I have a question about rational functions in precalculus trigonometry picture included
Answers
Given the following parameters
[tex]\begin{gathered} \text{Vertical asymptotes}\Rightarrow x=-2,x=1 \\ X-\text{intercepts}\Rightarrow x=-1,x=3 \\ \text{horizontal asymptote}\Rightarrow y=10 \end{gathered}[/tex]The vertical asymptotes indicates that denomenator will be (x+2)(x-1), i.e
[tex]\begin{gathered} x=-2 \\ x+2=0 \\ x=1 \\ x-1=0 \end{gathered}[/tex]The intercepts on the x axis indicates that the numerator will be ( x+ 1) (x-3)
i.e
[tex]\begin{gathered} x=-1\Rightarrow x+1=0 \\ x=3\Rightarrow x-3=0 \end{gathered}[/tex]The equation would be in this form
[tex]y=\frac{(x+1)(x-3)}{(x+2)(x-1)}[/tex]The equation above has a horizontal asymptote of y=1 as the degree of the numerator and denominator is equal , likewise their leading coefficient, i.e the ratio is 1:1
Thus, to make the equation have horizontal asymptote of y=10, multiply by 10/1
[tex]\begin{gathered} y=\frac{10}{1}\times\frac{(x+1)(x-3)}{(x+2)(x-1)_{}} \\ y=\frac{10(x+1)(x-3)}{(x+2)(x-1)} \end{gathered}[/tex]Hence, the rational function with the following parameters provided in the question is given by
[tex]y=\frac{10(x+1)(x-3)}{(x+2)(x-1)}[/tex]Related Questions
Use the table below to find each probability. The table gives information about students at one school.
Answers
Conditional Probabilities
The table gives us information about the favorite leisure activities of students at a school
We are required to find the probability of some events, given the occurrence of another event.
a) P(Sports | Female). It's the probability that a student likes sport if it's a female.
There are 334 female students, from which 39 like sports, thus:
[tex]P(S|F)=\frac{39}{334}=11.68\text{ \%}[/tex]b) P(Female | Sports). It's the probability that a student is known to like sports and it's also a female.
There are 106 students that like sports, from which 39 are female, thus:
[tex]P(F|S)=\frac{39}{106}=36.79\text{ \%}[/tex]c) P(Reading | Male). It's the probability that a male student likes reading.
There are 366 male students from which 76 like reading, thus:
[tex]P(R|M)=\frac{76}{366}=20.77\text{ \%}[/tex]d) P(Male | Reading). It's the probability that a student is known to like reading and it's also a male. There are 161 students who like reading out of which 76 are male, thus:also a male. d)
[tex]P(M|R)=\frac{76}{161}=47.20\text{ \%}[/tex]e) P(Hiking | Male). There are 366 male students out of which 58 like hiking, thus:H
[tex]P(H|M)=\frac{58}{366}=15.85\text{ \%}[/tex]f) P(Hiking | Female). There are 334 female students out of which 48 like hiking, thus:
[tex]P(H|F)=\frac{48}{334}=14.37\text{ \%}[/tex]g) P(Male | Shopping). There are 139 students who like shopping and from them, 68 are male, thus:39 students who like shopping and fr
[tex]P(M|S)=\frac{68}{139}=48.92\text{ \%}[/tex]h) P(Shopping | Female). There are 334 female students from which 71 like shopping, thus:
[tex]P(S|F)=\frac{71}{334}=21.26\text{ \%}[/tex]I) P(Phoning | Male). There are 366 male students from which 54 like phoning, thus:
[tex]P(P|M)=\frac{54}{366}=14.75\text{ \%}[/tex]ich 39 are female, thus:
S
im not smart at all and i need my grade up in math or im screwed if i can just get a direct answer thatd be poggers. anyways: chris is mailing his friend a poster that has been rolled up in a long tube. he has a box that measures 20 inches by 8 inches by 4 inches. what is the maximum length the rolled poster can be?
Answers
the maximum length the rolled poster can be is 21.9089 inches
(6,8) and (-12,-4) put in standard form
Answers
Problem
(6,8) and (-12,-4) put in standard form
Solution
The standard form for a line is given by:
y =mx +b
We can calculate the slope with this formula:
[tex]m=\frac{-4-8}{-12-6}=\frac{2}{3}[/tex]Then we can find the intercept using one point given for example x=6 and y=8 and we have:
8= 2/3*6 +b
b= 8- 4= 4
And our standard form would be given by:
y= 2/3 x +4
Find the surface area of each figure. Round answer to two decimal places. use 3.14 for pi
Answers
Solution
Step 1
Determine the number of shapes in figure 1
Figure 1 is made up of a cone and a hemisphere
Step 2
Write out the expression for the area of a cone and a hemisphere
[tex]\begin{gathered} \text{The area of a cone = }\pi\times r\times l \\ \text{The area of a hemisphere = 2}\times\pi\times r^2 \end{gathered}[/tex]Where
pi =3.14
r = 7 inches
l = 15 inches
Step 3
Substitute in the values and find the area of the shape
[tex]\begin{gathered} \text{Total area of the shape is given as} \\ \text{area of cone + area }of\text{ hemisphere} \\ =\text{ 2 x 3.14 }\times7^2+(3.14\text{ }\times7\times15) \\ =\text{ 307.72 + 329.7} \\ =637.42inch^2\text{ to 2 decimal places} \end{gathered}[/tex]Area = 637.42 square inches
Step 4
Determine the number of shapes in figure 2
Figure 2 is made up of a cone and a cylinder
Step 5
Write an expression for the area of a cylinder
[tex]\text{The area of a cylinder = 2}\times\pi\times r\times h\text{ + }\pi\times r^2[/tex]where h = 13yards
radius(r) = 11/2 = 5.5inches
l = ?
To find l, the slant height we use the Pythagoras theorem
so that
[tex]\begin{gathered} l^2=11^2+9^2 \\ l^2=202 \\ l\text{ =}\sqrt[]{202} \\ l\text{ = 14.2126704 in} \end{gathered}[/tex]Step 6
Substitute in the values and find the area of the shape
[tex]\begin{gathered} \text{Area of figure 2 is given as} \\ \pi\times r\times l\text{ + (2}\times\pi\times r\times h\text{ + }\pi\times r^2) \\ (3.14\text{ }\times\text{ 5.5}\times14.2126704)+((2\times3.14\times5.5\times13)+\text{ (3.14 }\times5.5^2) \end{gathered}[/tex][tex]\begin{gathered} 245.4528179+94.985+449.02=789.4578179Inches^2 \\ To\text{ 2 decimal places }\approx789.46inches^2 \end{gathered}[/tex]Area = 789.46 square inches
Alex jogs 1 miles east, 5 miles north, and then 10 miles west.How far is Alex from his starting position, to the nearest tenth of a miles.
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1 mils East + 5 miles North, and + 10 miles West.
It would be the same if I say the position of X = -9 and Y = 5
From Pythagorean theorem: a² + b² = c²
And in this case, c is the distance we want to calculate, a=X and b=Y
(-9)² + 5² = c²
81 + 25 = c²
106 = c²
c = √106 ≅ 10,3 miles
please help me please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please help me
Answers
a
Explanation
Step 1
remember some laws of exponents
[tex]\begin{gathered} a^m\cdot a^n=a^{m+n} \\ (a^m)^n=a^{m\cdot n} \\ a^{-m}=\frac{1}{a^m} \end{gathered}[/tex]then
[tex]2^{-5}\cdot2^8=2^{-5+8}=2^3=8[/tex]now, check the answer options
[tex]\begin{gathered} a)\frac{2^2}{2^{-1}}=2^2\cdot\frac{1}{2^{-1}}=4\cdot2=8 \\ b)(2^3)^{-1}=2^{3\cdot-1}=2^{-3}=\frac{1}{2^3}=\frac{1}{8} \\ c)\frac{2^{-2}}{2^{-1}}=\frac{2^1}{2^2}=\frac{2}{4}=\frac{1}{2} \\ d)(2^{-1})^3=2^{1-\cdot3}=2^{-3}=\frac{1}{2^3}=\frac{1}{8} \end{gathered}[/tex]so, the answer is a
Can you please help with questions 1&2. As they go together. Directions are with the pic below.
Answers
Given: Three balances A, B, and C
To Determine: Which block will balance C
Solution
Let a be number of sphere, b be the number of cylinder, and c the number of cube
So
[tex]\begin{gathered} BalanceA:2a+b+c=a+3b \\ BalanceB:3b=2c \\ BalanceC:a+2c=b+c \end{gathered}[/tex][tex]\begin{gathered} 2a+b+c=a+3b \\ 2a-a+b+c=3b \\ a+b+c=3b \\ Note \\ 3b=2c \\ Therefore \\ a+b+c=2c \\ 2c=a+b+c===compare\text{ with Balance C} \\ a+2c=b+c \\ a+2c=a+a+b+c \\ a+2c=2a+b+c \end{gathered}[/tex]Hence, 2 spheres will balance C
I figured this out by what will balance one of side of by combining balance A and balance B
Find the value of x.
44 °
Х
x = [? ]°
Answers
Given the figure of a circle:
As shown, there is a cyclic quadrilateral
Every opposite angle are supplementary angles
So, the sum of the angles x and 44 is 180
So,
[tex]\begin{gathered} x+44=180 \\ x=180-44 \\ \\ x=136 \end{gathered}[/tex]So, the answer will be x = 136
what is the greatest prime factor of 24 and 32 . the answer for the three same question is
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we know that
24=(2^3)(3)
32=2^5
Identify the equation of the circle that has its center at (-16, 30) and passes through the origin.A. (x−16)^2+(y+30)^2=34B. (x−16)^2+(y+30)^2=1156C. (x+16)^2+(y−30)^2=1156D. (x+16)^2+(y−30)^2=34
Answers
In order to determine the equation of the given circle, take into account the general formula for the equation of a circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]where the point (h,k) is the center of the circle.
You have tha the center of the circle is (-16,30). It means that:
h = -16
k = 30
then, the left side of the equation becomes:
[tex]\begin{gathered} (x-(-16))^2+(y-30)^2 \\ =(x+16)^2+(y-30)^2 \end{gathered}[/tex]next, consider that if the point passestroug the origin, it is necessary that for
x = 0 and y = 0 the equation coincides with one of the given options.
Replace x=0 and y=0 in the following expression:
[tex](0+16)^2+(0-30)^2=1156[/tex]Hence, it is necessary taht r^2 = 1156
The requierd equation of the circle is:
[tex](x+16)^2+(y-30)^2=1156[/tex]
What is the magnitude of the vector a = (-3,5)?
Answers
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
a = ( -3 , 5)
Step 02:
magnitude of the vector:
[tex]|a|\text{ = }\sqrt[]{x^2+y^2}=\sqrt[]{(-3)^2+5^2}=\sqrt[]{9+25}=\sqrt[]{34}[/tex]|a| = 5.83
The answer is:
|a| = 5.83
Which step contains and error and how would you correct that error? PLEASE HURRY FAST I WILL MARK BRAINIEST
Answers
1. Remove parenthesis: Distributive property (a(b+c)=ab+ac)
[tex]x-5x-5=3x+2[/tex]The error is in the line 1.2. Combine like terms:
[tex]-4x-5=3x+2[/tex]3. Substract 3x in both sides of the equation:
[tex]\begin{gathered} -4x-3x-5=3x-3x+2 \\ \\ -7x-5=2 \end{gathered}[/tex]4. Add 5 in both sides of the equation:
[tex]\begin{gathered} -7x-5+5=2+5 \\ \\ \\ -7x=7 \end{gathered}[/tex]5. Divide into -7 both sides of the equation:
[tex]\begin{gathered} \frac{-7}{-7}x=\frac{7}{-7} \\ \\ x=-1 \end{gathered}[/tex][tex] \frac{x - 1}{(x + 2)^{2} }[/tex]write the partial fraction decomposition.
Answers
Explanation
We are given the following expression:
[tex]\frac{x-1}{(x+2)^2}[/tex]We are required to determine the partial fraction decomposition of the given expression.
This is achieved thus:
We know that the partial fraction form of repeated roots is given as:
[tex]\frac{f(x)}{(x+a)^2}=\frac{A}{x+a}+\frac{B}{(x+a)^2}[/tex]Therefore, we have:
[tex]\frac{x-1}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}[/tex]Next, we take the LCD and simplify as follows:
[tex]\begin{gathered} \frac{x-1}{(x+2)^{2}}=\frac{A}{x+2}+\frac{B}{(x+2)^{2}} \\ \frac{x-1}{(x+2)^{2}}=\frac{A(x+2)+B}{(x+2)^2} \\ \Rightarrow x-1=A(x+2)+B\text{ ----- \lparen equation 1\rparen} \end{gathered}[/tex]Next, we determine the values of A and B as follows:
[tex]\begin{gathered} x-1=A(x+2)+B \\ \text{ Let x = -2} \\ -2-1=A(-2+2)+B \\ -3=B \\ \therefore B=-3 \\ \\ From\text{ }x-1=A(x+2)+B \\ \text{ Let x = 0} \\ 0-1=A(0+2)+B \\ -1=2A+B \\ \text{ Substitute for B} \\ -1=2A-3 \\ 2A=2 \\ A=1 \end{gathered}[/tex]Therefore, the partial fraction becomes:
[tex]\begin{gathered} \frac{x-1}{(x+2)^2}=\frac{1}{x+2}+\frac{-3}{(x+2)^2} \\ \Rightarrow\frac{x-1}{(x+2)^2}=\frac{1}{x+2}-\frac{3}{(x+2)^2} \end{gathered}[/tex]Hence, the answer is:
[tex]\begin{equation*} \frac{1}{x+2}-\frac{3}{(x+2)^2} \end{equation*}[/tex]Previous1Which expression is equivalent to 47 x 4-57412AB42C) 4-2D 4-35
Answers
Solution
We have the following expression
[tex]4^7\cdot4^{-5}=4^{7-5}=4^2[/tex]then the correct answer is:
B
For the triangle shown in the picture two angles are given find the third angle without using a protractor.
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Let's use x for the missing angle. The sum of the internal angles of a triangle is equal to 180°. Then we have:
[tex]\begin{gathered} 180^{\circ}=x+29^{\circ}+109^{\circ}=x+138^{\circ} \\ 180^{\circ}=x+138^{\circ} \end{gathered}[/tex]Then if we substract 138° from both sides we get:
[tex]\begin{gathered} 180^{\circ}-138^{\circ}^{}=x+138^{\circ}-138^{\circ} \\ x=42^{\circ} \end{gathered}[/tex]Then the answer is 42°.
Reduce the ratio to its lowest form. 63:9
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Consider function fand its graph.fte) =gsin(2)What is the amplitude of this function?
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1) Considering that the general formula for a sinusoidal function is:
[tex]y=A\sin(Bx+C)+D[/tex]2) And that the amplitude is halfway between the highest and the lowest point of that curve.
3) We can tell that the amplitude is:
[tex]\frac{1}{2}[/tex]Solve the linear equation using equivalent equations to isolate the variable. Write your solution as aninteger, as a simplified fraction, or as a decimal number.13x + 5x++ 5 - -2AnswerKeypadKeyboard ShortcutsX=
Answers
The equation is given to be:
[tex]3x+5x+\frac{1}{6}=\frac{1}{2}[/tex]Step 1: Multiply each term by 6 to eliminate the fractions
[tex]\begin{gathered} 3x(6)+5x(6)+\frac{1}{6}(6)=\frac{1}{2}(6) \\ 18x+30x+1=3 \end{gathered}[/tex]Step 2: Add the common terms
[tex]48x+1=3[/tex]Step 3: Subtract 1 from both sides
[tex]\begin{gathered} 48x+1-1=3-1 \\ 48x=2 \end{gathered}[/tex]Step 4: Divide both sides of the equation by 48
[tex]\begin{gathered} \frac{48x}{48}=\frac{2}{48} \\ x=\frac{2}{48} \end{gathered}[/tex]Step 5: Reduce the fraction by dividing through by 2
[tex]\begin{gathered} x=\frac{2\div2}{48\div2} \\ x=\frac{1}{24} \end{gathered}[/tex]ANSWER
The solution is:
[tex]x=\frac{1}{24}[/tex]The question and the triangle are in the image.For part B I just forgot the formula I use to find the length of the segments. If you give me the formula that would be awesome so I can do it by myself. But part C I'll need help with
Answers
Solution:
Given:
Two transversals with four line segments.
[tex]AC,CE,BD,DF[/tex]Part A:
For the line segment AC, the length is the distance between points A and C.
[tex]\begin{gathered} A=(5,7) \\ C=(6,4) \\ \text{where;} \\ x_1=5,y_1=7 \\ x_2=6,y_2=4 \\ \\ \text{The distance betwe}en\text{ two points is given by;} \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ \text{Hence,} \\ d=\sqrt[]{(6-5)^2+(4-7)^2} \\ d=\sqrt[]{1^2+(-3)^2} \\ d=\sqrt[]{1+9} \\ d=\sqrt[]{10} \\ To\text{ the nearest hundredth,} \\ d_{AC}\approx3.16 \end{gathered}[/tex]For the line segment CE, the length is the distance between points C and E.
[tex]\begin{gathered} C=(6,4) \\ E=(7,1) \\ \text{where;} \\ x_1=6,y_1=4 \\ x_2=7,y_2=1 \\ \\ \text{The distance betwe}en\text{ two points is given by;} \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ \text{Hence,} \\ d=\sqrt[]{(7-6)^2+(1-4)^2} \\ d=\sqrt[]{1^2+(-3)^2} \\ d=\sqrt[]{1+9} \\ d=\sqrt[]{10} \\ To\text{ the nearest hundredth,} \\ d_{CE}\approx3.16 \end{gathered}[/tex]For the line segment BD, the length is the distance between points B and D.
[tex]\begin{gathered} B=(17,7) \\ D=(16,4) \\ \text{where;} \\ x_1=17,y_1=7 \\ x_2=16,y_2=4 \\ \\ \text{The distance betwe}en\text{ two points is given by;} \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ \text{Hence,} \\ d=\sqrt[]{(16-17)^2+(4-7)^2} \\ d=\sqrt[]{(-1)^2+(-3)^2} \\ d=\sqrt[]{1+9} \\ d=\sqrt[]{10} \\ To\text{ the nearest hundredth,} \\ d_{BD}\approx3.16 \end{gathered}[/tex]For the line segment DF, the length is the distance between points D and F.
[tex]\begin{gathered} D=(16,4) \\ F=(15,1) \\ \text{where;} \\ x_1=16,y_1=4 \\ x_2=15,y_2=1 \\ \\ \text{The distance betwe}en\text{ two points is given by;} \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ \text{Hence,} \\ d=\sqrt[]{(15-16)^2+(1-4)^2} \\ d=\sqrt[]{(-1)^2+(-3)^2} \\ d=\sqrt[]{1+9} \\ d_{}=\sqrt[]{10} \\ To\text{ the nearest hundredth,} \\ d_{DF}\approx3.16 \end{gathered}[/tex]Part B:
On the first transversal, the ratio of the lengths of the line segments formed on it is;
[tex]\begin{gathered} \frac{AC}{CE}=\frac{3.16}{3.16} \\ \\ \text{Hence, the ratio is;} \\ AC\colon CE=1\colon1 \end{gathered}[/tex]On the second transversal, the ratio of the lengths of the line segments formed on it is;
[tex]\begin{gathered} \frac{BD}{DF}=\frac{3.16}{3.16} \\ \\ \text{Hence, the ratio is;} \\ BD\colon DF=1\colon1 \end{gathered}[/tex]From the ratio of the lengths of each transversal, it is noticed that they are the same.
Choose the correct reasons for the given statements that are already there
Answers
Definition of a Parallelogram
1) Let's fill in the 2nd row of that table since we know that
Statement Reason
1. ABCD is a parallelogram Given
2. AB ≅ CE Given
3. CD ≅ AB Definition of a Parallelogram
Note that a parallelogram has 2 pairs of congruent and parallel line segments.
2) Thus, the answer is:
"Definition of a Parallelogram".
Diana collected 5600 milliliters of rainwater on Saturday day. She collected 3.5 litter of rainwater on Sunday. How many total of milliliters did Daina collected on Saturday and Sunday?A. 910B. 4,060C. 4,600D. 9,100
Answers
Answer
D. 9,100
Explanation
What is given:
Diana collected 5600 milliliters of water on Saturday.
Diana collected 3.5 liters of water on Sunday.
What to find:
The total amount of water in milliliters Diana collected on Saturday and Sunday.
Step-by-step solution:
Step 1: Convert 3.5 liters of water collected on Sunday to milliliters.
Conversion factor: 1000 milliliters = 1 liter
Therefore, 3.5 liters = (3.5 liters/1 liter) x 1000 milliliters = 3500 milliliters
So 3500 milliliters of water were collected by Diana on Sunday
Step 2: Add the milliliters of water collected on Saturday and Sunday together.
Total milliliters collected = Milliliters of water collected on Saturday + Milliliters of water collected on Sunday
Total milliliters collected = 5600 milliliters + 3500 milliliters
Total milliliters collected = 9100 milliliters
Hence, the total milliliters of rainwater collected on Saturday and Sunday is 9100.
The correct answer is option D. 9,100
What is an equation of the line that passes through the point (6,3)(6,3) and is parallel to the line x+3y=24x+3y=24?
Answers
Hello there. To solve this question, we'll have to remember some properties about lines and properties of parallel lines.
We want to determine the equation of a line that passes through the point (6, 3) and is parallel to the line x + 3y = 24.
First, we'll rewrite the equation of this line, that was given in general form, into slope-intercept form:
For this, simply solve the equation for y
Subtract x on both sides of the equation
[tex]3y=24-x[/tex]Divide both sides of the equation by a factor of 3
[tex]y=-\dfrac{1}{3}x+8[/tex]We need this because in this form it is easier to find the slope of this line:
[tex]y=mx+b[/tex]So we find that
[tex]m=-\dfrac{1}{3}[/tex]Is the slope of this line.
A line that is parallel to this has the same slope, such that we can use the following equation to find the answer:
[tex]y=y_0+m(x-x_0)[/tex]Whereas (x0, y0) is the point the line passes through and m is the slope.
Plugging (x0, y0) = (6, 3) and m = -1/3 as we found, we get
[tex]y=3-\dfrac{1}{3}\cdot(x-6)[/tex]Multiply both sides of the equation by 3
[tex]\begin{gathered} 3y=9-(x-6) \\ \\ 3y=9-x+6 \\ \\ 3y=15-x \\ \end{gathered}[/tex]Add x on both sides of the equation
[tex]x+3y=15[/tex]This is the equation of the line passing through the desired point and is parallel to the line we had.
Question 81 pt 91 0 DetailsA die is rolled twice. What is the probability that a(n) 1 is rolled on the first roll and an even number on thesecond roll?The probability of rolling a(n) 1 on the first roll and an even number on the second roll is
Answers
Given:
A die is rolled twice.
To find:
What is the probability that 1 is rolled on the first roll and an even number on the second roll?
Explanation:
[tex]Probability\text{ of event= }\frac{favorable\text{ outcome}}{Total\text{ outcome}}[/tex]Solution:
A die is roll twice.
Each roll of the dice is an independent event. The probability of a 1 on one roll is 1/6 and the probability of an even number on the second roll is 3/6=1/2. Since, each roll is an independent event, the probability of a 1 on one roll and an even number on the second roll is
[tex]\frac{1}{6}\times\frac{1}{2}=\frac{1}{12}[/tex]Hence, these are the required probabilities.
are students desks a function of the classroom
Answers
Yes, because depending on the size of the classroom there may be more or less number of student's desks. In other words, the number of student's desks is a function of the number of m² in the classroom.
Place the steps for finding f^-1(x) in the correct order
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We are given the following function:
[tex]f(x)=\sqrt[]{7x-21}[/tex]We are asked to determine its inverse function:
[tex]f^{-1}(x)[/tex]To do that, we will do the following change:
[tex]y=f(x)[/tex]We get:
[tex]y=\sqrt[]{7x-21}[/tex]Now, since we want to determine the inverse we will swap variables, like this:
[tex]x=\sqrt[]{7y-21}[/tex]Now we will solve for "y", first by squaring both sides:
[tex]x^2=7y-21[/tex]Now we will add 21 to both sides:
[tex]x^2+21=7y[/tex]Now we will divide both sides by 7:
[tex]\frac{1}{7}x^2+3=y[/tex]This is the inverse function, therefore, we can do the following change:
[tex]y=f^{-1}(x)[/tex]We get:
[tex]\frac{1}{7}x^2+3=f^{-1}(x)[/tex]Evaluate: can you please help me with these and explain
Answers
Given:
a)
[tex](3^{-3})(3^{-2})[/tex]b)
[tex](5^{-6})(5^{-3})[/tex]Required:
We need to eveluate both
Explanation:
a)
[tex]\frac{1}{3^3}*\frac{1}{3^2}=\frac{1}{243}[/tex]b)
[tex]\frac{1}{5^6}*\frac{1}{5^3}=\frac{1}{5^9}[/tex]Final answer:
[tex]\begin{gathered} \frac{1}{3^5}=\frac{1}{243} \\ \\ \frac{1}{5^9} \end{gathered}[/tex]Select the property of equality that is used to generate each set of equations. Division Multiplication Addition Subtraction 4m = 60 4m + 2 = 60 = 2 7b + 9 = 22 7b +9-6 = 22-6 DO 9x - 4 = 33 9x - 4 + 3 = 33 + 3 у 3 у x 8 = 21 x 8 3
Answers
the given expression is,
4m = 60
4m ÷ 2 = 60 ÷ 2 ...........(Division)
7b + 9 = 22
7b + 9 - 6 = 22 - 6 ......... (subtraction)
9x - 4 = 33
9x - 4 + 3 = 33 + 3 ..............( Addition)
y/3 = 21
(y/3) x 8 = 21 x 8 ...............(multiplication)
PQ is tangent to •C at P. If PQ = 5 and CQ = 6, find CP and m
Answers
Answer:
[tex]\begin{gathered} CP=\sqrt{11} \\ m\operatorname{\angle}C=56.44 \end{gathered}[/tex]Explanation:
Step 1. The information that we have is that
• PQ=5
,• CQ=6,
and that PQ is tangent to circle C.
Since PQ is a tangent line, it forms a 90° angle with the circumference, and the triangle is a right triangle.
We need to find CP and the measure of angle C (m
Step 2. To find CP we use the Pythagorean theorem:
In this case:
[tex](CQ)^2=(CP)^2+(PQ)^2[/tex]Substituting the known values:
[tex]6^2=(CP)^2+5^2[/tex]Solving for CP:
[tex]\begin{gathered} 6^2-5^2=(CP)^2 \\ 36-25=(CP)^2 \\ 11=(CP)^2 \\ \sqrt{11}=CP \end{gathered}[/tex]The value of CP is:
[tex]\boxed{CP=\sqrt{11}}[/tex]Step 3. To find the measure of angle C, we use the trigonometric function sine:
[tex]sinC=\frac{opposite\text{ side}}{hypotenuse}[/tex]The opposite side to angle C is 5 and the hypotenuse is 6:
[tex]sinC=\frac{5}{6}[/tex]Solving for C:
[tex]C=sin^{-1}(\frac{5}{6})[/tex]Solving the operations:
[tex]\begin{gathered} C=s\imaginaryI n^{-1}(0.83333) \\ C=56.44 \\ \downarrow \\ \boxed{m\operatorname{\angle}C=56.44} \end{gathered}[/tex]Answer:
[tex]\begin{gathered} CP=\sqrt{11} \\ m\operatorname{\angle}C=56.44 \end{gathered}[/tex]Find the annual rate of interest.Principal = 4600 rupeesPeriod5 yearsTotal amount =6440 rupeesAnnual rate of interest =%Stuck? Review related articles/videos or use a hint.
Answers
The compound interest is given, in general, by the next formula:
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]Where A is the amount, P is the initial amount, r is the rate of interest, n is the number of times the interest rate is applied in a period, and t is the numbers of periods.
Then, in our problem:
[tex]A=6440,P=4600,t=5,n=1[/tex]Solving the formula for r:
[tex]\begin{gathered} n=1 \\ \Rightarrow A=P(1+r)^t \\ \Rightarrow(1+r)^t=\frac{A}{P} \\ \Rightarrow r=-1+\sqrt[t]{\frac{A}{P}} \end{gathered}[/tex]Then, using the numerical values above:
[tex]r=-1+\sqrt[5]{\frac{6440}{4600}}=-1+\sqrt[5]{\frac{7}{5}}\approx-1+1.0696\approx0.07[/tex]Then, the rate of interest is approximately equal to 7%
Use the given information to find the unknown value:y varies jointly as x, z, and w. When x = 2, z = 6, and w = 6. then y = 144. Find y when x = 6, z = 1, and w = 1.y =
Answers
Joint variation:
[tex]y=k\cdot x\cdot z\cdot w[/tex]Procedure:
0. We have to calculate the value of ,k ,with the first given values.
[tex]144=k\cdot2\cdot6\cdot6[/tex]2. Simplifying:
[tex]144=k\cdot2\cdot36[/tex][tex]144=k\cdot72[/tex][tex]k=\frac{144}{72}[/tex][tex]k=2[/tex]3. Finally, we find y using the second values given now that we now k.
[tex]y=2\cdot6\cdot1\cdot1[/tex]Answer: y = 12