From the Question given, we are able to write the following relationship:
[tex]\begin{gathered} a_1=2 \\ a_{n+1}=(a_n)²-4 \end{gathered}[/tex]By substituting the values 1, 2, and 3, we are able to calculate as follows:
[tex][/tex]the beginning of the question is "if the slope of..." please help
we know that
If two lines are parallel, then their slopes are equal
In this problem
mAB=mCD
substitute the given values
3/4=(x-2)/12
solve for x
multiply by 12 both sides
(x-2)=12(3/4)
x-2=9
x=9=9+2
x=11The perimeter of a triangle ABC is 100 cm.The length of AB is 45 cm and the length of BC is 32 cm.What is the length ofCA?
Recall that the perimeter of the triangle ABC is given by the following formula:
[tex]Perimeter=AB+BC+CA\text{.}[/tex]Substituting the given data we get:
[tex]100\operatorname{cm}=45\operatorname{cm}+32\operatorname{cm}+CA\text{.}[/tex]Solving the above equation for CA we get:
[tex]\begin{gathered} CA=100\operatorname{cm}-45\operatorname{cm}-32\operatorname{cm} \\ =23\operatorname{cm}\text{.} \end{gathered}[/tex]Answer: The length of CA is 23cm.
In Mr. Johnson’s third and fourth period classes, 30% of the students scored a 95% or higher on a quiz.Let be the total number of students in Mr. Johnson’s classes. Answer the following questions, and showyour work to support your answer.If 15 students scored a 95% or higher, write an equation involving that relates the number ofstudents who scored a 95% or higher to the total number of students in Mr. Johnson’s third andfourth period classes. Of the students who scored below 95% 40% of them are girls. How many boys scored below 95%?
Total number of students = scored below 95% + scored above 95% (I)
______________________________
Students scored below 95%
Scored below 95%* 0.40 = girls
Boys = total scored below 95% (100% - 40%)
Boys = total scored below 95% (60%)
Boys = total scored below 95% (0.6) (II)
__________________________________________
Can you see the updates?
_______________________________
30% of the students scored a 95% or higher on a quiz and 15 students scored a 95% or higher
Total number of students* 30 = 15
30%*n = 15
n= 15/ 0.3
n= 50
_____________________________
Replacing in (I)
Total number of students = scored below 95% + scored above 95%
50 = 15 + 35
Replacing in (II)
Boys = 15 (0.60) = 9
__________________________________________
Answer
9 of the students who scored below 95% are boys
Correctnomial function with the stated properties. Reduce all fractions to lowest terms.Third-degree, with zeros of - 3, - 1, and 2, and passes through the point (3, 5).
We must construct a polynomial with the following characteristics:
0. degree: 3,
,1. zeros: x₁ = -3, x₂ = -1 and x₃ = 2,
,2. passes through the point (3, 5).
The general form for this polynomial is:
[tex]p(x)=a*(x-x_1)(x-x_2)(x_{}_{}-x_3).[/tex]Where a is a constant factor and x₁, x₂ and x₃ are the zeros of the polynomial.
Replacing the values of the zeros, we have:
[tex]p(x)=a*(x+3)(x+1)(x-2).[/tex]Using the condition that the polynomial passes through (3, 5), we have:
[tex]y=a*(3+3)(3+1)(3-2)=a*24=5.[/tex]Solving for a, we get a = 5/24. Replacing this value in the equation above, we get:
[tex]p(x)=\frac{5}{24}(x+3)(x+1)(x-2).[/tex]Answer[tex]p(x)=\frac{5}{24}(x+3)(x+1)(x-2)[/tex]221/34 simplifyed........................................
Consider the given expression,
[tex]\frac{3}{x+1}-\frac{1}{x-1}-\frac{2x}{x^2-1}[/tex]Consider the algebraic identity,
[tex]a^2-b^2=(a+b)(a-b)[/tex]Simplify the expression as,
[tex]\begin{gathered} \frac{3}{x+1}-\frac{1}{x-1}-\frac{2x}{(x-1)(x+1)} \\ \frac{3(x-1)-1(x+1)-2x(1)}{(x-1)(x+1)} \\ \frac{3x-3-x-1-2x}{(x-1)(x+1)} \\ \frac{-4}{(x-1)(x+1)} \\ \frac{-4}{x^2-1} \end{gathered}[/tex]Thus, the given expression is in simplified form gives,
[tex]\frac{3}{x+1}-\frac{1}{x-1}-\frac{2x}{x^2-1}=\frac{-4}{x^2-1}[/tex]Use the GCF to factor this expression.40x + 24y - 56
The given expression is,
[tex]40x+24y-56[/tex]The factors of 40, 24 and 56 are,
[tex]\begin{gathered} 40\colon2,4,5,8,10 \\ 24\colon2,4,3,8 \\ 56\colon2,4,7,8 \end{gathered}[/tex]The greatest common factor is therefore, 8.
Therefore, the given expression can be written as,
[tex](8\times5)x+(8\times3)y-(8\times7)[/tex]Taking 8 as common, we have,
[tex]8(5x+3y-7)[/tex]Find the surface area of a cylinder with a base radius of 6 ft and a height of 9 ft.Use the value 3.14 for n, and do not do any rounding.Be sure to include the correct unit.
The surface area of the cylinder can be found below
[tex]\text{surface area=}2\pi r(r+h)[/tex]h = 9 ft
r = 6 ft
Therefore,
[tex]\begin{gathered} \text{surface area=2}\times3.14\times6(6+9) \\ \text{surface area=}37.68(15) \\ \text{surface area=}565.2ft^2 \end{gathered}[/tex]Identify the transformations for the function below. Check all that applyf (x) = 2(x – 3)^3 + 2DilationHorizontal ShiftVertical ShiftReflection
The given function is,
[tex]f(x)=2(x-3)^3+2[/tex]The parent function of the given function can be identified as,
[tex]f(x)=x^3[/tex]A transformed function can be represented as,
[tex]f(x)=a(bx-h)^3+k[/tex]If k is a positive or a negative number, then function is shifted k units vertically.
So, comparing the equations, we find that in the given function k=2.
Hence, the function is vertically shifted.
A function f(x) is shifted h units horizontally if h is a positive or a negative number.
So, in the given function h=3.
Hence, the function is horizontally shifted.
If |a| >1 or 0<|a|<1, the function f(x) is dilated vertically by a scale factor of a units and if a is a negative number , the function is also reflected across the x axis.
In the given function, a=2.
So, f(x) is dilated, but not reflected.
If |b| >1 or 0<|b|<1, the graph of function f(x) is dilated by a scale factor of b units horizontally and if b is a negative number, the function is also reflected across the y axis.
In the given function, b=1.
So, f(x) is not dilated or reflected.
Hence, f(x) has undergone the transformations:
Dilation
Horizontal Shift
Vertical Shift
Given Point A, what is the coordinate for A' after the following transformation has occurred?A (5,7)AC
Using the following given,
[tex]\begin{gathered} (x,y)\Rightarrow(x-5,-y+2) \\ (5,7) \end{gathered}[/tex]substitute the given coordinates to the new coordinates.
[tex](5,7)\Rightarrow(5-5,-7+2)[/tex]Simplify the coordinates.
[tex](5,7)\Rightarrow(0,-5)[/tex]Thus, the coordinates of A' is (0, -5).
Pleaseeeeee help
2x+7y=-5;(a,1)
Answer:
x= −5a/2 - 7y/2
I think. correct if wrong!!!
Alex surveyed 60 student about their Vera zoo animals and made the circle graph of the results shown below
we can use the cross multiplication
we know that the full angle of a circle is 360° so the total angle corresponds to 60 students
so, what is 72 degrees?
[tex]\begin{gathered} 360\longrightarrow60 \\ 72\longrightarrow x \end{gathered}[/tex]where x is the number of students than said giraffes
[tex]\begin{gathered} x=\frac{72\times60}{360} \\ \\ x=12 \end{gathered}[/tex]the students than said giraffes are 12
Find the magnitude of the vector (-4,-4).Write your answer in simplified radical form.030/0 (0,0)ХX6?
Given the vector < -4, 4 >
The magnitude of the vector =
[tex]\sqrt[]{(-4)^2+(4)^2}=\sqrt[]{2\cdot4^2}=4\sqrt[]{2}[/tex]so, the answer will be:
[tex]4\sqrt[]{2}[/tex]Same took out a loan for 6200 that charges an annual rate of 8.6% compounded quarterly. Answer each part. Sent picture
Given:
Sam took out a loan for 6200 that charges an annual rate of 8.6% compounded quarterly.
Required:
Find effective annual interest rate.
Explanation:
a).
We know compound interest formula
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]Now,
[tex]undefined[/tex]b).
We know the effective annual interest rate
[tex]EAR=(1+\frac{i}{m})^m-1[/tex]EAR = Effective annual interest
i = Annual nominal rate of interest
m = No. of compounding periods in a year.
[tex]\begin{gathered} EAR=(1+\frac{0.086}{4})^4-1 \\ EAR=0.088813 \\ \text{ To find percentage multiply by 100 } \\ =0.088813\times100 \\ =8.8813\% \end{gathered}[/tex]Answer:
answered the question.
What are the zeroes of f(x) = x^2 + 5x + 6? (4 points)A) x = -2, -3B) x = 2,3C) x= -2,3D) x = 2, -3
You have the following function:
[tex]f(x)=x^2+5x+6[/tex]in order to find the zeros of the previous function, use the quadratic formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]where a, b and c are the coefficients of the polynomial. In this case:
a = 1
b = 5
c = 6
replace the previous values of the parameters into the formula for x:
[tex]\begin{gathered} x=\frac{-5\pm\sqrt[]{5^2-4(1)(6)}}{2(1)} \\ x=\frac{-5\pm\sqrt[]{25-24}}{2}=\frac{-5\pm1}{2} \end{gathered}[/tex]hence the solution for x are:
x = (-5-1)/2 = -6/2 = -3
x = (-5+1)/2 = -4/2 = -2
A) x = -2 , -3
Find the equation of the line that has a slope of -2 and passes through point (-3 ,4)
Let's use the slope-point form to find the equation:
[tex]\begin{gathered} y-4=-2(x-(-3)) \\ \rightarrow y-4=-2(x+3) \\ \rightarrow y-4=-2x-6 \\ \rightarrow y=-2x-2 \\ \end{gathered}[/tex]Thereby, the equation of the line is:
[tex]y=-2x-2[/tex]Karmahhaze09 Can i have your number
Answer: irdk u yet so no
Step-by-step explanation:
Solve the equation for X. Round the answer to three decimal places. 4^x = 6
Answer:
c. x =1.293
Explanation:
To solve the expression, we will apply the properties of the logarithms, so
[tex]\begin{gathered} 4^x=6 \\ \log 4^x=\log 6 \\ x\log 4=\log 6 \\ x=\frac{\log 6}{\log 4} \\ x=1.293 \end{gathered}[/tex]Therefore, the value for x is
c. x =1.293
maurice read a research 10 pages that is 50 percent of the paper lenght what i the paper lenght
we know that
10 pages -------> represent 50%
so
Multiply by 2 both sides
20 pages --------> 100%
therefore
the paper length is 20 pagesApply proportionRemember that the paper length represent the 100%
10/50=x/100
solve for x
x=10*100/50
x-20 pages100/50x-20 pages⁰which of the following is the volume of a hemisphere with a radius of 8 inches?
The general expression for the volume of hemisphere is :
[tex]\text{ Volume = }\frac{2}{3}\Pi\text{ }\times radius^3\text{ }^{}^{}[/tex]In the given question we have radius = 8 inches
Substitute the value of r = 8 in the expression for the volume of hemisphere :
[tex]\begin{gathered} \text{ Volume = }\frac{2}{3}\Pi\text{ }\times radius^3\text{ }^{} \\ \text{Volume}=\frac{2}{3}\times3.14\times8^3 \\ \text{Volume =}1071.78666 \\ Volume\text{ = 1071.70 inches cubed} \end{gathered}[/tex]B) 1071.79 inches cube
Answer
Reason
A library has 144 books. A long
shelf can fit 100 books. A short shelf can fit 10 books.
The books that are left over can be put in a bin.
Draw two ways to sort books on a shelf.
100 books can be kept in the long shelf and the 10 books can be kept in the short shelf.
The remaining 34 books can be kept in the bin.
Given, a library has 144 books. A long shelf can fit 100 books. A short shelf can fit 10 books.
The books that are left over can be put in a bin.
Now, we have to find the way to sort the books on a shelf.
So, we can put the books in this fashion,
100 books can be kept in the long shelf and the 10 books can be kept in the short shelf.
The remaining 34 books can be kept in the bin.
Hence, 100 books can be kept in the long shelf and the 10 books can be kept in the short shelf.
The remaining 34 books can be kept in the bin.
Learn more about Unitary Method here https://brainly.com/question/28627316
#SPJ9
A study compared five different methods for teaching descriptive statistics. The five methods were traditional lecture and discussion, programmed textbook instruction, programmed text with lectures, computer instruction, and computer instruction with lectures. 45 students were randomly assigned, 9 to each method. After completing the course, students took a 1-hour exam.
a. What are the null and alternative hypotheses for addressing the research question, "are average test scores different between the different teaching methods?"
b. What are the degrees of freedom associated with the F distribution for evaluating these hypotheses?
c. Suppose the p-value for this test is 0.0168. What would you conclude? (Be sure to specify your significance level.)
a. The hypotheses are:
Null hypothesis: the average test scores are the same for the different teaching methods.
Alternative hypothesis: the average test scores are different for the different teaching methods.
b. To determine the degree of freedom for the F test: we must find two sources of variation such that we have two variances. The two sources of variation are: Factor (between groups) and the error (within groups) and add this up. Or use (N - 1). N is number in sample
c. With a p value of of 0.0168 and using a standard significance level of 0.05, we will reject the null hypothesis as 0.0168 is less than 0.05 and conclude that the average test scores are different for the different teaching methods.
Hence we get the required answer.
Learn more about Hypothesis here:
brainly.com/question/15980493
#SPJ1
pls help I've had a bad day and I've been trying to figure this out forever
the given expression is
4Ix-2I - 3
According to a pencil company's advertising campaign, 2 out of 3 students prefer the company's pencils to their competitor's pencils. A representative of the company went to a high school with 1479 students for alq & A session, and one of the students asked, "If this is true, in this high school, how many more students are there who prefer your pencils than students who prefer your competitor's pencils?" Help the company's representative come up with an answer. Assume that the company's claim is true.Part 1: How many of the students in the high school prefer the company's pencils? Part II: Out of 3 students, how many prefer the pencils of the company's competitor?Part III: How many of the students in the high school prefer the pencils of the company's competitor? Part IV: In the high school, how many more students are there who prefer the company's pencils than students who prefer their competitor's pencils?
Data:
2 of 3 students prefer the company's pencils to their competitor's pencils
high school with 1479 students
Part 1:You have 1479 students in total and know that 2 of 3 students prefer the company's pencils, then you multiply the toatl number of students by the factor 2/3
[tex]1479\cdot\frac{2}{3}=\frac{2958}{3}=986[/tex]Then, 986 students in the high school prefer the company's pencilsPart 2:Of 3 students 1 prefer the pencils of the company's competitorPart 3:Substract the number of students that prefer the company's pencils from the total of students:
[tex]1479-986=493[/tex]Then, 493 students in the high school prefer the pencils of the company's competitorPart 4:Substract the number of students who prefer the competitor's pencils fom the students that prefer the company's pencils:
[tex]986-493=493[/tex]Then, in the high school there are 493 students more that prefer the company's pencils than students who prefer their competitor's pencilsState if the triangles in each pair are similar. If so, state how you know they are similar andcomplete the similarity statement.1) 2)
Triangles Similarity
For two shapes to be similar, two conditions must be satisfied:
* They must have the same angles.
* The side lengths must be in proportion
Let's focus on the image provided in problem 1.
We must try to find if the length sides of ABU and VWU are in proportion.
To do it, we find the ratio of the sides. If we find the same ratio of two pairs of sides, then the second condition is met.
just need help and a simple way to solve this
ANSWER
The length of the third leg is
STEP-BY-STEP EXPLANATION:
The figure given is a right-angled triangle.
To find the third length of the triangle, we need to apply Pythagora's theorem
It states that
[tex]\begin{gathered} (Hypotenuse)^2=(opposite)^2+(adjacent)^2 \\ \end{gathered}[/tex]The third length of the triangle is the hypotenuse because it is the longest
[tex]\begin{gathered} (Hypotenuse)^2=4^2+2^2 \\ (Hypotenuse)^2\text{ = 16 + 4} \\ (Hypotenuse)^2\text{ = 20} \\ \text{ Take the squareroots of both sides} \\ \text{ }\sqrt[]{(Hypotenuse)^2\text{ }}\text{ = }\sqrt[]{20} \\ \text{Hypotenuse = }4.472 \\ \text{Hypotenuse }\approx\text{ 4.5} \end{gathered}[/tex]Hence, the length of the third leg is 4.5
A card is drawn randomly from a standard deck of cards. You win $5 if the card is adiamond or a king. What is the probability that you will win 5 dollars?
EXPLANATION
Let name event A as the event of drawing a Diamond and let us name event B as the event of drawing an King.
Now, we are required to find P(A union B)
We know that P(A union B) = P(A) + P(B) - P(A intersection B) … (i)
Let's suppose there are 52 cards in a standard deck.
A standard deck of cards has 13 diamonds and 4 kings
We have P(A) = 13/52 = 1/4, P(B) = 4/52 = 1/13
A intersection B denotes the case of the King of Diamonds whose probability = 1/52
Now plugging in these values to equation (i) and simplifying, we obtain the required probability as P(A union B) = 1/4 + 1/13 - 1/52 = 4/13
The probability is 4/13 or 0.307 or 30.7%
A student has 5 pairs of plants, 2 shirts and 7 necklaces. He chooses one shirt, one pair of pants, and one necklace. How many different outfits could he make?
Answer:
70
Explanation:
By the rule of multiplication, we can calculate the number of different outfits as
5 x 2 x 7 = 70
pants shirts necklaces
Therefore, there are 70 different outfits
The volume of a sphere is a function of it's radius, V=4/3* πr^3. evaluate the function for the volume of a volleyball with radius of 11.3 cm.Round to the nearest tenth.
Volume of sphere = 6040.9cm³
Explanation:
Volume of sphere = 4/3* πr³
radius = r = 11.3cm
if π = 3.14
[tex]\begin{gathered} \text{volume = }\frac{4}{3}\times3.14\times11.3^3 \\ \end{gathered}[/tex][tex]\begin{gathered} V\text{ = }\frac{4}{3\text{ }}\times3.14\times1442.897 \\ V=6040.929cm^3 \end{gathered}[/tex]Rounding to the nearest tenth:
Volume of sphere = 6040.9cm³
The function g(x) approaches positive infinity as x approaches positive infinity. The zeros of the function are -1,2 and 4. Which graph best represents g(x)?
Explanation
We are asked to select the correct option for which g(x) approaches positive infinity as x approaches positive infinity.
Also, the zeros of the function are -1,2 and 4.
The correct option will be
Can you help me with this assignment
Those are vertical angles, therefore:
[tex]\begin{gathered} m\angle ONB=m\angle MNK \\ so\colon \\ m\angle ONB=85 \end{gathered}[/tex]