we know that
Distance=ratex Time
where
Rate=50 mph
Time=15 min
Convert the time to hours
1 hour=60 min
so
15 min=15/60=0.25 hours
substitute in the formula
Distance=50x0.25
Distance=12.50 milesWhich property of equality would you use to solve the equation 5m = 12?
We would have to use the division (and/or multiplication) property in order to solve, and that would be:
[tex]5m=12\Rightarrow m=\frac{12}{5}[/tex]Write the expression in the standard form a + bi.
SOLUTION
Write out the expression
[tex]i^{22}[/tex][tex]\begin{gathered} i^{22} \\ \text{can be written as} \\ (i^2)^{11} \end{gathered}[/tex]Recall that
[tex]i^2=-1[/tex]Replace into the expression above
[tex](-1)^{11}=-1[/tex]Hence
[tex]i^{22}=-1[/tex]Therefore
The first option is Right
what are the solutions of the equation 0 equals x ^ 2 + 3x - 10
The given expression is :
[tex]0=x^2+3x-10[/tex]Factorize the expression :
[tex]\begin{gathered} 0=x^2+3x-10 \\ x^2+3x-10=0 \end{gathered}[/tex]Find the pair of number such that : the product of two numbers are equal = (-10)
and thier summation is equal to 3
i.e. 5 x ( -2) = -10 and 5 + (-2) = 3
So,
[tex]\begin{gathered} x^2+3x-10=0 \\ x^2+5x-2x-10=0 \end{gathered}[/tex]Take x common from the first two terms and (-2) from last two terms :
[tex]\begin{gathered} x^2+5x-2x-10=0 \\ x(x+5)-2(x+5)=0 \\ \text{Now, take (x+5) common :} \\ (x-2)(x+5)\text{ =0} \end{gathered}[/tex]Now equate each factor with zero :
[tex]\begin{gathered} (x-2)(x+5)=0 \\ x-2=0\Rightarrow x=2 \\ x+5=0\Rightarrow x=-5 \end{gathered}[/tex]Answer : C) x = -5, 2
Consider the following rational expression:2 – 2y / 2y - 2Step 1 of 2: Reduce the rational expression to its lowest terms.Answer
Factor out 2 on both numerator and denominator
[tex]\begin{gathered} \frac{2-2y}{2y-2} \\ =\frac{2(1-y)}{2(y-1)} \\ \\ \text{cancel out }2\text{ on both numerator and denominator} \\ =\frac{\cancel{2}(1-y)}{\cancel{2}(y-1)} \\ =\frac{(1-y)}{(y-1)} \\ \\ \text{factor out }-1\text{ on numerator},\text{ and rearrange to cancel out common binomial} \\ =\frac{(1-y)}{(y-1)} \\ =\frac{-1(-1+y)}{(y-1)} \\ =\frac{-1(y-1)}{(y-1)} \\ =\frac{-1\cancel{(y-1)}}{\cancel{(y-1)}} \\ =-1 \\ \\ \text{Therefore,} \\ \frac{2-2y}{2y-2}=-1 \end{gathered}[/tex]Part 2:
Since the given expression is in fraction, we cannot let the denominator equal to zero. Find values of y that makes the denominator by zero
[tex]\begin{gathered} \text{Denominator: }2y-2 \\ \\ \text{Equate to zero} \\ 2y-2=0 \\ 2y-2+2=0+2 \\ 2y\cancel{-2+2}=2 \\ \frac{2y}{2}=\frac{2}{2} \\ y=1 \\ \\ \text{If }y=1,\text{ the denominator }2y-2\text{ becomes zero therefore}, \\ y\neq1 \end{gathered}[/tex]A table of 5 students has 2 seniors and 3 juniors. The teacher is going to pick 2 students at random from this group to present homework solutions. Find the probability that both students selected are juniors
ANSWER
[tex]\text{ P\lparen both students are junior\rparen = }\frac{1}{10}[/tex]EXPLANATION
Given information
The total number of junior students = 2
The total number of senior students = 3
The total number of students = 5
To determine the probability of picking two junior students, follow the steps below
Step 1: Define probability
[tex]\text{ Probability = }\frac{\text{ possible outcome}}{\text{ total outcome}}[/tex]Step 2: Find the probability of picking the first junior students
[tex]\begin{gathered} \text{ Probability = }\frac{possible\text{ outcome}}{total\text{ outcome}} \\ \text{ Probability of picking the first junior students is} \\ \text{ P\lparen Junior student\rparen = }\frac{2}{5} \end{gathered}[/tex]Assuming the first picking was successful, then, we will be left with 1 junior student and 3 senior students.
Therefore, the new total outcome can be calculated below
1 + 3 = 4 students
Step 3: Find the probability that the second picking will be a junior student
[tex]\begin{gathered} \text{ Probability = }\frac{\text{ possible outcome}}{\text{ total outcome}} \\ \text{ P\lparen picking the second junior student\rparen = }\frac{1}{4} \end{gathered}[/tex]Step 4: Find the probability that both students are junior students
[tex]\begin{gathered} \text{ P\lparen both students are junior students\rparen = }\frac{2}{5}\times\frac{1}{4} \\ \text{ P\lparen both students are junior students\rparen = }\frac{2}{20} \\ \text{ P \lparen both students are junior students \rparen = }\frac{1}{10} \end{gathered}[/tex]Hence, the probability that both students selected are juniors is 1/10
A carpenter cuts a 5-ft board in two pieces. One piece must be three times as longas the other. Find the length of each piece.
3.75 ft and 1.25 ft
Explanation
Step 1
Diagram
Step 2
set the equations
let x represents the longest piece
lety represents the smaller piece
so
a)A carpenter cuts a 5-ft board in two pieces, hence
[tex]x+y=5\Rightarrow equation(1)[/tex]b)One piece must be three times as long as the other,then
[tex]x=3y\Rightarrow equation(2)[/tex]Step 3
finally, solve the equations:
a) replace the x value from equation (2) into equation(1)
[tex]\begin{gathered} x+y=5\Rightarrow equation(1) \\ (3y)+y=5 \\ add\text{ like terms} \\ 4y=5 \\ divide\text{ both sides by 4} \\ \frac{4y}{4}=\frac{5}{4} \\ y=1.25 \end{gathered}[/tex]b) now, replace the y value into equation (2) to find x
[tex]\begin{gathered} x=3y\Rightarrow equation(2) \\ x=3(1.25) \\ x=3.75 \end{gathered}[/tex]therefore, the lengths of the pieces are
3.75 ft and 1.25 ft
I hope this helps you
1. P, Q and R are three buildings. A car began its journey at P, drove to Q, then to R and returned to P. The bearing of Q from P is 058º and R is due east of Q. PQ = 114 km and QR = 70 km. © Draw a clearly labelled diagram to represent the above informationen on the diagram TƏRund (a) the north/south direction (b) the bearing 058° (c) the distances 114 km and 70 km. (ii) Calculate (a) the measure of angle POR (b) the distance PR [3] (c) the bearing of P from R [3]
Step 1
Given;
[tex]\begin{gathered} The\text{ bearing of Q from P is 058}^o\text{ } \\ R\text{ is due east of Q} \\ PQ=114km \\ QR=70km \end{gathered}[/tex]Step 2
Draw the diagram
Step 2
Calculate the measure of angle PQR
[tex]\angle PQR=58+90=148^o[/tex]This is because using alternate exterior angles are equal theorem, the first part of angle Q 58 degrees. Since R is due east of Q, then the other part must be 90 degrees, when summed we get 148 degrees
Step 3
Calculate the distance PR. To do this we will use the cosine rule
[tex]\begin{gathered} PR^2=PQ^2+QR^2-2PQ\left(QR\right?cosQ \\ PR^2=114^2+70^2-2\left(114\right)\left(70\right)cos\left(148\right) \\ PR^2=17896+13534.84761 \\ PR=\sqrt{31430.84761} \\ PR=177.2874717 \\ PR\approx177.3km\text{ to the nearest tenth} \end{gathered}[/tex]Step 4
Calculate the bearing of P from R.
Use sine rule and find angle R
[tex]\frac{sin\text{ 148}}{177.2874717}=\frac{sinR}{114}[/tex][tex]\begin{gathered} 114sin148=177.2874717sinR \\ R=\sin^{-1}\frac{\mleft(114sin148\mright)}{177.2874717} \\ R=19.92260569 \end{gathered}[/tex]The bearing of P from R = (90-angle R)+90+90=250 degrees approximately to the nearest whole number
[tex]\begin{gathered} =\left(90-19.92260569\right)+90+90 \\ =250.07739 \\ \approx250^o \end{gathered}[/tex]The bearing of P from R =250 degrees approximately to the nearest whole number
what digit is in the
Rounding each number to the nearest ten:
• 96 = 100
,• 63 = 60
,• 27 = 30
,• 76 = 80
Sum with rounded numbers:
[tex]100\text{ + 60+30+80=270}[/tex]Answer = 270
identify the horizontal asymptotes, if they exist, for the following function…
You have the following function:
[tex]f(x)=\frac{5x^4-2x}{x^4+32}[/tex]Take into account that if the expression for the numerator has the same degree that the expression at the denominator, the horizontal asymptote is given by the quotient between the leading coefficient of each polynomial.
In this case, leading coeffcicient of numerator is 5 and from the denominator we get a leading coeeficient equal to 1.
Then, the horizontal asymptote is:
y = 5/1 = 5
The points ( 0.5 , 1/10 ) and ( 7 , 1 2/5)The points are on the graph of a proportional relationshipIt is required to find the constant of proportionally
Given: the points ( 0.5 , 1/10 ) and ( 7 , 1 2/5)
The points are on the graph of a proportional relationship
It is required to find the constant of proportionally
so, the graph of the points will be as shown in the following image:
as shown there a proportional relationship
Because the line is pass through zero
So, the constant of proportionality = y/x
It can be calculated using any point from the given points
So ,
Using the point ( 0.5 , 1/10)
the constant = 0.1/0.5 = 0.2
We can check the answer using the other point ( 7, 1 2/5)
The constant = (1 2/5)/7 = 1.4/7 = 0.2
So, the constant of proportionality = 0.2
If you have a 77.2% and you got 34% on a test and it’s worth 60% of your grade, what would you grade be now?
Answer:
51.28%
Step-by-step explanation:
since the test is worth 60% of your grade, the rest is worth 40%
calculate your new grade by multiplying each grade percent (as written) by the percent of your grade (as a decimal):
77.2(0.4) = 30.88
34(0.6) = 20.4
then add them together: 30.88 + 20.4
17. In trapezoid FGJK, what is the value of x? N CO 18.6 L K 23.6 9.3 11.8 ET O 13.6
Given data:
The given figure is shown.
The expression for the trapezium is,
[tex]\begin{gathered} \frac{x}{18.6}=\frac{18.6}{23.6} \\ 23.6x=18.6^2 \\ x=14.6 \\ =15 \end{gathered}[/tex]Thus, thi
I need help with triangles
Which choices are equivalent to the expression below? Check all that apply.A.B.C.72D.E.F.
GIven:
[tex]3\sqrt{8}[/tex]Required:
We need to find the equivalent expression
Explanation:
let
[tex]\begin{gathered} x=3\sqrt{8} \\ x^2=72 \end{gathered}[/tex]now just we need to check that which square is 72
1)
[tex]\begin{gathered} a=\sqrt{3}\sqrt{12} \\ a^2=36\text{ not possible} \end{gathered}[/tex]2)
[tex]\begin{gathered} b=\sqrt{6}\sqrt{12} \\ b^2=72\text{ possible} \end{gathered}[/tex]3)
[tex]\begin{gathered} c=72 \\ c^2=5184\text{ not possible} \end{gathered}[/tex]4)
[tex]\begin{gathered} d=\sqrt{3}\sqrt{24} \\ d^2=72\text{ possible} \end{gathered}[/tex]5)
[tex]\begin{gathered} e=\sqrt{6}\sqrt{24} \\ e^2=144\text{ not possible} \end{gathered}[/tex]6)
[tex]\begin{gathered} f=\sqrt{9}\sqrt{8} \\ f^2=72\text{ possible} \end{gathered}[/tex]Final answer:
[tex]\begin{gathered} \sqrt{6}\sqrt{12} \\ \sqrt{3}\sqrt{24} \\ \sqrt{9}\sqrt{8} \end{gathered}[/tex]
are equivalent to given expression
Determine the equation of the graphed circle below!Equation should look like the example below!
Step 1:
Write the formula for the equation of a circle.
[tex]\begin{gathered} (x-a)^2+(y-b)^2=r^2 \\ \text{Center = ( a , b )} \\ \text{Radius = r} \end{gathered}[/tex]Step 2:
Locate and write the center and radius of the circle.
Step 3:
Write the equation of the circle with center (-7, -2) and radius r = 2
[tex]\begin{gathered} (x-a)^2+(y-b)^2=r^2 \\ (x-(-7))^2+(y-(-2))^2=2^2 \\ (x+7)^2+(y+2)^2=\text{ 4} \end{gathered}[/tex]Final answer
[tex](x+7)^2+(y+2)^2=\text{ 4}[/tex]Creating and solving equationstwo-thirds a number plus 4 is 7
Given:
two-thirds a number plus 4 is 7
First Part: Converting the statement into equation
Let x be the number in the given statement.
The phrase "two-thirds a number" can be expressed as
[tex]\frac{2}{3}x[/tex]Pair it with "... plus 4" and we get
[tex]\frac{2}{3}x+4[/tex]Finally, it is stated it is equal to 7, and we complete the equation
[tex]\frac{2}{3}x+4=7[/tex]Second Part: Solving for the number
Now, that we have the equation, we can now solve for the missing number x.
Subtract both sides by 4, to remove the constant 4 on the left side of the equation
[tex]\begin{gathered} \frac{2}{3}x+4=7 \\ \frac{2}{3}x+4-4=7-4 \\ \frac{2}{3}x\cancel{+4-4}=3 \\ \frac{2}{3}x=3 \end{gathered}[/tex]Multiply both sides by 3/2, and we get
[tex]\begin{gathered} \frac{2}{3}x=3 \\ \frac{2}{3}x\cdot\frac{3}{2}=3\cdot\frac{3}{2} \\ \frac{\cancel{2}}{\cancel{3}}x\cdot\frac{\cancel{3}}{\cancel{2}}=\frac{9}{2} \\ x=\frac{9}{2} \end{gathered}[/tex]Therefore, the number is 9/2 or nine-halves.
a) What were the ranges of typing speeds for the two groups?Group 1: Group 2:b) Which group had more typing speed in the 40s1) Group 1 2)Group 2 3) Each had the samec) Which group had the greater median typing speed?1) Group 1 2)Group 2 3) Each had the same
The given stem and leaf plot shows the typing speeds of two groups of students.
Group 1 has n1= 20 students
Group 2 has n2= 19 students
The stem and leaf plot is two sided, meaning that they share the same stem.
The observed values are the number of words per minute.
In the steam the ten of each value is placed and in the leafs you find the units:
This way you can determine the observations for both samples. I'll do so and arrange them form least to greatest:
Group 1:
33, 34, 37, 42, 44, 44, 45, 47, 48, 49, 49, 50, 51, 52, 52, 55, 55, 63, 66, 67
Group 2:
33, 36, 41, 42, 44, 46, 46, 51, 52, 52, 52, 53, 53, 56, 59, 60, 66, 67, 69
Part a
The range is calculated as the difference between the maximum and minimum observations of a sample. To determine those values you need the sample ordered from least to greatest.
For group 1:
Minimum value: 33 words/min
Maximum value: 67 words/min
Range= maximum-minimum=67-33=34words/min
For group 2:
Minimum value: 33words/min
Maximum value: 69 words/min
Range: 69-33=36words/min
→ the range for group 1 is 34words/min while the range for group 2 is 36words/min
Part b
To determine which group had more typing speeds in the fourties you have to count said observations for both of them.
You can do it directly from the stem and leaf plot, go to the row correpsonding to the 4 in the plot and count or use the values:
For group 1: in the second row there are 8 leafs, corresponding to the observations: 42, 44, 44, 45, 47, 48, 49, 49,
For group 2: in the second row there are 5 leafs, corresponding to the observations: 41, 42, 44, 46, 46
→There are more typing speeds in the 40s in group 1.
Part c:
The median is a measure of center that divides the sample in two halves. To calculate it you have to determine its position and then look for the corresponding value in the sample that was previously ordered from least to greatest.
To determine the position of the mean you have to use the following fomula:
For even samples: n/2
For odd samples: (n+1)/2
Median of group 1
n1=20 students
The sample is even, calculate its position using the first formula:
Position: n/2 = 20/2= 10
The median is in the tenth position, look in the sample for the tenth observation:
33, 34, 37, 42, 44, 44, 45, 47, 48, 49, 49, 50, 51, 52, 52, 55, 55, 63, 66, 67
→The median typing speed for group 1 is 49 words/min
Median of group 2
n2=19 students
The sample is odd, you have to use the second formula to find its position:
Position: (n+1)/2= (19+1)/2= 20/2= 10
The median of this group is the 10th observation:
33, 36, 41, 42, 44, 46, 46, 51, 52, 52, 52, 53, 53, 56, 59, 60, 66, 67, 69
→The median typing speed for group 2 is 52 words/min
→Group 2 has the greater median typing speed.
Solve. Show all your work!The digits of a positive two-digit integer N are interchanged to form an integer K. Find allpossibilities for N if N is even and exceeds K by more than 50.
Let the units place digit be U and the tens place digit be T.
The number N is given by:
[tex]N=10T+U\ldots(i)[/tex]The number K is given by:
[tex]K=10U+T\ldots(2)[/tex]It is given that N is even that means U can be only from 0,2,4,6,8.
It is also given that N exceeds K by more than 50 so it follows:
[tex]\begin{gathered} N-K\ge50 \\ 10T+U-(10U+T)\ge50 \\ 9T-9U\ge50 \end{gathered}[/tex]So it can be said that:
[tex]T-U\ge\frac{50}{9}\approx5.5556\approx6[/tex]Since the value of T-U will always be an integer and it should be greater than or equal to 6.
The number T can be 1 to 9 and U can be only 0,2,4,6,8 so it follows:
[tex]\begin{gathered} T=9,U=0\Rightarrow T-U=9 \\ T=9,U=2\Rightarrow T-U=7 \\ T=8,U=0\Rightarrow T-U=8 \\ T=7,U=0\Rightarrow T-U=7 \\ T=6,U=0\Rightarrow T-U=6 \\ T=8,U=2\Rightarrow T-U=6 \end{gathered}[/tex]Hence the possible values for integer N are 90,92,80,70,60,82 and the respective integer K will be 09,29,08,07,06,28.
In all cases the difference is more than 50 as you can check.
how do i solve 13-3/2x=37
you are selling snacks at the border trade fair. you are selling nachos and lemonade. each nachos costs $2.50 and each lemonade cost $2.25. at the end of the night you made a total of $112.50. you sold a total of 94 nachos and lemonade combined. how many nachos and lemonades were sold?
In order to determine the number of nachos and lemonade sold, you first write the given situation in an algebraic way.
If x is the number of nachos and y the number of lemonades, then, you have:
2.50x + 2.25y = 112.50 cost of the nachos and lemonade sold
x + y = 94 nachos and lemonade sold
Next, solve the previous system.
Multiply the second equation by 2.50. Next, subtract the equation to the first one:
(x + y = 94)(2.50)
2.50x + 2.50y = 235
2.50x + 2.25y = 112.50
-2.50x - 2.50y = -235
-0.25y = -122.5
solve the previous equation for y:
y = -122.5/(-0.25)
y = 490
Next, replace the previous value of y into the expression x + y = 94 and solve for x:
x + y = 94
x + 490 =
On a piece of paper, graph y+25**-1. Then determine which answer choicematches the graph you drew.ABСD0.9.-3)0,-)(0-3)69,-2)(4-2)(4.23(2)O A. Graph AB. Graph BO C. Graph CO D. Graph D
Let's graph the given inequality:
As we can see, it matches graph A from the options we were given.
Simplify. Assume that all variables result in nonzero denominators.
2n^3 y−8n^2 y/3y^4 * 12/n-4
The simplified form of given expression 2n^3 y−8n^2 y/3y^4 * 12/n-4 is 8n^2/y^3
In this question, we have been given an expression.
2n^3 y−8n^2 y/3y^4 * 12/n-4
We need to simplify given expression.
2n^3 y − 8n^2 y/3y^4 * 12/n-4
= [2n^2y (n - 4)] / 3y^4 * 12/(n - 4)
= 4 * (2n^2y)/y^4
= 8n^2/y^3
Therefore, the simplified form of given expression 2n^3 y−8n^2 y/3y^4 * 12/n-4 is 8n^2/y^3
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• 21 Theodore inherited two different stocks whose yearly income was $2100. The total appraised value of the stocks was $40,000 and one was paying 4% and one 690 per year. What was the value of each stock? o
hello
the yearly income was $2100
the appraised value = $40,000
one of the stocks pays 4% annually
the other pays $690 yearly
let's find how much the 4% stock pays annually
to do this, let's subtract the income of one of the stocks from the total income. i.e 690 from 2100
[tex]2100-690=1410[/tex]the other stock pays $1410 annually
now we can simply find the value of each stock
[tex]\begin{gathered} 4\text{\% of x gives 1410 annually} \\ \frac{4}{100}=\frac{1410}{x} \\ \text{cross multiply both sides } \\ 4\times x=100\times1410 \\ 4x=141000 \\ \text{divide both sides by 4} \\ \frac{4x}{4}=\frac{141000}{4} \\ x=35250 \end{gathered}[/tex]the value of one of the stock is $35250
we can proceed to find the value of the other stock by subtracting 35250 from 40000 which is the value of the two stock
[tex]40000-35250=4750[/tex]from the calculations above, the value of the stocks is $35250 and $4750
(X^-3y^2/x^3)^-2
Simplify the expression. Your final answer should use positive exponents.
Answer:
y^-4
here you are
,........
If 20 assemblers can complete a certain job in 6 hours, how long will the same job take if the number of assemblers is cut back to 8?
ANSWER
[tex]15[/tex]EXPLANATION
For 1 assembler, it will take;
[tex]\begin{gathered} 20\times R\times6=1 \\ R=\frac{1}{120} \end{gathered}[/tex]For 8 assemblers;
[tex]8\times R\times T=1[/tex]Substitute R
[tex]\begin{gathered} 8\times R\times T=1 \\ 8\times\frac{1}{120}\times T=1 \\ \frac{8T}{120}=1 \\ 8T=120 \\ T=\frac{120}{8} \\ =15 \end{gathered}[/tex]An account earns an annual rate of 5.4% compounded monthly. If $3,000 is deposited into this account, then after 3 years there is $___. Round your answer to two decimals.
Given:
rate (r) = 5.4% or 0.054 in decimal form
Principal (P) = $3,000
time in years (t) = 3 years
number of conversions per year (m) = 12 (because it says monthly)
Find: future value or maturity value
Solution:
The formula for getting the future value of a compound interest is:
[tex]F=P(1+\frac{r}{m})^{mt}[/tex]Let's plug in the given data to the formula above.
[tex]F=3,000(1+\frac{0.054}{12})^{12\times3}[/tex]Then, solve for F or future value.
[tex]\begin{gathered} F=3,000(1.0045)^{36} \\ F=3,000(1.17532999) \\ F\approx3,526.30 \end{gathered}[/tex]Answer: After 3 years, the deposited money will become $3, 526.30.
which measurement could create more than one triangle measuring 20 cm / 9 cm and 10cm be a triangle with sides measuring 10 cm and 20 cm and included angle measurement 65 C a right angle with acute angles measuring 45 and 45 d a triangle with sides measuring 15 in 20 in and 25 in
Input data
The triangles created by the measurements of options A, B and D have specific side lengths. Therefore, you cannot create more than one triangle.
However, for a triangle with acute angles measuring 45° and 45°, a countless number of similar triangles (triangles with the same shape but different sizes) can be created.
The correct choice is C.
the perimeter of the original rectangle is 16 ft. what is the perimeter of the enlarged rectangle? round to the nearest tenth if necessary.
To calculate the perimeter of the enlarged rectangle, we need to get the breadth first.
The original rectangle is similar to the enlarged rectangle, so the ratio of thier corresponding sides must be equal
Let b represent the breadth of the enlarged rectangle,
[tex]\begin{gathered} \text{ Ratio of corresponding sides is given as } \\ \frac{1.8}{b}=\frac{6.2}{12.4} \\ \text{cross multiply} \\ 6.2b=12.4\text{ x 1.8} \\ 6.2b=22.32 \\ b=\frac{22.32}{6.2} \\ b=3.6\text{ ft} \end{gathered}[/tex]Now, the length of the enlarged rectangle is 12.4ft while the breadth is 3.6ft
The perimeter = 2(L+B)
= 2(12.4 + 3.6)
=2(16)
=32ft
The perimeter of the enlarged rectangle is 32 ft
Algebraically determine whether each of the following functions is even, odd or neither. then graph it B. y = x^3 – 3 C. y = 2x^3 - x
According to the even and odd function rules, we found out that the function [tex]y=x^{3}-3[/tex] is neither even nor odd and the function [tex]y=2x^{3}-x[/tex] is an odd function.
It is given to us that the functions are -
B. [tex]y=x^{3}-3[/tex]
C. [tex]y=2x^{3}-x[/tex]
We want to determine each of the following functions is even, odd or neither.
To see if the function is even, we have to check if [tex]f(-x)=f(x)[/tex]
To see if the function is odd, we have to check if [tex]f(-x)=-f(x)[/tex]
B. Here, we have
[tex]y=x^{3}-3\\= > f(x)=x^{3}-3\\= > f(-x)=(-x)^{3}-3\\= > f(-x)=-x^{3}-3[/tex]
We see that [tex]f(-x)\neq f(x)[/tex]. This implies that the function is not even.
Also, [tex]f(-x)\neq -f(x)[/tex]. This implies that the function is not odd.
Therefore, this function is neither even nor odd.
C. Here, we have
[tex]y=2x^{3}-x\\= > f(x)=2x^{3}-x\\= > f(-x)=2(-x)^{3}-(-x)\\= > f(-x)=-2x^{3}+x[/tex]
We see that [tex]f(-x)\neq f(x)[/tex]. This implies that the function is not even.
However,
[tex]f(-x)=-2x^{3}+x\\= > f(-x)= -(2x^{3}-x)\\ = > f(-x)=-f(x)[/tex]
This implies that the function is odd.
Therefore, this function is odd.
Thus, the function [tex]y=x^{3}-3[/tex] is neither even nor odd and the function [tex]y=2x^{3}-x[/tex] is an odd function.
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What is the equation for the linear model in the scatterplot obtained by choosing the two points closest to the line
consider two points closest to the line. say ,
[tex]\begin{gathered} (x_1,y_1)=(6,0) \\ (x_2_{}_{}_{},y_2)=(8,1) \end{gathered}[/tex]let us find the slope, m by the formula
[tex]m=\frac{y_2-y_1}{x_2_{}_{}-x_1}[/tex]subsitute the points in the formula,
[tex]\begin{gathered} m=\frac{1-0}{8-6} \\ m=\frac{1}{2} \end{gathered}[/tex]let us find the y - intercept.
[tex]y=mx+b\ldots(1)[/tex]subsitute the one of the point (6,0) in the above equation.
[tex]\begin{gathered} 0=\frac{1}{2}\times6+b \\ 0=3+b \\ b=-3 \end{gathered}[/tex]thus,
subsitute m= 1/2 and b = - 3 in the equation (1),
[tex]y=\frac{1}{2}x-3[/tex]