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SOLUTION
The questions is outside scope
Related Questions
You go to a candy store and want to buy a chocolate
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To find the amount of servings, we just need to divide the entire bar weight by the serving weight. Solving this calculation, we have
[tex]\frac{14.8}{2.4}=\frac{37}{6}=6.166666666..\text{.}[/tex]the length of a screwdriver is 0.75 cm is how many screws can be placed to the end to make a road that's 18 cm long show yours
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Length of screwdriver = 0.75
Length of road = 18cm
Number of screws that can be placed on a road
[tex]\begin{gathered} =\text{ }\frac{18}{0.75} \\ =\text{ 24} \end{gathered}[/tex]Use the graph shown to the right to find each of the following
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The x intercept is the value of x at the point where the curve touches the x axis of the graph. Looking at the graph,
x intercept = - 1
It is written as (- 1, 0)
The zeros of the quadratic function is the same as the x intercept. Since the curve touches the x axis at only x = - 1, the zeros would be
x = - 1 twice
In a survey of 200 college students it is found that:61 like cooking32 like reading73 like video games19 like both cooking and reading23 like cooking and video games92 like reading or video games6 like all 3 hobbiesa. How many do not like any of these hobbiesb how many like reading onlyc how many like reading and video gamesd how many do not like cooking or video games
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Given:
The number of total students = 200
The number of students like cooking = 61
The number of students who like reading = 32
The number of students who like both cooking and reading= 19
The number of students who like video games = 73
The number of students who like cooking and video games= 23
The number of students who like reading and video games = 92
The number of students who like all 3 hobbies = 6
Required:
(a)
The probability of failing a test is 0.115 if you consider a group of 12 people taking a test on a given day, what is the probability that two or more of them will fail the test
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If the probability of failing a test is 0.115 if you consider a group of 12 people taking a test on a given day, then the probability that two or more of them will fail the test is 0.41
The probability of failing a test = 0.115
Total number of people = 12
We have to find the probability that two or more of them will fail the test
We know the binomial distribution
P(X≥2) = 1 - P(X<2)
= 1 - P(X=0) - P(X=1)
P(X≥2)= 1 - [tex](12C_{0}) (0.115^0)(1-0.115)^{12}[/tex] - [tex](12C_{1}) (0.115^1)(1-0.115)^{11}[/tex]
= 0.41
Hence, if the probability of failing a test is 0.115 if you consider a group of 12 people taking a test on a given day, then the probability that two or more of them will fail the test is 0.41
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If triangle JKL = triangle TUV , which of the following can you NOT conclude as being true? __ ___JK = TU
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If two triangles are said to be congruent, then they must have equal side lengths and equal angle measures.
See a sketch of triangles JKL and TUV below:
As shown in the sketch above:
- The side JK is equal in length as with the side TU
- The angle L is equal in measure as with the angle V
- The side LJ is equal in length as with the side VT
- The angle K is equal in measure as with the angle U
Therefore, we can NOT conclude that the angle J is equal in measure as with the angle V: Option B
A ladder resting on a vertical wall makes an angle whose tangent is 2.5 with the ground of the distance between the foot of the ladder and the wall is 60cm what is the length on the ladder
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If AC denote the ladder and B be foot of the wall the length of the ladder AC be x metres then the length of the ladder exists 5 m.
What is meant by trigonometric identities?Trigonometric Identities are equality statements that hold true for all values of the variables in the equation and that use trigonometry functions. There are numerous distinctive trigonometric identities that relate a triangle's side length and angle.
Let AC denote the ladder and B be foot of the wall. Let the length of the ladder AC be x metres.
Given that ∠ CAB = 60° and AB = 2.5 m In the right Δ CAB,
cos 60° = AB / AC
simplifying the above equation, we get
⇒ AC = AB / (cos 60°)
x =2 × 2.5 = 5 m
Therefore, the length of the ladder exists 5 m.
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How do you determine a relation is function on a GRAPH?
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To Determine: How to determine a relation is function on a GRAPH
In other to achieve this, we will use the vertical test
If a vertical line is moved across the graph and, at any time, touches the graph at only one point, then the graph is a function. If the vertical line touches the graph at more than one point, then the graph is not a function
Check the image below for a better clarification
With the use of the vertical line test, the graph in OPTION C and OPTION D are functions and the graph in OPTION A and B are not functions
In summary, you determine a relation is a function on a graph by using a vertical line test
Victoria, Cooper, and Diego are reading the same book for theirlanguage arts class. The table shows the fraction of the bookeach student has read. Which student has read the leastamount? Explain your reasoning.
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Given:
Completion of reading in fractions:
[tex]\text{Victoria}=\frac{2}{5};\text{Cooper}=\frac{1}{5};\text{Diego}=\frac{3}{5}[/tex]Since the denominators,
[tex]\text{The least value of the three given values is }\frac{1}{5}[/tex]Therefore, Cooper has read the least amount.
Write an equation of the line that passes through (-4,-5) and is parallel to the line defined by 4x +y = -5. Write the answer inslope-intercept form (if possible) and in standard form (Ax+By=C) with smallest integer coefficients. Use the "Cannot bewritten" button, if applicable.The equation of the line in slope-intercept form:
Answers
Answer: y = -4x - 21 OR 4x + y = -21
The given line is 4x + y = -5
Given point = (-4, -5)
Step 1: find the slope of the line
The slope intercept form of equation is given as
y = mx + b
Re -arrange the above equation to slope - intercept form
4x + y = -5
Isolate y
y = -5 - 4x
y = -4x - 5, where m = -4
Since the point is parallel to the equation
Therefore, m1 = m2
m2 = -4
For a given point
(y - y1) = m(x - x1)
Let x1 = -4, and y1 = -5
[(y - (-5)] = -4[(x - (-4)]
[y + 5] = -4[x + 4]
Open the parentheses
y + 5 = -4x - 16
y = -4x - 16 - 5
y = -4x - 21
The equation is y = -4x - 21 or 4x + y = -21
I need help with the work question Find area of regular polygon.Round to nearest tenth
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Given:
The number of sides in a given polygon is n = 5.
The length of each side is s = 8
The length of the apothem is a = 5.5
To find:
The area of the regular polygon
Explanation:
The formula of the area of the regular polygon is,
[tex]\begin{gathered} A=\frac{1}{2}\times n\times s\times a \\ A=\frac{1}{2}\times5\times8\times5.5 \\ A=110\text{ units}^2 \end{gathered}[/tex]Thus, the area of the given regular polygon is 110 square units.
Final answer:
The area of the regular polygon is 110 square units.
A scientist needs 270 milliliters of a 20% acid solution for an experiment. The lab has available a 25% and a 10% solution. How many milliliters of the 25% solution and how many milliliters of the 10% solution should the scientist mix to make the 20% solution?
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Given:
A scientist has 5% and a 10% acid solution in his lab.
He needs 270 milliliters of a 20% acid solution.
To find the amount of 25% solution and how many milliliters of the 10% solution should the scientist mix to make the 20% solution:
Here,
The dearer percentage is 25%.
The cheaper percentage is 10%.
The mean percentage is 20%.
Using the mixture and allegation method,
The ratio of the litters of cheaper (10% solution) to dearer value (25% solution) is,
[tex]\begin{gathered} (\text{Dearer value-mean): (Mean-Ch}eaper\text{ value)} \\ (25-20)\colon(20-10) \\ 5\colon10 \\ 1\colon2 \end{gathered}[/tex]So, the number of liters to be taken from 10% solution is,
[tex]\frac{1}{3}\times270=90\text{ liters}[/tex]So, the number of liters to be taken from 25% solution is,
[tex]\frac{2}{3}\times270=180\text{ liters}[/tex]Hence, the answer is
Wilson paints 40% of a bookcase in 20 minutes.How much more time will it take him to finish the bookcase?1. Write an equation using equal fractions to represent this situation. Use a box to represent the time it takes to paint the whole bookcase. 2 Use your equation to find the amount of time it will take Wilson to paint the whole bookcase. Explain how you found this answer. 3. How much time will it take Wilson to finish painting the bookcase? Explain.
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We can start that, by rewriting 40% as a fraction:
[tex]\frac{40}{100}=\frac{2}{5}[/tex]So let's find how long it will take to finish this painting, by writing the following fractions, and from them an equation:
1)
[tex]\begin{gathered} \frac{2}{5}---20 \\ \frac{3}{5}---x \\ \frac{2}{5}x=\frac{3}{5}\cdot20 \\ \frac{2}{5}x=12 \end{gathered}[/tex]So this is the equation, let's find the time to complete the painting:
[tex]\begin{gathered} \frac{2}{5}x=12 \\ 5\times\frac{2}{5}x=12\times5 \\ 2x=60 \\ \frac{2x}{2}=\frac{60}{2} \\ x=30 \end{gathered}[/tex]So it will take plus 30 minutes for to Wilson finish the bookcase. Note that
5/5 is equivalent to the whole bookcase or 100%
2) The amount of time to paint this whole bookcase, is found taking the initial 20 minutes and adding to them the 30 minutes we can state that the painting overall takes 50 minutes
3) Sorting out the answers:
[tex]\begin{gathered} 1)\frac{2}{5}x=\frac{3}{5}\cdot20 \\ 2)50\min \\ 3)30\min \end{gathered}[/tex]
A coordinate grid is shown from negative 6 to 6 on both axes at increments of 1. Figure ABCD has A at ordered pair negative 4, 4, B at negative 2, 2, C at negative 2, negative 1, D at negative 4, 1. Figure A prime B prime C prime D prime has A prime at ordered pair 4, 0, B prime at 2, negative 2, C prime at 2, negative 5, D prime at 4, negative 3.
Part B: Are the two figures congruent? Explain your answer.
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The two figures ABCD and A'B'C'D' are congruent .
In the question ,
it is given that the coordinates of the figure ABCD are
A(-4,4) , B(-2,2) , C(-2,-1) , D(-4,1) .
Two transformation have been applied on the figure ABCD ,
First transformation is reflection on the y axis .
On reflecting the points A(-4,4) , B(-2,2) , C(-2,-1) , D(-4,1) on the y axis we get the coordinates of the reflected image as
(4,4) , (2,2) , (2,-1) , (4,1) .
Second transformation is that after the reflection the points are translated 4 units down .
On translating the points (4,4) , (2,2) , (2,-1) , (4,1) , 4 units down ,
we get ,
A'(4,0) , B'(2,-2) , C'(2,-5) , D'(4,-3).
So , only two transformation is applied on the figure ABCD ,
Therefore , The two figures ABCD and A'B'C'D' are congruent .
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f(x)=x^6+10x^4 - 11x^2
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You can notice that the given function is symmetric respect to the y-axis.
It means that the value of the function for both x and -x is the same:
[tex]f(-x)=f(x)[/tex]This is the characteristic of a even function.
Hence, the answer is B
drag and drop the matching inequality from the left into the box on the right
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The first problem is modeled by the following inequality:
[tex]40+5x\ge95-4x[/tex]The second problem is represented by
[tex]95+4x<40+5x[/tex]The third problem is represented by
[tex]95-4x<40+5x[/tex]Observe that, "spending" refers to subtraction, "earnings" refers to addition. Also, the variables represent time. Additionally, "less than" is expressed as "<", "as much as or more than" is expressed as >=.
Take a look at the graph below. In the text box provided, describe to the best of yourability the following characteristics of the graph:Domain & RangeIs it a function?InterceptsMaximum/Minimum• Increasing/Decreasing intervals
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we know that
The domain is the set of all possible values of x and the range is the set of all possible values of y
so
In this problem
The domain is the interval {-6,5}
[tex]-6\leq x\leq5[/tex]The range is the interval {-2,1}
[tex]-2\leq y\leq1[/tex]Intercepts
we have
x-intercepts (values of x when the value of y is equal to zero)
x=-5,x=0 and x=3
y-intercepts (values of y when the value of x is equal to zero)
y=0
Maximum value y=1
Minimum value y=-2
Increasing intervals
{-6,4}, {2,3}
Decreasing intervals
{-1,2} and {3,5}
A third friend wants to offer Rebecca andSteve some of the animal models she hasalready made. The model she has of thegiant squid is 5 inches tall. Using thesame scale (2 in:5ft), how tall would thegiant squid be in real life?
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From the present question, it is said that the scale of a model is equal to:
[tex]e=\frac{2in}{5ft}[/tex]It means that the ratio of the size of the model and the real size of the giant squid must be always this same value. It was given that the size of the model is 5 in. Because we don't know the size of the real-life giant squid, we will use it as x. From this, we can write the following relation:
[tex]\frac{5in}{x}=\frac{2in}{5ft}[/tex]Now, we just need to isolate x in the present relation to find how tall would be a giant squid in real life.
[tex]\begin{gathered} \to2in\times x=5in\times5ft \\ x=\frac{5in\times5ft}{2in}=\frac{25}{2}ft=12.5ft \end{gathered}[/tex]From the solution developed above, we conclude that the real-life giant squid would be 12.5 ft tall.
The long-distance calls made by South Africans are normally distributed with a mean of 16.3 minutes and a standard deviation of 4.2 minutes for 1500 south Africans what is the expected number of callers whose calls last less than 15 minutes?
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The question provides the following parameters:
[tex]\begin{gathered} \mu=16.3 \\ \sigma=4.2 \end{gathered}[/tex]For 15 minutes, the z-score is calculated using the formula:
[tex]z=\frac{x-\mu}{\sigma}[/tex]At x = 15:
[tex]z=\frac{15-16.3}{4.2}=-0.3[/tex]The probability is calculated using the formula:
[tex]P(X<15)=Pr(z<-0.3)=Pr(z<0)-Pr(0From tables, we have:[tex]\begin{gathered} Pr(z<0)=0.5 \\ Pr(0Therefore, the probability is given to be:[tex]\begin{gathered} P(X<15)=0.5-0.1179 \\ P(X<15)=0.38 \end{gathered}[/tex]The expected number of callers will be calculated using the formula:
[tex]\begin{gathered} E=xP(x) \\ At\text{ }x=1500 \\ E=1500\times0.38 \\ E=570 \end{gathered}[/tex]Therefore, the expected number of callers whose calls last less than 15 minutes is 570 callers.
what is the factored form of his expression ? 2x^3+5x^2+6x+15
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The given expression is:
[tex]2x^3+5x^2+6x+15[/tex]It is required to write the expression in factored form.
[tex]\begin{gathered} \text{ Factor out }x^2\text{ in the first two terms of the expression:} \\ x^2(2x+5)+6x+15 \end{gathered}[/tex]
Next, factor out 3 in the last two terms of the expression:
[tex]x^2(2x+5)+3(2x+5)[/tex]Factor out the binomial 2x+5 in the expression:
[tex](2x+5)(x^2+3)[/tex]The expression in factored form is (2x+5)(x²+3).Find the maximum and minimum values of the function g(theta) = 2theta - 4sin(theta) on the interval Big[0, pi 2 Bigg\
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Hello there. To solve this question, we have to remember some properties about polar curves and determining maximum and minimum values.
In this case, we have the function in terms of the angle θ:
[tex]g(\theta)=2\theta-4\sin(\theta)[/tex]We want to determine its minimum and maximum values on the closed interval:
[tex]\left[0,\,\dfrac{\pi}{2}\right][/tex]We graph the function as follows:
Notice on the interval, it has a maximum value of 0.
We can determine its minimum value using derivatives, as follows:
[tex]g^{\prime}(\theta)=2-4\cos(\theta)[/tex]Setting it equal to zero, we obtain
[tex]\begin{gathered} 2-4\cos(\theta)=0 \\ \Rightarrow\cos(\theta)=\dfrac{1}{2} \\ \\ \Rightarrow\theta=\dfrac{\pi}{3} \end{gathered}[/tex]Taking its second derivative, we obtain
[tex]g^{\prime}^{\prime}(\theta)=4\sin(\theta)[/tex]And notice that when calculating it on this point, we get
[tex]g^{\prime}^{\prime}\left(\dfrac{\pi}{3}\right)=4\sin\left(\dfrac{\pi}{3}\right)=2\sqrt{3}[/tex]A positive value, hence it is a minimum point of the function.
Its minimum value is then given by
[tex]g\left(\dfrac{\pi}{3}\right)=2\cdot\dfrac{\pi}{3}-4\sin\left(\dfrac{\pi}{3}\right)=\dfrac{2\pi}{3}-2\sqrt{3}[/tex]Of course we cannot determine that 0 is a maximum value of this function using derivatives because it is a local maxima on a certain interval, and derivatives can only gives us this value when the slope of the tangent line is equal to zero.
7. Julie has $250 to plan a party. There is a one-time fee of $175 to reserve a room. It also cost $1.25 perperson for food and drinks. What is the maximum number of people that can come to the dance?
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Julie has $250 to plan the party.
The room costs $175 to reserve plus $1.25 per person for food and drinks.
Let "x" represent the number of people she can invite, you can express the total cost for the party as follows:
[tex]175+1.25x\leq250[/tex]From this expression, we can determine the number of people she can invite, without exceeding the $250 budget.
The first step is to pass 175 to the right side of the expression by applying the opposite operation "-175" to both sides of it:
[tex]\begin{gathered} 175-175+1.25x\leq250-175 \\ 1.25x\leq75 \end{gathered}[/tex]Next, divide both sides of the equation by 1.25 to reach the value of x:
[tex]\begin{gathered} \frac{1.25x}{1.25}\leq\frac{75}{1.25} \\ x\leq60 \end{gathered}[/tex]She can invite up to 60 people to the party
los números que faltan.
What comes after
3.0
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Answer:
3.1? 4.0? 3.0000001?
I did not know what are you mean sorry but uf i correct say me it
15=g/7 what does g equal to
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Answer:
g = 105
Explanation:
We want to find the value of g if
[tex]15=\frac{g}{7}[/tex]We multiply both sides of the equation by 7
[tex]\begin{gathered} 15\times7=\frac{g}{7}\times7 \\ \\ 105=g \end{gathered}[/tex]Therefore, the value of g is 105
Answer:
[tex]15=g/7[/tex]
We can get the value of g by multiplying the denominator, which in this case is 7.
So,
[tex]g = 15 x 7\\ g=105[/tex]
Help with number one a and b is both parts of number one
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Solving the operation_
We are given two figures that represent a garden. We are asked to determine its areas.
The shape of figure A is a rectangle of 9 ft by 12 ft. The area of a rectangle is the product of its dimensions therefore, we have:
[tex]A_A=\left(9ft\right)\left(12ft\right)[/tex]Solving the operations:
[tex]A_A=108ft^2[/tex]The shape of figure B is a circle of radius 5ft. The area of a circle is:
[tex]A_B=\pi r^2[/tex]Where "r" is the radius. Substituting we get:
[tex]A_B=\pi\left(5ft\right)^2[/tex][tex]A_B=25\pi ft^2[/tex]In decimal notation, the area is:
[tex]A_B=78.54ft^2[/tex]factor the following by taking on the greatest common factor 14a^3 + 35a^2 +42a
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Let's break apart each term into its factors:
[tex]\begin{gathered} 14a^3=2\cdot7\cdot a\cdot a\cdot a \\ 35a^2=5\cdot7\cdot a\cdot a \\ 42a=2\cdot3\cdot7\cdot a \end{gathered}[/tex]The common factors are
7 * a
That is,
[tex]7\cdot a=7a[/tex]Now, factorizing the expression, we have:
[tex]\begin{gathered} 14a^3+35a^2+42a \\ =7a(2a^2+5a+6) \end{gathered}[/tex]Answer[tex]7a(2a^2+5a+6)[/tex]How do you solve this??
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A mathematical statement comprehended as an equation exists created up of two expressions joined together by the equal sign.
If the equation be 12 - 2x = x - 3 then the value of x = 5.
What is meant by an equation?The definition of an equation in algebra is a mathematical statement that demonstrates the equality of two mathematical expressions.
A mathematical phrase with two equal sides and an equal sign is called an equation. A formula that expresses the connection between two expressions on each side of a sign. Typically, it has a single variable and an equal sign.
Let the equation be 12 - 2x = x - 3
Subtract 12 from both sides
12 - 2x - 12 = x - 3 - 12
Simplifying the above equation, we get
-2x = x - 15
Subtract x from both sides
-2x - x = x - 15 - x
Simplifying the above equation, we get
-3x = -15
Divide both sides by -3
[tex]$\frac{-3 x}{-3}=\frac{-15}{-3}[/tex]
Therefore, the value of x = 5.
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Johnathan works on IXL 5 nights per week. One week, he masters 7 skills. If he makes the sameamount of progress each night, how many skills does he master per night?Linear Equation:Solve:
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In this problem
Divide total skills by the total night per week
so
7/5=1.4 skills per night
therefore
Let
x ----> number of night
y ----> total skills
so
y=(7/5)x ------> y=1.4xUse inductive reasoning to find a pattern then make a reasonable conjecture for the next three items in the pattern p g q h r I
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Consider the first, third, and fifth terms of the sequence: p,q,r; these are consecutive letters starting with p.
Similarly, as for the second, fourth, and sixth terms: g,h, i; these are consecutive letters starting with g.
Thus, the seventh term has to be the letter that follows r; this is, s.
Analogously, the eighth and ninth terms are
[tex]\begin{gathered} \text{ eighth}\to\text{letter that follows i}\to j \\ \text{ ninth}\to\text{ letter that follows s}\to t \end{gathered}[/tex]Thus, the missing terms are: s, j, and t.
use the second derivative test to classify the relative extrema if the test applies
Answers
Answer
The answer is:
[tex](x,f(x))=(0,256)[/tex]SOLUTION
Problem Statement
The question gives us a polynomial expression and we are asked to find the relative maxima using the second derivative test.
The function given is:
[tex](3x^2+16)^2[/tex]Method
To find the relative maxima, there are some steps to perform.
1. Find the first derivative of the function
2. Equate the first derivative to zero and solve for x.
3. Find the second derivative of the function.
4. Apply the second derivative test:
This test says:
[tex]\begin{gathered} \text{ If }a\text{ is one of the roots of the equation from the first derivative, then,} \\ f^{\doubleprime}(a)>0\to\text{There is a relative minimum} \\ f^{\doubleprime}(a)<0\to\text{There is a relative maximum} \end{gathered}[/tex]5. Find the Relative Minimum
Implementation
1. Find the first derivative of the function
[tex]\begin{gathered} f(x)=(3x^2+16)^2 \\ \text{Taking the first derivative of both sides, we have:} \\ f^{\prime}(x)=6x\times2(3x^2+16) \\ f^{\prime}(x)=12x(3x^2+16) \end{gathered}[/tex]2. Equate the first derivative to zero and solve for x.
[tex]\begin{gathered} f^{\prime}(x)=12x(3x^2+16)=0 \\ \text{This implies that,} \\ 12x=0\text{ OR }3x^2+16=0 \\ \therefore x=0\text{ ONLY} \\ \\ \text{Because }3x^2+16=0\text{ has NO REAL Solutions} \end{gathered}[/tex]This implies that there is ONLY ONE turning point/stationary point at x = 0
3. Find the second derivative of the function:
[tex]\begin{gathered} f^{\prime}(x)=12x(3x^2+16) \\ f^{\doubleprime}(x)=12(3x^2+16)+12x(6x) \\ f^{\doubleprime}(x)=36x^2+192+72x^2 \\ \therefore f^{\doubleprime}(x)=108x^2+192 \end{gathered}[/tex]4. Apply the second derivative test:
[tex]\begin{gathered} f^{\doubleprime}(x)=108x^2+192 \\ a=0,\text{ which is the root of the first derivative }f^{\prime}(x) \\ f^{\doubleprime}(a)=f^{\doubleprime}(0)=108(0)^2+192 \\ f^{\doubleprime}(0)=192>0 \\ \\ By\text{ the second derivative test,} \\ f^{\doubleprime}(0)>0,\text{ thus, there exists a relative minimum at }x=0\text{ } \\ \\ \text{ Thus, we can find the relative minimum when we substitute }x=0\text{ into the function }f(x) \end{gathered}[/tex]5. Find the Relative Minimum:
[tex]\begin{gathered} f(x)=(3x^2+16)^2 \\ \text{substitute }x=0\text{ into the function} \\ f(0)=(3(0)^2+16)^2 \\ f(0)=16^2=256 \\ \\ \text{Thus, the minimum value of the function }f(x)\text{ is }256 \\ \\ \text{The coordinate for the relative minimum for the function }(3x^2+16)^2\text{ is:} \\ \mleft(x,f\mleft(x\mright)\mright)=\mleft(0,f\mleft(0\mright)\mright) \\ \text{But }f(0)=256 \\ \\ \therefore(x,f(x))=(0,256) \end{gathered}[/tex]Since the function has ONLY ONE turning point, and the turning point is a minimum value, then THERE EXISTS NO MAXIMUM VALUE
Final Answer
The answer is:
[tex](x,f(x))=(0,256)[/tex]