we use distributive property
[tex](4y\times3y^2)+(4y\times-2y)+(4y\times-7)+(11\times3y^2)+(11\times-2y)+(11\times-7)[/tex][tex]\begin{gathered} (12y^3)+(-8y^2)+(-28y)+(33y^2)+(-22y)+(-77) \\ 12y^3-8y^2+33y^2-28y-22y-77 \\ 12y^3+25y^2-50y-77 \end{gathered}[/tex]mond Baware Infinits Piscais Angles and Angle Measure Name 5.2
If we want to find the reference angle, you have to find the smallest possible angle formed by the x-axis and the terminal line, going either clockwise or counterclockwise.
In this case, the angle 290° is in the fourth quadrant, so the reference angle can be drawn and calculated as:
The reference angle can be calculated as:
[tex]360-290=70\degree[/tex]Answer: the reference angle for 290° is 70°.
If h(x)-(fog)(x) and h(x) = 4(x+1)*, find one possibility for 5 %) and g(x).f(x) = x +1O A.8(x) = 4x2O B. M(x)=(x+1)8(x)=4x2O c.f(x) = 4x2g(x) = x +1D.f(x) = 4x28(x)= (x+1)
It is given that h(x)=fog(x) and h(x)=4(x+1)^2.
So it follows:
[tex]\text{fog(x)}=4(x+1)^2[/tex]For option A, f(x)=x+1,g(x)=4x^2
So the value of fog(x) is given by:
[tex]f(g(x))=g(x)+1=4x^2+1[/tex]So A is incorrect.
For option B, f(x)=(x+1)^2,g(x)=4x^2
So the value of fog(x) is given by:
[tex]f(g(x))=g(x)+1=(g(x)+1)^2=(4x^2+1)^2[/tex]So B is incorrect.
For option C, f(x)=4x^2,g(x)=x+1
So the value of fog(x) is given by:
[tex]f(g(x))=4\lbrack g(x)\rbrack^2=4(x+1)^2[/tex]So C is correct.
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Answer:
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A projectile is launched from the ground at 128 feet per second and will hit the ground after a certain amount of time. It models the function g(x) = -16x2 + 128x, where x represents the time of the of the flight (in seconds) of the projectile. What is an appropriate domain for this model? 00 0
time can only take values greater than or equal to zero, so the domain starts on 0
the las point of the domain is when the projectile hit the ground, this means when g=0
now we will replace g=0 o find the last value of the domain
[tex]\begin{gathered} g=0 \\ -16x^2+128x=0 \end{gathered}[/tex]solve x to know the last time
[tex]\begin{gathered} 16x(-x+8)=0 \\ -x+8=\frac{0}{16x} \\ -x+8=0 \\ -x=-8 \\ x=8 \\ \end{gathered}[/tex]the last time in seconds of the model is 8
so the domain is
[tex]\lbrack0,8\rbrack[/tex]Translate the sentence into an equation Three times the sum of a number and 2 is equal to 9 Use the variable y for the unknown number
Three times
multiply a value by 3
[tex]3\times()[/tex]The sum of a number and 2
We named the number "Y", then inside the parenthesis be the sum of x and 2
[tex]3\times(y+2)[/tex]Is equal to 9
we equal the equation to 9
[tex]3\times(y+2)=9[/tex]1. [2/3 Points)DETAILSPREVIOUS ANSWERSSALGTRIG45.4.001.MY NOTESASK YOUR TEACHERFor a function to have an inverse, it must be one-to-oneTo define the Inverse sine function, we restrict the domainDof the sine function to the IntervalXNeed Help? Paad
In the case of sine function the domain of the function is given by the limits of the range of sine function.
The limits of the range of sine function are [-1,1].
Hence, the domain of the inverse sin function is [-1,1].
¿Cual es el resultado de efectuar (2x+5)³ + (x-2)(x-2)?
The equation (2x+5)³ + (x-2)(x-2) we get 8x³ + 60x² + 150x +129.
What is meant by binomial equation?A binomial number is an integer that can be produced by evaluating a homogeneous polynomial with two terms in mathematics, more specifically in number theory. It is a Cunningham number that has been generalized.
Let the equation be (2x + 5)² + (x -2)(x- 2)
By using binomial formula we get,
(a + b)³ = a³ + 3a²b + 3ab² + b³
The coefficient multipliers are located in row 3 of Pascal's triangle.
(2x + 5)³ + (x - 2)(x - 2)
= 8x³ + 60x² + 150x +125(x - 2)(x - 2)
8x³ + 60x² + 150x + 125 + x(x - 2) - 2(x-2)
8x³ + 60x² + 150x + 125 + x² - 2x - 2x +4
simplifying the above equation, we get
8x³ + 60x² + 150x + 125 + x² - 4x +4
8x³ + 60x² + 150x +129 + x² - 4x
8x³ + 60x² + 150x + 129 - 4x
8x³ + 60x² + 146x + 129
= 8x³ + 60x² + 150x +129
Therefore, by simplifying the equation (2x+5)³ + (x-2)(x-2) we get 8x³ + 60x² + 150x +129.
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Rewrite each equation in slope-intercept form, if necessary, then determine whether the lines are parallel , perpendicular, or neither.A.) y=2×+1B.)2x+y=7The slope line A is _The slope of line B is _Lines A and B are _
You have the following equation of two lines:
A) y = 2x + 1
B) 2x + y = 7
the general form of an equation of a line is given by:
y = mx + b
where m is the slope and b is the y-intercept.
The equation A is already written in the slope-intercep form. By comparing the equation with the general form you can notice that the slope is:
mA = 2
Next, you rewrite the equation B:
2x + y = 7 subtract 2x both sides
y = -2x + 7
by comparing with the general fom you have that the equation B has the following slope:
mB = -2
In order to determine if the lines are parallel,perpendicular, or neither, you calculate the quotien between the slopes of the lines.
mA/mB = 2/(-2) = -1
The quotient between the slopes is -1, this means that the lines are perpendicular
what is 24 as a fraction or mixed number
Answer: 24/1
Step-by-step explanation:
luann is playing a math game . she chose three cards - first card: -12 - second card: 3- thrid card: -5 what is the sum of the value ?
we sum the values
[tex]-12+3+(-5)=-12+3-5=-14[/tex]answer: -14
4) (-21)³ multiplying complex numbers
-9257
explanation..........................................................................................................
Plot the points. Then identify the polygon formed.a) A(4, 1), B(4, 6), C(-1, 6), D(-1, 1)b) A(2, -2), B(5, -2), C(7, -4), D(0, -4)
a)
The points for polygon a are shown below:
From the graph we notice that this is a square.
b)
The points for polygon b are shown below:
From the graph we conclude that this is a trapezoid.
500 books were sold the first day it went on sale. 150 books were sold each day after that. Write an equation to represent the total number of books sold. How many books were sold after 50 days?
Let x represent the number of days after release and y represent the number of books sold.
The first day there were 500 books sold, after that, 150 books were solf each passing day.
This means that for the first day y=500 ann each passing day 150 books were added, the equation is:
y=500+150x
Using this equation you have to calculate the number of books solf after x=50 days.
To do so replace in the equation above:
y=500+150*50
y=8000
After 50 days 8000 books were sold
ratio of number of boys to girl is 5 to 4 there are 60 girls in choir how many boys are there
Since the ratio of boys to girls is 5:4 and there are 60 girls, let b be the number of boys and g be the number of girls. Then:
[tex]\begin{gathered} \frac{b}{g}=\frac{5}{4} \\ \Rightarrow\frac{b}{60}=\frac{5}{4} \\ \Rightarrow b=\frac{5}{4}\times60 \\ \Rightarrow B=75 \end{gathered}[/tex]Therefore, there are 75 boys in the choir.
can you please help me
Given the equation:
[tex]\text{ 2x + 2y = -4}[/tex]Two Column Proof. If you could write it on a piece of paper and send a picture, that would be great.
Let's suppose
∠1 =∠5 and ∠2 = ∠4
then
180° - ∠1 - ∠2 = ∠3
180° - ∠1 - ∠2 =
For each coefficient choose whether it is positive or negative.Choose the coefficient with the least value.Choose the coefficient closest to zero.
From the given graph
a)
If the coefficient is positive, then the graph is upward
If the coefficient is negative, then the graph is downward
In the first 2 figures, the graphs are upward, then
A and B are positive
In the last 2 figures, the graphs are downward, then
C and D are negative
b)
The least coefficient is the coefficient of the graph which nearest to the x-axis
Since graph C is the nearest graph to the x-axis, but we have negative values, then
The least coefficient is the coefficient of the downward graph and farthest from the x-axis, then
D is the least coefficient
c)
The closest coefficient to zero is the graph C
Math answers and how you got the answer to solve
Hello there. To solve this question, we'll have to remember some properties about functions.
Given the functions:
[tex]\begin{gathered} f(x)=x^3 \\ g(x)=6x^2+11x-2 \end{gathered}[/tex]We have to determine:
[tex]\begin{gathered} (f+g)(x) \\ (f-g)(x) \\ (fg)(x) \\ (ff)(x) \\ \left(\frac{f}{g}\right)(x) \\ \left(\frac{g}{f}\right)(x) \end{gathered}[/tex]And their domain.
Let's do each separately:
(f + g)(x)
In this case, this function is the same as adding f(x) and g(x):
[tex](f+g)(x)=f(x)+g(x)=x^3+6x^2+11x-2[/tex]And as it is a polynomial function, it has no holes or asymptotes, therefore its domain is all the real line. We write:
(f - g)(x)
In the same sense, it is equal to the difference between f and g:
[tex](f-g)(x)=f(x)-g(x)=x^3-(6x^2+11x-2)=x^3-6x^2-11x+2[/tex]Again, as it is a polynomial function, its domain is all the real line, just as before.
(fg)(x)
In this case, it is the same as the product of f and g:
[tex](fg)(x)=f(x)\cdot g(x)=x^3\cdot(6x^2+11x-2)=6x^5+11x^4-2x^3[/tex]Once again, its domain is all the real line.
(ff)(x)
In this case, it is the product of f and itself:
[tex](ff)(x)=f(x)\cdot f(x)=x^3\cdot x^3=x^6[/tex]As before, its domain is entire real line.
(f/g)(x)
In this case, it is the quotient between f and g, respectively:
[tex]\mleft(\frac{f}{g}\mright)(x)=\frac{x^3}{6x^2+11x-2}[/tex]But in this case, its domain is not the entire real line. We have to get rid of the holes and vertical asymptotes of the function.
This function has no holes, since we cannot simplify any terms in the fraction, but it has at least two vertical asymptotes (that we'll find by taking the roots of the denominator).
In fact, the name vertical asymptote stands for the values of x in which the function would not exist (its limit goes to either infinity, -infinity or would not exist).
These roots are given by:
[tex]6x^2+11x-2=0[/tex]Using the quadratic formula, we get:
[tex]\begin{gathered} x=\frac{-11\pm\sqrt[]{11^2-4\cdot6\cdot(-2)}}{2\cdot6}=\frac{-11\pm\sqrt[]{121+48}}{12}=\frac{-11\pm\sqrt[]{169}}{12} \\ \Rightarrow x=\frac{-11\pm13}{12} \\ \Rightarrow x_1=\frac{-11+13}{12}=\frac{2}{12}=\frac{1}{6} \\ x_2=\frac{-11-13}{12}=\frac{-24}{12}=-2 \end{gathered}[/tex]The roots are 1/6 and -2. They are the vertical asymptotes of the function.
The domain of (f/g)(x) is then given by subtracting these values from the real line:
Or also in interval notation:
We do the same to (g/f)(x):
It is equal to the quotient between g and f, respectively, thus
[tex]\left(\frac{g}{f}\right)(x)=\frac{g(x)}{f(x)}=\frac{6x^2+11x-2}{x^3}[/tex]And again in this case, we have no holes, but we do have a vertical asymptote.
Taking the roots of the denominator:
[tex]x^3=0[/tex]The only solution to it is:
[tex]x=0[/tex]And the domain is then given by:
36 inches 23 inches is what fraction of a yard
If a yard contains 36 inches, then 21 inches is the 7/12 fraction of a yard
A yard contains 36 inches
We know
One yard = 36 inches
Here we have to use the unitary method for the conversion
The unitary method is the method of the value of a single unit.
36 inches = 1 yard
Then one inch = 1 / 36 yard
21 inches = (1 / 36) × 21
Multiply the terms
= 21 / 36
Divide both numerator and denominator by 3
= (21 / 3) / (36 / 3)
Divide the terms
= 7 /12 yard
Hence, if a yard contains 36 inches, then 21 inches is the 7/12 fraction of a yard
The complete question is
A yard contains 36 inches. 21 inches is what fraction of a yard ?
Learn more about unitary method here
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a and b are supplementary angles. if ma=(2x-24)°and mb=(5x-27)°, find the measure for b
Supplementary angles add up 180°.
So:
M
(2x-24) + (5x-27) = 180
Solve for x:
2x - 24 + 5x - 27 = 180
Combine like terms:
2x + 5x - 24 -27 = 180
7x -51 = 180
7x = 180 +51
7x = 231
x= 231/7
x= 33
Replace x on m
m< B = 5x-27 = 5 (33) - 27 = 165-27 = 138
m
∣+8∣−5=2Group of answer choicesv = -1 and v = -15v = -1 and v = -5No Solutionv = -15 and v = 15
Given:
[tex]|v+8|-5=2[/tex][tex]|v+8|=2+5[/tex][tex]|v+8|=7[/tex]case (1)
[tex]v+8=7[/tex][tex]v=7-8[/tex][tex]v=-1[/tex]Case (2)
[tex]-(v+8)=7[/tex][tex]-v-8=7[/tex][tex]v=-8-7[/tex][tex]v=-15[/tex]Therefore,
[tex]v=-1,-15[/tex]1st option is the correct answer.
Answer the statistical measures and create a box and whiskers plot for the following set of data.1, 1, 2, 2, 5, 6, 11, 11, 12, 13, 14, 16, 17, 19
DEFINITIONS
A boxplot is a way to show the spread and centers of a data set.
The box and whiskers chart shows you how your data is spread out. Five pieces of information (the “five-number summary“) are generally included in the chart:
1) The minimum (the smallest number in the data set). The minimum is shown at the far left of the chart, at the end of the left “whisker.”
2) First quartile, Q1, is the far left of the box (or the far right of the left whisker).
3) The median is shown as a line in the center of the box.
4) Third quartile, Q3, shown at the far right of the box (at the far left of the right whisker).
5) The maximum (the largest number in the data set), is shown at the far right of the box.
SOLUTION
From the data set given, we have the following information:
1) Minimum Value: 1
2) First Quartile: The position for the first quartile is given by the formula
[tex]\Rightarrow\frac{n+1}{4}[/tex]where n is the number of data.
In the problem, there are 14 data values. Therefore, the position is:
[tex]\Rightarrow\frac{14+1}{4}=3.75th\text{ position}[/tex]Using the 3.75th position, we have
[tex]\begin{gathered} 3rd\Rightarrow2 \\ 4th\Rightarrow2 \\ \therefore \\ Q1=2 \end{gathered}[/tex]3) Median: The median position is given by the formula
[tex]\Rightarrow\frac{n+1}{2}[/tex]Therefore, the median position will be:
[tex]\Rightarrow\frac{14+1}{2}=\frac{15}{2}=7.5th\text{ position}[/tex]The 7.5th position will give:
[tex]\begin{gathered} 7th\Rightarrow11 \\ 8th\Rightarrow11 \\ \therefore \\ Med=11 \end{gathered}[/tex]4) Third Quartile: The third quartile's position is gotten using the formula:
[tex]\Rightarrow\frac{3}{4}(n+1)_{}[/tex]Therefore, the Q3 position will be:
[tex]\Rightarrow\frac{3}{4}\times15=11.25th\text{ position}[/tex]Therefore, the 11.25th position will give:
[tex]\begin{gathered} 11th\Rightarrow14 \\ 12th\Rightarrow16 \\ \therefore \\ Q3=14(0.75)+16(0.25)=14.5 \end{gathered}[/tex]5) Maximum: 19
Therefore, the boxplot is shown below:
Find the circumference and area of each. Round for the nearest tenth:
1) a circle has a radius of 2 meters
2) a circle has a diameter of 16 cm
3) a circle has a radius of 8ft
4) a circle has a diameter of 11 cm
Answers:
1) C = 12.6 m
A = 12.6 m²
2) C= 50.3 cm
A = 201.1 cm²
3) C= 50.3 ft
A = 201.1 ft²
4) C= 34.6 cm
A = 95.0 cm²
Explanation:
The circumference and area of a circle with radius r can be calculated as:
[tex]\begin{gathered} \text{Circumference = 2}\pi r \\ Area\text{ = }\pi r^2 \end{gathered}[/tex]Where π is approximately 3.1416
Then, for each option, we get:
1) Replacing the radius by 2 m, we get:
[tex]\begin{gathered} \text{Circumference}=2(3.1416)(2m)=12.6m \\ \text{Area}=(3.1416)(2m)^2=(3.1416)(4m^2)=12.6m^2 \end{gathered}[/tex]2) If the diameter is 16 cm, the radius is 8 cm because the radius is half the diameter. So, replacing r by 8 cm, we get:
[tex]\begin{gathered} \text{Circumference = 2(3.1416)(8cm) = 50.3cm} \\ \text{Area = (3.1416)(8cm)}^2=(3.1416)(64cm^2)=201.1cm^2 \end{gathered}[/tex]3) Replacing r by 8 ft, we get:
[tex]\begin{gathered} \text{Circumference = 2(3.1416)(8ft) = 50.3ft} \\ \text{Area = (3.1416)(8ft)}^2=(3.1416)(64ft^2)=201.1ft^2 \end{gathered}[/tex]4) If the diameter is 11 cm, the radius is 11/2 = 5.5 cm, so:
[tex]\begin{gathered} \text{Circumference = 2(3.1416)(5.5cm) = 34.6 cm} \\ \text{Area = (3.1416)(5.5cm)}^2=(3.1416)(30.25cm^2)=95.0cm^2 \end{gathered}[/tex]
Solve the system algebraically 5 x - y = 0
Answer:
To solve the system of equations,
[tex]\begin{gathered} 5x-y=0 \\ \frac{y^2}{90}-\frac{x^2}{36}=1 \end{gathered}[/tex]Solving 1st equation we get,
[tex]y=5x[/tex][tex]\frac{y^2}{90}-\frac{x^2}{36}=1[/tex]Substitute y=5x in the above equation, we get
[tex]\frac{(5x)^2}{90}-\frac{x^2}{36}=1[/tex][tex]\frac{25x^2}{90}-\frac{x^2}{36}=1[/tex][tex]\frac{5x^2}{18}-\frac{x^2}{36}=1[/tex][tex]\frac{10x^2-x^2}{36}=1[/tex][tex]\frac{9x^2}{36}=1[/tex][tex]\frac{x^2}{4}=1[/tex][tex]x^2=4[/tex][tex]x=\pm2[/tex]when x=2, we get y=5x=5(2)=10
when x=-2, we get y=5x=5(-2)=-10
There are two solution for the given system.
[tex](2,10),(2,-10)[/tex]Answer is: x=2,y=10 and x=2,y=-10
2.) Which equation represents the balance scale shown? 3x = 7 X-3 = 7 x/3 = 7 x + 3 = 7
As you can see from the figure,
There are 7 dots on the right side of the scale.
On the left side of the scale, there are 3 dots + x
So, we write these numbers on the left and the right side of the equality sign.
[tex]x+3=7[/tex]Therefore, the equation x + 3 = 7 represents the balance scale shown in the figure.
Graph the parabola.Y = -2x^2 - 16x - 34Plat five points on a parable the vertex ,two points to the left of the vertex ,and two points to the right of the vertex . then click on the graph a function button.
The graph is shown below:
• The point (-4, -2) is the vertex
,• The points to the left are (-5, -4) and (-6, -10)
,• The points to the right are (-3, -4) and (-2, -10)
⦁ A vine called the mile-a-minute weed is known for growing at a very fast rate. It can grow up to 0.5 ft per day. How fast in inches per hour can the mile-a-minute weed grow up to? Show your work using the correct conversion factors.
Answer:
"0.5 ft * 12 = 6 inches (because there are 12 inches in 1 foot)1 day = 24 hours0.5 ft per day = 6 inches in 24 hoursInches grown in one hour = 6/24 = 0.25"
Step-by-step explanation:
For each of the following pairs of rational numbers, place a greater than symbol, >, a less than symbol, <, or an equality symbol, =, in the square to make the statement true.
I chow you how to solve for (d) and (i) and you could do the rest by yourself:
The best way to solve this operations is convert the numbers to a one form and then compare.
For (d)
[tex]\begin{gathered} \frac{7}{3}=\frac{6+1}{3}=\frac{6}{3}+\frac{1}{3}=2+\frac{1}{3}=2\frac{1}{3}=2.333 \\ \frac{13}{5}=\frac{10+3}{5}=\frac{10}{5}+\frac{3}{5}=2+\frac{3}{5}=2\frac{3}{5}=2.6 \\ So, \\ \frac{7}{3}<\frac{13}{5} \end{gathered}[/tex]Now for (i), take into account that this numbers are negative:
[tex]\begin{gathered} -11.5=-11.5\cdot\frac{4}{4}=-\frac{11.5\cdot4}{4}=-\frac{46}{4} \\ So,\text{ } \\ -\frac{46}{4}<-\frac{31}{4} \end{gathered}[/tex]Note that 46/4 is greater than 31/4, but -46/4 is lower than -31/4.
Also note that in this example I find to equalize the denominator of the numbers adn then you can compare the numerators.
The notation __1____ reads the probability of Event B given that Event A has occurred. If Events A and B are independent, then the probability of Event A occurring ___2___the probability of Event B occurring. Events A and B are independent if_3____1.A. P(AlB)B. P(BlA)C. P (A and B)2. A. Doesn't affectB. Affects3. A. P(BIA) = P(B)B. P(BIA) = P(A)C. P (BIA)= P(A and B).
P(B|A) (option B)
Doesn't affect (option A)
P(B|A) = P(B) (option A)
Explanation:1) Conditional probabilities could be in the form P(A|B) or P(B|A)
P(B|A) is a notation that reads the probability of event B given that event A has occurred.
P(B|A) (option B)
2) Independent events do not affect the outcome of each other
For event A and B to be independent, the probability of event A occurring doesn't affect the the probability of event B occurring
Doesn't affect (option A)
3) Events A and B are independent if the following are satisfied:
P(A|B) = P(A)
P(B|A) = P(B)
The ones that appeared in the option is P(B|A) = P(B) (option A)
2,047÷41=sloveadd expression
The given expression is,
[tex]\frac{2047}{41}[/tex]On solving, we have,
[tex]\frac{2047}{41}=49\frac{38}{41}=49.93[/tex]Thus, 2,047÷41=49.93.