would look something like this
So, we want to determine the length's side of this square. We can see that in this case the length side corresponds to adding twice the length of one of the small cube. So the side's length of the new cube is
[tex]\frac{1}{3}+\frac{1}{3}=\frac{2}{3}[/tex]So, the volume of the bigger cube is
[tex](\frac{2}{3})^3^{}=\frac{2^3}{3^3}=\frac{8}{27}[/tex]So the volume of the bigger cube is 8/27 cubic centimeters
The math class is selling raffle tickets to raise money for the fall ball the raffle tickets cost them $0.08 each at the office supply store and they spent $55 making T-shirts for the club members to wear. they are selling the raffle tickets for two dollars each if X represents the number of raffle tickets they sell ,the following functions can be used to present the situation where c(x) is the clubs cost and r(x) is the amount of money they make selling the tickets
We have the next information
x is the number of raffle tickets
the Profit function is R(x) function minus C(x)
C(x)=0.08x+55
R(x)=2x
P(x)=R(x)-C(x)=2x-(0.08x+55)=2x-0.08x-55
We sum similar terms
P(x)=1.92x-55
Find two points on the line.4y + 2x = 7
Given the equation of a line:
4y +2x = 7
It's required to find two points on the line.
Solving for y:
[tex]y=\frac{7-2x}{4}[/tex]We give x any value and calculate the corresponding value of y.
For example, for x = 1:
[tex]y=\frac{7-2(1)}{4}=\frac{5}{4}[/tex]The point is (1, 5/4)
Now for x = -4:
[tex]y=\frac{7-2(-4)}{4}=\frac{15}{4}[/tex]The point is (-4, 15/4)
A jellybean company produces 1,680 jellybeans per second. One box of jellybeans contains 50 jellybeans. What is this company’s production rate, in boxes of jellybeans per minute?
Answer:
2016
Step-by-step explanation:
1680:50=33.6
33.6x60=2016
Farmer Anders' pumpkin patch is shown below.
19 yards ,
11 yards ,
48 yards and
22 yards.
He purchases fertilizer that costs $2.68 per square foot of coverage, but he has a coupon for half off.
How much will he spent to apply fertilizer to the entire patch.
Please help me answer!
If the cost of fertilizer is $2.68 per square foot of coverage then the farmer need to spend an amount of $5085.3 to apply fertilizer to the entire patch.
Given that the farmer Anders' pumpkin patches are such that they are in rectangle and the lengths are 19 yards and 48 yards and the breadths are 11 yards and 22 yards.
We are required to find the amount of money that the farmer need to spend to apply the fertilizer to the entire patch.
Area which is covering the entire patch=11*19+22*48
=209+1056
=1265 square yards
We know that 1 yard=3 foot
Area in foot=1265*3
=3795 square foot
Amount for fertilizer=3795*3
=$10170.6
Since 1/2 is off so it will be $5085.3 ($10170.6/2).
Hence if the cost of fertilizer is $2.68 per square foot of coverage then the farmer need to spend $5085.3 to apply fertilizer to the entire patch.
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Solve the quadratic equation by completing the square.x^2-14x+46=0First choose the appropriate form and fill in the blanks with the correct numbers. Then solve the equation. If there’s more than one solution, separate them with commas.
x² - 14x + 46 = 0
subtract 46 from both-side
x² - 14x = -46
Add the square of the half of the co-efficient of x
x² - 14x + (-7)² = -46 + 7²
(x-7)² = -46 + 49
(x-7)² =3
Take the square root of both-side
x-7 = ±√3
x-7= ±1.732
Add 7 to both-side of the equation.
x= 7 ± 1.732
Eithe x= 7 + 1.732 or x= 7 - 1.732
x=8.732 or x=5.268
Therefore x = 8.732 , 5.268
Stores on the verbal Graduate Record Exam (GRE) have a mean of 462 and a standard deviation of 119. Scores on the quantitative GRE have a mean of 584 and a standard deviation of 151. Assuming the scores are normally distributed, what quantitative scores are required for an applicant to score at or above the 90th percentile
We need Z-score here.
The formula is:
[tex]z=\frac{x-\mu}{\sigma}[/tex]A normal curve (with 90th percentile), looks like the one below:
We need a standard normal table to move further.
When we go to the table, we find that the value 0.90 is not there exactly, however, the values 0.8997 and 0.9015 are there and correspond to Z values of 1.28 and 1.29, respectively
(i.e., 89.97% of the area under the standard normal curve is below 1.28).
The exact Z value holding 90% of the values below it is 1.282.
--------------- Now, we work backwords and find the value of x:
Verbal GRE:
[tex]\begin{gathered} \mu=462 \\ \sigma=119 \\ z=\frac{x-\mu}{\sigma} \\ 1.282=\frac{x-462}{119} \\ x-462=1.282(119) \\ x=614.558 \end{gathered}[/tex]So, a 90th percentile on Verbal GRE is a score above 614.56
Quantitative GRE:
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ 1.282=\frac{x-584}{151} \\ x=777.582 \end{gathered}[/tex]So, a 90th percentile on Quantitative GRE is a scoer above 777.58
the cost price of 24 articles is the selling price of 15 articles find the gain percentage
Answer:
Step-By step explanation:
Two rows of fluorescent lamps are installed in an office on the same branch circuit, with each row drawing 12.5 amperes. The source voltage is 277 volts, and the line resistance is 0.5 Ω. The wire used has a constant (k) of 12.6. The voltage drop will be within the recommendation of the NEC.
What is type of angle is it vertical complementary supplementary or none of the above I will reward you greatly if you answer
In the given figure,
∠a and ∠b are complementary angles.
This is because ∠CXB + ∠a + ∠b = 180° .........(as they form linear pairs)
∠CXB = 90° ..........(given)
Hence,
∠a + ∠b = 90°
What are Complementary Angles?Two angles whose total is 90 degrees are referred to as complementary angles in geometry. To put it another way, complimentary angles are two angles whose sums equal 90 degrees. 60° and 30°, for illustration. By adding up the two angles' measurements, one can determine the complement and supplement of the two angles. Complementary angles are those where the sum of the two angles equals the measurement of a right angle. They are referred to as adjacent complementary angles if they share a vertex and an arm. It is referred to as non-adjacent complementary angles when two complementary angles are NOT contiguous.To learn more about Complementary angles, refer to:
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how to write the standard equation for the hyperbola that is on the graph
The standard form of the hyperbola:
[tex]\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}\text{ = }1[/tex]The center of the hyperbola is at (h, k) = The point where the lines intersect
(h, k) = (0, 0)
For hyperbola:
c² = a² + b²
b is gotten by tracing the value of a to the intersecting line. Trace 3 down to the x axis. you get 5.
[tex]\begin{gathered} a\text{ = distance from center to vertex} \\ \text{the hyperbola coordinate = (0, a) = }(0,\text{ 3)} \\ \text{the other one = (0, -a) = (0, -3)} \\ \text{Hence, a = 3} \end{gathered}[/tex][tex]\begin{gathered} c\text{ = distance from the center to the focus point} \\ b\text{ = 5 and -b = 5} \\ c^2=3^2+5^2 \end{gathered}[/tex][tex]\begin{gathered} c^2\text{ = 9+25 = 34} \\ c\text{ = }\sqrt[]{34} \\ c\text{ = 5.83} \end{gathered}[/tex]The equation becomes:
[tex]\begin{gathered} \frac{(y-k)^2}{5^2}-\frac{(x-h)^2}{3^2}=\text{ }\frac{(y-0)^2}{5^2}-\frac{(x-0)^2}{3^2} \\ =\text{ }\frac{y^2}{25}-\frac{x^2}{9} \end{gathered}[/tex]