Given the inverse function
[tex]f(x)\text{ = 2arccosx}[/tex]A function g is the inverse of f if for y = f(x) , x = g(y)
[tex]\begin{gathered} y\text{ = 2arccosx} \\ \arccos x\text{ = }\frac{y}{2} \\ \arccos a\text{ = b} \\ a=\cos (b) \end{gathered}[/tex][tex]\begin{gathered} x=\text{ cos(}\frac{y}{2}) \\ \text{substitute y = x} \\ y=\cos (\frac{x}{2}) \end{gathered}[/tex]Hence the correct answer is Option B
I know the answer to fill i is -24 but how do I simplify ??
[tex]a^{-24\text{ }}=\frac{1}{a^{24}}[/tex]
The first expression es the simplified expression using the property of negative exponents.
Convert the Cartesian equation x^2 + y^2 + 3y = 0 to a polar equation.r^2 = -3 sin θr = √3 sin θr = -3 sin θ
SOLUTION
From the question
[tex]x^2+y^2+3y=0[/tex]This becomes
[tex](x^2+y^2)+3y=0[/tex]In polar,
[tex]\begin{gathered} x^2+y^2=r^2 \\ \\ \text{and } \\ \\ y=r\sin \theta \end{gathered}[/tex]So, this becomes
[tex]\begin{gathered} r^2+3r\sin \theta=0 \\ \\ \frac{r^2}{r}=\frac{-3r\sin \theta}{r} \\ \\ r=-3\sin \theta \end{gathered}[/tex]Fill in the following values for a 45-45-90 triangle Leg Leg Hypotenuse 5 А B C С D 32 Fill in the following values for a 30-60-90 triangle Short Leg Long Leg Hypotenuse 6 E H 20 G
First Part 45-45-90 Triangle
first triangle
where the two angles different to 90° are same, the measure of the legsof the triangle are the same
then
[tex]A=5[/tex]and to calculate B or the hypotenuse we use pythagoras
[tex]a^2+b^2=h^2[/tex]where a and b are legs and h the hypotenuse
replacing
[tex]\begin{gathered} 5^2+5^2=h^2 \\ 25+25=h^2 \\ 50=h^2 \\ h=\sqrt[]{50} \\ h=5\sqrt[]{2} \end{gathered}[/tex]the hypotenuse or B is
[tex]B=5\sqrt[]{2}[/tex]Second triangle
legs of the triangle have the same value then if we apply pythagoras
[tex]a^2+b^2=h^2[/tex]and replace the legs with the same value(a)
[tex]\begin{gathered} a^2+a^2=h^2 \\ 2a^2=h^2 \end{gathered}[/tex]we can replace the hypotenuse and solve a
[tex]\begin{gathered} 2a^2=(3\sqrt[]{2})^2 \\ 2a^2=18 \\ a^2=\frac{18}{2} \\ \\ a=\sqrt[]{9} \\ a=3 \end{gathered}[/tex]value of each leg is 3 units, then
[tex]C=D=3[/tex]Second part 30-60-90 triangle
First triangle
we use trigonometric ratios to solve, for example I can use tangent for the angle 60 to find E
[tex]\tan (\alpha)=\frac{O}{A}[/tex]where alpha is the angle, O the oppiste side of the angle and A the adjacet side of the angle
using angle 60°
[tex]\begin{gathered} \tan (60)=\frac{E}{6} \\ \\ E=6\tan (60) \\ \\ E=6\sqrt[]{3} \end{gathered}[/tex]now using sine we calculate F or the hypotenuse
[tex]\sin (\alpha)=\frac{O}{H}[/tex]where alpha is the angle, O the opposite side from the angle and H the hypotenuse
using angle 60°
[tex]\begin{gathered} \sin (60)=\frac{E}{F} \\ \\ F=\frac{E}{\sin (60)} \\ \\ F=\frac{6\sqrt[]{3}}{\sin (60)} \\ \\ F=12 \end{gathered}[/tex]Second triangle
we use sine with 60° to find H
[tex]\begin{gathered} \sin (\alpha)=\frac{O}{h} \\ \\ \sin (60)=\frac{H}{20} \\ \\ H=20\sin (60) \\ H=10\sqrt[]{3} \end{gathered}[/tex]use cosine with 60° to find G
[tex]\begin{gathered} \cos (\alpha)=\frac{A}{h} \\ \\ \cos (60)=\frac{G}{20} \\ \\ G=20\cos (60) \\ \\ G=10 \end{gathered}[/tex]Final Values
[tex]\begin{gathered} A=5 \\ B=5\sqrt[]{2} \\ C=3 \\ D=3 \\ E=6\sqrt[]{3} \\ F=12 \\ G=10 \\ H=10\sqrt[]{3} \end{gathered}[/tex]Keico is selling rattle tickets to raise money for the sancol band. The odds againet winning a prize in the raffleare 121. What is the probability of winning a prize? Express your anewer as a decimal. if necessary, round youranower to the nearest thousandth.0 130 0.923O 0.083O 0.0777
Given:
The odds against winning a prize in the raffle are 12:1
[tex]\begin{gathered} \text{Probability of winning a prize=}\frac{1}{13} \\ \text{Probability of winning a prize=}0.077 \end{gathered}[/tex]0.077 is the probability of winning a prize.
Evaluate the input/output table for the expression x - 9.Хy-101
x - 9
when x = -1
out put is -1 -9 = -10
when x = 0
Output is 0-9 = -9
when x = 1
output = 1-9 = -8
x y
-1 -10
0 -9
1 -8
simplify 2a x a x 3a + b x 4b
Explanation:
[tex]\begin{gathered} 2a\text{ *a *a * 3a = 2 * 3 * a *a * a } \\ 6\text{ * a}^3\text{ = 6a}^3 \end{gathered}[/tex][tex]\begin{gathered} \text{b * 4b = 4 *b * b } \\ \text{4 * b}^2\text{ = 4b}^2 \end{gathered}[/tex]Put them together
[tex]2a*a*a*3a\text{ + b * 4b = 6a}^3+4b^2[/tex]The graph of a function f is given. Use the graph to estimate the following. (Enter your answers using interval notation) PLEASE HELP!! confused on whole problem
From the graph
Domain = [ -1 , 4]
Range = [ -1 , 3 ]
b) When you look at the graph, the function f
Increasing = [ -1, 1 ] , [ 2, 4 ]
Decreasing = [ 1 , 2 ]
A local road rises 33 feet for every 423 feet of pavement. What is the slope of the road? Simplify your answer.
If a local road rises 33 feet for every 423 feet of pavement, the slope of the road will be rate of change of pavement with respect to the road. This is expressed as;
slope of the road = 423/33
Slope of the road = 12.82
Hence the slope of the road is 12.82
5/3×(-3/4) what's the answer
using the definition of multiplication between two fractions
[tex]\frac{a}{b}\times\frac{c}{d}=\frac{a\cdot c}{b\cdot d}[/tex]we obtain,
[tex]\frac{5}{3}\times-\frac{3}{4}=-\frac{5\cdot3}{3\cdot4}=-\frac{5}{4}[/tex]
Rita earns scores of 83, 87, 85, 88, and 90 on her five chapter tests for a certain class and a grade of 82 on the class project.
The overall average for the course is computed as follows: the average of the five chapter tests makes up 30% of the course
grade; the project accounts for 30% of the grade; and the final exam accounts for 40%. What scores can Rita earn on the final
exam to earn a "B" in the course if the cut-off for a "B" is an overall score greater than or equal to 80, but less than 90? Assume>that 100 is the highest score that can be earned on the final exam and that only whole-number scores are given.>To obtain a "B", Rita needs to score between>and>inclusive
Given:
Rita earns scores of 83, 87, 85, 88, and 90 on her five-chapter tests for a certain class.
And a grade of 82 on the class project.
First, we will find the average of the scores of the five tests
[tex]5-tests\text{ }average=\frac{83+87+85+88+90}{5}=\frac{433}{5}=86.6[/tex]The overall average for the course is computed as follows:
30% of the course grade ⇒ Rita get 86.6
30% of project grade ⇒ Rita get 82
40% of the final exam ⇒ let Rita get x
We will find the value of x provided that Rita will earn a "B" score
a "B" is an overall score greater than or equal to 80, but less than 90
So, we will find (x) as follows:
[tex]\frac{30*86.6+30*82+40*x}{100}\ge80[/tex]Solve the inequality to find (x):
[tex]\begin{gathered} 5058+40x\ge8000 \\ 40x\ge8000-5058 \\ 40x\ge2942 \\ x\ge\frac{2942}{40} \\ \\ x\ge73.55 \end{gathered}[/tex]And the upper limit will be as follows:
[tex]\frac{30\times86.6+30\times82+40x}{100}<90[/tex]Solve to find (x):
[tex]\begin{gathered} 5058+40x<9000 \\ 40x<9000-5058 \\ 40x<3942 \\ x<\frac{3942}{40} \\ \\ x<98.55 \end{gathered}[/tex]So, the answer will be:
To obtain a "B", Rita needs to score between 73.55 and 98.55
Which expression is a factor of 9r2 – 4r + 1? F3r-1 Gr 1 H 9741 3 There are no real factors.
There are no real factors
Here, we want to factorize the given expression
To check if there are real factors, we will need to get the determinant
mathematically, we have this as;
[tex]D=b^2-4ac[/tex]where a is 9, b is -4 and c is 1
Substituting these values;
[tex]D=(-4)^2-4(9)(1)\text{ = 16-36 = -20}[/tex]As we can see, the value of the determinant is negative
Whenever the determinant value is negative, there are no real roots
John is a salesman for a company. he earns a straight commission at a rate of 4 and 1/2% . last month his total says were $82,969. what is his gross monthly income for last month?
hello
his gross income was = $82,969
commission = 4 1/2% or 4.5%
since we have the gross income, we can use that data to find his actual salary for the month.
all we need to do is find 4.5% of 82969 and subtract the value from it
[tex]\begin{gathered} 4.5\text{ \% of 82969} \\ \frac{4.5}{100}=\frac{x}{82969} \\ \text{cross multiply both sides and solve for x} \\ 100\times x=4.5\times82969 \\ 100x=373360.5 \\ \text{divide both sides by 100} \\ \frac{100x}{100}=\frac{373360.5}{100} \\ x=3733.605 \end{gathered}[/tex]the commission pay was $3733.605
to find his actual salary, subtract 3733.605 from 82969
[tex]\text{ income}=82969-3733.605=79235.395[/tex]from the calculations above, his income for last month was $79235.395
Triangle JKL is rotated 270° counterclockwise about the origin to form triangle J'K'L'. What is the y-coordinate of point J'? y 6 5 4 K 3 2. 1 -6 -5 -4 -3 -2 1 2 3 4 0 -1 6 Jo -2 3 4 -5 L -6 Type the answer in the box.
Rotation of 270° ,counterclockwise
Find center point of triangle
JK = √ 9^2 +5^2
. = √ 106
JL = √9^2 + 3^2
. = √90
KL = 8
Then now find point J'
Distance from J to origin = JO = √4^2+2^2 =√20
Angle is arctan (2/4) = 26.56°
Then new coordinates J' are
Angle 270+26.56 = 296.56°
From origin trace a line perpendicular to JO
then
y = √20• Cos 26°
y = 4
Answer:
-4
Step-by-step explanation:
good luck kiddos :)
help please and thankyou Graph the line y = 2x + 3
the line will be:
we can find the method of finding the intercepts and then link them with a unique line:
the x-intercept is when y=0 then 2x+3=0 then x=-3/2=-1.5
the y-intercept is when x=0 then y=3
then we can link the points (-1.5,0) and (0,3) and we have the graph of the given line.
Write an inequality to match the statement.The difference of a number x and 7 is less than or equal to the sum of the same number x and 5
The difference of a number x and 7 is less than or equal to the sum of the same number x and 5
The difference means subtraction.
A number is x
Less than or equal is represented with the symbol ≤
So:
The difference of a number x and 7
x-7
is less than or equal to the sum of the same number x and 5
≤ x+5
Altogether:
x-7≤ x+5
determine whether the triangle with the given side lengths is a right triangle4 ,7, 11
For the a triangle to be a right angle triangle, the length of the three sides must form a pythagorean triple. This means that if we apply pythagoras theorem, the square of the length of the longest side must equal the sum of the squares of the length of the other sides. This means that
11^2 must be equal to 4^2 + 7^2
11^ = 121
4^2 + 7^2 = 16 + 49 = 65
Since they are not equal, then the triangle with the given sides, 4, 7, 11 cannot form a right angle triangle.
Matt drew the two rectangles shown in thediagram below.ABD16 in.ABD12 inсMatt dilated Rectangle ABCD to createRectangle A'B'CD'.What scale factor did Matt use todilate Rectangle ABCD?
Which lists all of the x-intercepts of the graphed function? A. (0,6)B. (1,0) and (2,0)C. (1,0), (2, 0), and (-3,0)D. (1,0), (2,0), (-3, 0), and (0, 6)
From the graph, the points at which the curve touch the x-axis is the intercepts on x
looking at the graph, the curve cut through the x axis at points -3, 1 and 2
hence the points will be
(-3, 0), (1, 0) and (2, 0)
Thus, the correct option is C;
(1, 0), (2, 0) and (-3, 0)
Hi, I need help with this problem please. No explanations/steps required. Just need the final answer
To simplify the expression we fist need to express each trigonometric function in terms of sines and cosines; to do this we need to remember that:
[tex]\begin{gathered} \sec x=\frac{1}{\cos x} \\ \csc x=\frac{1}{\sin x} \end{gathered}[/tex]With this in mind we have:
[tex]\begin{gathered} \frac{\csc x}{\sec x}=\frac{\frac{1}{\sin x}}{\frac{1}{\cos x}} \\ =\frac{\cos x}{\sin x} \end{gathered}[/tex]Finally we need to remember that:
[tex]\cot x=\frac{\cos x}{\sin x}[/tex]Therefore, we have that:
[tex]\frac{\csc x}{\sec x}=\cot x[/tex]find the volume for the spear shown to The Right. Use 3.14 for pi sphere is 5.8
Semicircle perimeter is 5.8 m
THEN
2π • R + 2R = 5.8
R = 5.8/(2π + 2) = 0.7 meters
Now find Volume
Volume V = (4/3)•π•R^3
V= (4/3)•π•(0.7^3) = 1.438 cubic meters
V=. 1.438 m^3
Is the ordered pair (3,-4) a solution to : 5x-2y=17
Given
[tex]5x-2y=17[/tex]Set x=3 and solve for y, as shown below
[tex]\begin{gathered} x=3 \\ \Rightarrow5*3-2y=17 \\ \Rightarrow15-2y=17 \\ \Rightarrow-2y=2 \\ \Rightarrow y=-1 \end{gathered}[/tex]Therefore, a solution to the equation is the ordered pair (3,-1).
(3,-4) cannot be a solution to the equation.Use the given information to create the equation for the rational function. The function is written in factored form to help you see how the given information shapes our equation. If the leading coefficient is not an integer enter the value as a fraction.Vertical asymptote at x=-1, double zero at x=2, y-intercept at (0,2).The numerator is: Answer (x-Answer )(x-Answer )The denominator is: (x+Answer )
Given:
• Vertical asymptote at : x = -1
,• Double zero at: x = 2
,• y-intercept at: (0, 2)
Let's create the equation for the rational function using the given properties.
Since the vertical asymptote is at x = -1, to find the deominator of the equation, equate the vertical asymptote to zero.
Add 1 to both sides:
[tex]\begin{gathered} x+1=-1+1 \\ x+1=0 \end{gathered}[/tex]Therefore, the denominator of the function is ==> x + 1
Since it has a double zero at x = 2, we have the factors:
[tex]\Longrightarrow(x-2)(x-2)[/tex]We now have the equation:
[tex]y=\frac{a(x-2)(x-2)}{x+1}[/tex]Also, the y-intercept is at: (0, 2)
To find the value o a, substitute 2 for y and 0 for x then evaluate:
[tex]\begin{gathered} 2=\frac{a(0-2)(0-2)}{0+1} \\ \\ 2=\frac{a(-2)(-2)}{1} \\ \\ 4a=2 \\ \\ a=\frac{2}{4} \\ \\ a=\frac{1}{2} \end{gathered}[/tex]Therefore, the rational function is:
[tex]y=\frac{\frac{1}{2}(x-2)(x-2)}{x+1}[/tex]ANSWER:
[tex]y=\frac{\frac{1}{2}(x-2)(x-2)}{x+1}[/tex][tex]\begin{gathered} \text{Numerator: }\frac{1}{2}(x-2)(x-2) \\ \\ \\ \text{Denominator: (x + 1)} \end{gathered}[/tex]What is the output value for the following function, f(x) = 5x - 2 if the input value is 3?options:1751131
Solution:
Given the function below
[tex]f(x)=5x-2[/tex]Where
[tex]\begin{gathered} x\text{ is the input value} \\ f(x)\text{ is the output value} \end{gathered}[/tex]If the input value is 3, i.e. x = 3, the output value will be
[tex]\begin{gathered} f(x)=5x-2 \\ f(3)=5(3)-2=15-2=13 \\ f(3)=13 \end{gathered}[/tex]Hence, the output value is 13
One month Susan rented 5 movies and 6 video games for a total of $57. The next month she rented 3 movies and 2 video games for a total of $25. Find the rental cost for each movie and each video game?
The rental cost of each movie is $4.50
The rental cost of each video game is $5.75
Explanation:let the cost of one movie = m
let the cost of one video games = v
1st month:
Number of movies = 5
number of video games = 6
Total cost of both = $57
Total cost = cost of one movie(Number of movies ) + cost of one video (Number of video games)
57 = m(5) + v(6)
57 = 5m + 6v ...equation 1
2nd month:
Number of movies = 3
number of video games = 2
Total cost of both = $25
Total cost = cost of one movie(Number of movies ) + cost of one video (Number of video games)
25 = m(3) + v(2)
25 = 3m + 2v ...equation 2
57 = 5m + 6v ...equation 1
25 = 3m + 2v ...equation 2
Using elimination method:
To eliminate v, we would multiply equation 2 by 3:
3(25) = 3(3m) + 3(2v) ...equation 2
75 = 9m + 6v ...equation 2
57 = 5m + 6v ...equation 1
subtract equation 1 from 2:
75 - 57 = 9m - 5m + 6v - 6v
18 = 4m + 0
18 = 4m
m = 18/4
m = 4.5
Substitute for m in any of the equation:
Using eqauation1: 57 = 5m + 6v
57 = 5(4.5) + 6v
57 = 22.5 + 6v
57 - 22.5 = 6v
34.5 = 6v
v = 34.5/6
v = 5.75
The rental cost of each movie is $4.50
The rental cost of each video game is $5.75
Not sure how to figure this out.Options:A. 49 square feet B. 7 square feet. C. 28 square feet. D.14 square feet
According to the information given in the exercise, the pen for his rabbit is square and each side is 7 feet long.
You know that the formula for calculating the area of a square is:
[tex]A=s^2[/tex]Where "s" is the length of each side of the square.
In this case:
[tex]s=7ft[/tex]Then, you need to substitute this length into the formula and then evaluate, in order to find the area. You get:
[tex]\begin{gathered} A=(7ft)^2 \\ A=49ft^2 \end{gathered}[/tex]Hence, the answer is: Option A.
Solve the quadratic equation by completing the square.4a²- 48a +52 = 0a= _,_
Given
[tex]4a^2-48a+52=0[/tex]Solution
[tex]\begin{gathered} 4a^2-48a+52=0 \\ So\text{ divide both sides by 4} \\ \frac{4a^2}{4}-\frac{48a}{4}+\frac{52}{4}=0 \\ which\text{ gives } \\ a^2-12a+13=0 \\ Keep\text{ a on LHS} \\ a^2-12a=-13 \\ Take\text{ the half of coefficient of a and square it} \\ (-\frac{12}{2})^2=36 \\ \\ rewrite\text{ as perfect square} \\ \\ (a-6)^2=-13+36 \\ (a-6)^2=23 \\ \end{gathered}[/tex][tex]\begin{gathered} a-6=\pm\sqrt{23} \\ a=6\pm\sqrt{23} \end{gathered}[/tex][tex]\begin{gathered} a=10.79583\text{ ,}a=1.20416 \\ or \\ a=6+\sqrt{23},\:a=6-\sqrt{23}\quad \\ \\ \end{gathered}[/tex]Find the mean, median, and mode for the data set. If there is no mode, write none. If there is more than one mode,write your solutions from least to greatest, separated by a comma.50,30,40,10,20,80,60,90,10,30,110, 70mean:median:mode:
Answer:
• Mean: 50
,• Median: 45
,• Mode: 10,30
Explanation:
Given the data set:
[tex]50,30,40,10,20,80,60,90,10,30,110,70[/tex]Before we begin, arrange the numbers from the least to the greatest.
[tex]10,10,20,30,30,40,50,60,70,80,90,110[/tex](a)Mean
To find the mean, add up the numbers and divide by the number of items (12 in this case).
[tex]\begin{gathered} Mean=\frac{10+10+20+30+30+40+50+60+70+80+90+110}{12} \\ =\frac{600}{12} \\ Mean=50 \end{gathered}[/tex]The mean of the dataset is 50.
(b)Median
The median is the number in the middle of the dataset when arranged in ascending order.
• There are two numbers in the middle: 40 and 50
,• Take the average to find the median.
[tex]Median=\frac{40+50}{2}=\frac{90}{2}=45[/tex]The median of the dataset is 45.
(c)Mode
The mode is/are the number(s) that appear the most number of times..
[tex]10,10,20,30,30,40,50,60,70,80,90,110[/tex]In the dataset:
• 10 appears twice
,• 30 appears twice
The modes of the dataset are 10 and 30.
What is the total percentage of college graduates who have found that their degree very helpful to develop specific skills and knowledge for the workplace?
From the graph given in the question, we can find out that
49% of the college graduates say that their college education was very useful for helping develop specific skills and knowledge for the workplace.
From the given question, the total percentage of college graduates who have found that their degree very helpful to develop specific skills and knowledge for the workplace is 49%
Complete the explanation of whether the graph represents a proportional oa neno relationship 5 5 5 relationship The graph represents a (select) (select) proportional nonproportional
We are given the graph of a line, and we are asked to determine if it is a proportional or non-proportional relationship. Let's remember the general form of the equation of a line, that is:
[tex]y=mx+b[/tex]where "m" is the slope and "b" the y-intercept. The y-intercept is the value where the line touches the y-axis. According to the graph, the value of "b" is b = 1, therefore, the equation of the line would be:
[tex]y=mx+1[/tex]A proportional relationship is of the form:
[tex]y=kx[/tex]since the value of "b" is different from zero the relationship is non-proporsional.
The relationship is non-proportional
1Convert without rounding:8into a decimal (no need to show work)Τ Τ Τ ΤParagraphVArialV3 (12pt)V
Given: the fraction 1/8.
Objective is to convert this fraction in to decimal.
[tex]\frac{1}{8}=0.125[/tex]Answer: 0.125