The zeros of the polynomial are -5i, √5i, -√5i.
The given polynomial is: f(x)=x⁴+16x²-225.
What is Zeros of a polynomial?Zeros of a polynomial are the values of x that make the polynomial equal to zero. These will be the x-intercepts of the polynomial's graph. To find the zeros of a polynomial, use the factored form of the polynomial and set each factor equal to zero. The solutions of this equation are the zeros of the polynomial.
The given zero is: -5i
To find the remaining zeros of the polynomial, we need to factor the polynomial.
We can factorize the polynomial as:
f(x) = (x² + 5i)(x² - 5i)
Finding Zeros:
Therefore, the remaining zeros of the polynomial are:
x = ±√5i
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The arithmetic mean of two numbers is half their sum, and the geometric mean is the square root of their product. using log properties, show how we can think of the log of the geometric mean of v and w as the arithmetic mean of two related numbers, p and q. expand and simplify
We have shown that the log of the geometric mean of two numbers is the arithmetic mean of the logs of those two numbers.
We can use the properties of logarithms to show that the log of the geometric mean of two numbers is the arithmetic mean of the logs of those two numbers.
First, recall that the arithmetic mean of two numbers is half their sum:
arithmetic mean = (v + w)/2
And the geometric mean is the square root of their product:
geometric mean = √(v*w)
Now, let's take the log of the geometric mean:
log(geometric mean) = log(√(v*w))
Using the property of logarithms that log(a*b) = log(a) + log(b), we can expand the expression:
log(√(v*w)) = log(√v) + log(√w)
Using the property of logarithms that log(a^b) = b*log(a), we can simplify the expression:
log(√v) + log(√w) = (1/2)*log(v) + (1/2)*log(w)
Now, we can see that the log of the geometric mean is the arithmetic mean of the logs of the two numbers:
log(geometric mean) = (log(v) + log(w))/2
So, if we let p = log(v) and q = log(w), we can see that the log of the geometric mean of v and w is the arithmetic mean of p and q:
log(geometric mean) = (p + q)/2
Therefore, we have shown that the log of the geometric mean of two numbers is the arithmetic mean of the logs of those two numbers.
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please answer as fast as possible
Answer:
5 2/3
Step-by-step explanation:
all you need is to add because the fraction is the same.
2566 concert tickets were sold for a total of $10,348.
if students paid $3 and nonstudents paid $5, how many student
tickets were sold?
1241 student tickets were sold. Below, you will learn how to solve the problem.
2566 concert tickets were sold for a total of $10,348, and if students paid $3 and nonstudents paid $5, then the number of student tickets sold can be calculated.
To solve this problem, we can set up an equation using the given information.
X + Y = 2566
3X + 5Y = 10348
Find the value of X:
Y = 2566 - X
3X + 5(2566 - X) = 10348
3X + 12830 - 5X = 10348
2X = 12830 - 10348
2X = 2482
X = 1241
Therefore, 1241 student tickets were sold.
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Carlos constructed 3 parallel lines as part of an art project. He also drew a line passing through each of the parallel lines. Some of the angles formed by the intersection of line t and lines l, m, and n are numbered in the diagram below.
Which conjecture can Carlos make about the angles formed by line t and lines l, m, and n?
a) Angles 1, 2, and 3 are congruent.
b) Angles 1, 3, and 5 are congruent.
c) Angles 2 and 4 are supplementary.
d) Angles 1 and 5 are supplementary.
A conjecture which Carlos can make about the angles formed by line t and lines l, m, and n include the following: B. angles 1, 3, and 5 are congruent.
What is corresponding angles postulate?In Mathematics, corresponding angles postulate simply refers to a theorem which states that corresponding angles are always congruent (equal) if the transversal intersects two parallel lines.
This ultimately implies that, the corresponding angles would always be congruent (equal) if a transversal intersects two (2) parallel lines.
By applying corresponding angles postulate to both lines l, m, and n, we can reasonably infer and logically deduce that the following angles are congruent:
∠1 ≅ ∠3
∠3 ≅ ∠5
∠1 ≅ ∠5
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Help me please I’m confused I can’t figure this out
The required function is g(x) = f(x)-7
What are functions?A function is a relation from a set of inputs to a set of possible outputs, where each input is related to exactly one output.
Given is a graph of two functions, the functions are in a way related, we need to find the relation between the two,
f(x) = x²+5
g(x) = x²-2
We can write,
g(x) = x²+5-7
g(x) = f(x)-2
We can clearly see a vertical shift of 7 units down, of f(x),
Therefore, when we subtract 7 from the function f(x) we will get g(x) [graph attached]
Hence, the required function is g(x) = f(x)-7
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From a population that is normally distributed with unknown mean µ and a standard deviation of 1.25, a sample of size 36 is taken.
(i) What is the standard error of the sample mean? (2 marks)
(ii) What is the probability that the sample mean is within 0.5 units of the population mean? (4 marks)
(iii) How large a sample size must we take to be 98% confident that the sample mean is within one-half of the population mean? (4 marks)
Thhe standard error of the sample mean is 0.208333. The probability that the sample mean is within 0.5 units of the population mean is 0.9918. We need to take a sample size of at least 7 to be 98% confident that the sample mean is within one-half of the population mean.
(i) The standard error of the sample mean is calculated by dividing the standard deviation of the population by the square root of the sample size. In this case, the standard error of the sample mean is 1.25 / √36 = 1.25 / 6 = 0.208333.
(ii) To find the probability that the sample mean is within 0.5 units of the population mean, we need to use the standard normal distribution. We can find the z-score for 0.5 units away from the mean by dividing 0.5 by the standard error of the sample mean, which is 0.5 / 0.208333 = 2.4.
Using a standard normal table, we can find the probability that the sample mean is within 2.4 standard deviations of the population mean, which is 0.9918.
(iii) To be 98% confident that the sample mean is within one-half of the population mean, we need to find the sample size that corresponds to a z-score of 2.33 (the z-score for a 98% confidence interval). We can use the formula for the standard error of the sample mean to solve for the sample size:
0.5 = 1.25 / √n
√n = 1.25 / 0.5
n = (1.25 / 0.5)²
n = 6.25
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One type of fertilizer has 30% nitrogen and
a second type has 15% nitrogen. If a farmer
needs 600 kg of fertilizer that is 20%
nitrogen, how much of each type should the
farmer mix together?
[tex]x=\textit{kgs of solution at 30\%}\\\\ ~~~~~~ 30\%~of~x\implies \cfrac{30}{100}(x)\implies 0.3 (x) \\\\\\ y=\textit{kgs of solution at 15\%}\\\\ ~~~~~~ 15\%~of~y\implies \cfrac{15}{100}(y)\implies 0.15 (y) \\\\\\ \textit{60 kgs of solution at 20\%}\\\\ ~~~~~~ 20\%~of~60\implies \cfrac{20}{100}(60)\implies 0.2 (60)\implies 12 \\\\[-0.35em] ~\dotfill[/tex]
[tex]\begin{array}{lcccl} &\stackrel{kgs}{quantity}&\stackrel{\textit{\% of kgs that is}}{\textit{nitrogen only}}&\stackrel{\textit{kgs of}}{\textit{nitrogen only}}\\ \cline{2-4}&\\ \textit{1st Fert.}&x&0.3&0.3x\\ \textit{2nd Fert.}&y&0.15&0.15y\\ \cline{2-4}&\\ mixture&60&0.2&12 \end{array}~\hfill \begin{cases} x + y = 60\\\\ 0.3x+0.15y=12 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{x+y=60}\implies y=60-x \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{\textit{using the 2nd equation}}{0.3x+0.15y=12}\implies \stackrel{\textit{substituting from above}}{0.3x+0.15(60-x)=12} \\\\\\ 0.3x+9-0.15x=12\implies 0.15x=3\implies x=\cfrac{3}{0.15} \\\\\\ \boxed{x=20}\hspace{5em}\stackrel{ 60~~ - ~~20 }{\boxed{y=40}}[/tex]
You earned 75% on your badminton practical. If the practical was out of 88, what mark did you earn?
Answer:
66 (at least if that is an answer choice)
Step-by-step explanation:
[tex]\frac{75*88}{100} = 66[/tex]
66 is your answer!
Solve the following system of equations using SUBSTITUTION METHOD.
SHOW YOUR WORK
Write your answer as an (x,y) point.
x= 4y - 1
3x + 2y = 25
The solution to the system of equations using the substitution method is (7,2).
Can the substitution approach be used to solve all equation systems?No, the substitution approach cannot be used to solve every system of equations. It can be challenging to replace one equation with another in some systems because they feature equations that are challenging or impossible to solve for one variable in terms of the other. Other approaches, such graphing or elimination, could be more useful in such circumstances.
Given that, the two equations are:
x= 4y - 1
3x + 2y = 25
Substitute x = 4y - 1 into the second equation:
3(4y - 1) + 2y = 25
12y - 3 + 2y = 25
14y - 3 = 25
14y = 28
y = 2
Substitute the value of y in equation:
x = 4y - 1
x = 4(2) - 1
x = 7
Hence, the solution to the system of equations using the substitution method is (7,2)
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If the product of (2x^(3)+4x-3) and (4x-1) is written in standard form, what will the first term be?
The first term of the product of (2x^(3)+4x-3) and (4x-1) in standard form will be 8[tex]x^{4}[/tex].
Since 4x is multiplied by 2x³, their product is 8x⁴. Then, distributing 4x to the terms inside the first bracket and -1 to the terms inside the second bracket, we have:
8x⁴ - 2x³ + 16x² - 13x + 3.
To write the above expression in standard form, we just need to arrange the terms of the polynomial in descending order of their degree.
Thus, the first term will be 8[tex]x^{4}[/tex].
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Let X be a random variable whose characteristic function ∅ satisfies ∫▒〖|∅(t)|dt< [infinity].〗 . Show that (2π)^(-1) ∫▒〖e^(-√(-1) xt)∅(t)dt〗 is the Lebesgue density of X.
(2π)^(-1)E[∫▒〖e^(-√(-1) xy)]dy〗
The characteristic function of a random variable X is defined as ∅(t) = E[e^(√(-1)tx)]. To show that (2π)^(-1) ∫▒〖e^(-√(-1) xt)∅(t)dt〗 is the Lebesgue density of X, we need to demonstrate that it is a probability density function.
We first calculate the integral to obtain (2π)^(-1) ∫▒〖e^(-√(-1) xt)∅(t)dt〗 = (2π)^(-1) ∫▒〖E[e^(-√(-1) xt +√(-1) tx)]dt〗.
Since the expectation is a constant, we can pull it out of the integral to get (2π)^(-1)E[∫▒〖e^(-√(-1) xt +√(-1) tx)]dt〗.
Now, if we substitute t = y-x and rewrite the integral, we obtain (2π)^(-1)E[∫▒〖e^(-√(-1) x(y-x))e^(√(-1) xy)]dy〗.
This simplifies to (2π)^(-1)E[∫▒〖e^(-√(-1) xy)]dy〗.
Because the integrand is 1, the integral is simply y. Then, the expectation becomes E[y] which is the mean of the random variable X. Thus, the Lebesgue density of X is (2π)^(-1)E[y], which is the probability density function of X.
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Info Sheet Math Definition
Power –
Base –
Exponent –
Coefficient –
Exponential Form –
Expanded Form –
Standard Form –
Exponent Laws –
Answer:
Power – A power is an expression that represents repeated multiplication of the same number or variable, such as a raised to the power of n, denoted by aⁿ.
Base – The base is the number or variable that is raised to a power or exponent in a power expression.
Exponent – The exponent is the number that indicates how many times the base is to be multiplied by itself in a power expression.
Coefficient – The coefficient is the numerical factor of a term that contains a variable in an algebraic expression.
Exponential Form – The exponential form is a way of writing a number using a base and an exponent, such as aⁿ, where a is the base and n is the exponent.
Expanded Form – The expanded form is a way of writing an expression as the sum or difference of its individual terms, such as (a+b)² = a² + 2ab + b².
Standard Form – The standard form is a way of writing a number using digits, such as 1234 or 1.234 x 10³.
Exponent Laws – The exponent laws are a set of rules that describe how exponents can be manipulated in algebraic expressions, including the product law, quotient law, power law, and negative exponent law. These laws help simplify and solve equations involving exponents.
In February, Alec had 1,513 visitors to his website. In March, he had 1,400 visitors to his website. What is the percentage decrease of the number of visitors to Alec's website from February to March? If necessary, round to the nearest tenth of a percent. A. 9. 3%
B. 7. 5%
C. 8. 1%
D. 92. 5%
solution
1531-1400=113
decrease/original ×100
113/1513×100
7.46
round off to 7.5
Which of the following statements is true? Please explain why.
(a) Regression coefficients estimated using least squares are always BLUE.
(b) If the omitted variable has positive effect on the explained variable and is negatively correlated with another explanatory variable we have negative bias.
(c) R2 j in the expression: V ar(βˆ j ) = σ 2 SSTj (1−R2 j ) is obtained from the regression of the dependent variable on the rest of explanatory variables.
(d) None of the above
The correct statement is (a) Regression coefficients estimated using least squares are always BLUE.
Therefore, statement (a) is true. The other statements are not correct. Statement (b) is incorrect because the omitted variable bias can be positive or negative, depending on the direction of the correlation between the omitted variable and the other explanatory variables. Statement (c) is incorrect because R2 j is obtained from the regression of the jth explanatory variable on the rest of the explanatory variables, not the dependent variable. Statement (d) is incorrect because statement (a) is true.
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Let l, m, and n be three lines; if n Im and m ll then 1 || n. Let l, m, and n be three lines; if I || m and m || n then 1 || n. Let k, l, m, and n be four lines; if k 11,11 m and m In then k \ n.
This property can be applied to any number of lines as long as they are all parallel to the same line.
The statement "if n I| m and m ll then 1 || n" is not valid as the symbols used are not correct. The correct statement should be "if l || m and m || n then l || n". This means that if line l is parallel to line m and line m is parallel to line n, then line l is also parallel to line n. This is known as the transitive property of parallel lines.
Similarly, the statement "if k 11,11 m and m In then k \ n" is not valid as the symbols used are not correct. The correct statement should be "if k || l, l || m, and m || n then k || n". This means that if line k is parallel to line l, line l is parallel to line m, and line m is parallel to line n, then line k is also parallel to line n. This is also an application of the transitive property of parallel lines.
In conclusion, the transitive property of parallel lines states that if two lines are parallel to the same line, then they are also parallel to each other. This property can be applied to any number of lines as long as they are all parallel to the same line.
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Triangle XYZ has vertices X(0,2), Y(4,4), and Z(3,-1). Graph \triangle XYZ△XYZ and its image after a rotation of 180 degrees about (2,-3).
The image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.
What is Geometry?It deals with the size of geometry, region, and density of the different forms both 2D and 3D.
Triangle XYZ has vertices X(0, 2), Y(4, 4), and Z(3, –1).
If the triangle is ΔXYZ. Then the image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.
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Determine the slope and intersection with the y-axis of each graph
y=-2x-1
Step-by-step explanation:
when x=2
y=2(2)-1
y=4-1
Answer:
slope = -2, Y intercept = -1
Step-by-step explanation:
[tex]y = mx + c[/tex]
[tex]where \: m \: is \: the \: slope \: and \: c \: is \: the \: intercept[/tex]
So compare the equation with y= -2x-1
Therefore m = -2, c = -1
Use the properties of logarithms to expand the expression as a sum, difference, and/or constant multiple of logarithms. (Assume all variables are positive)
The expression can be expanded as a difference of two logarithms:
[tex]ln(\sqrt\frac{x^2}{y^5})[/tex] = [tex]2 ln(x) - (5/2) ln(y)[/tex]
What is the logarithms?
Logarithms are mathematical functions that help to simplify the representation of very large or very small numbers. They are the inverse functions of exponential functions.
We can use the properties of logarithms to expand the expression as follows:
[tex]ln(\sqrt\frac{x^2}{y^5})[/tex] = [tex]ln(x^2) - ln(y^5/2)[/tex]
Next, we can simplify each logarithm using the property that log(a^b) = b log(a):
[tex]ln(x^2) - ln(y^5/2)[/tex] = [tex]2 ln(x) - (5/2) ln(y)[/tex]
Hence, the expression can be expanded as a difference of two logarithms:
[tex]ln(\sqrt\frac{x^2}{y^5})[/tex] = [tex]2 ln(x) - (5/2) ln(y)[/tex]
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Be Precise Two cylinders have the same volume of 845π cubic inches. The radius of Cylinder A is 13 inches and the radius of Cylinder B is 10 inches. Which cylinder is taller? How much taller? Express the difference in heights as a decimal.
Cylinder B is taller which is 3.45 inches more than cylinder A.
Volume of the cylinder:
A cylinder is a three-dimensional geometric shape that consists of two parallel circular bases that are connected by a curved surface.
The formula for the volume of a cylinder is given by
V = πr²hWhere V is the volume of the cylinder, r is the radius of the circular base of the cylinder, and h is the height of the cylinder.
Here we have
Two cylinders have the same volume of 845π cubic inches.
The radius of Cylinder A is 13 inches
Let 'h₁' be the height of th cylinder
By using the formula, V = π r² h₁
Volume of the cylinder A = π (13)² h₁ = 169πh₁
From the data, 169πh₁ = 845π
=> 169h₁ = 845
=> h₁ = 845/169
=> h₁ = 5
The radius of Cylinder B is 10 inches.
Let h₂ be the height
Volume of the Cylinder B = π (10)² h₂ = 100πh₂
From the data, 100πh₂ = 845π
=> 100h₂ = 845
=> h₂ = 8.45
The difference between the heights of the two cylinders
= 8.45 inch - 5 inch = 3.45 inch
Therefore,
Cylinder B is taller which is 3.45 inches more than cylinder A.
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How much distance will a tein cover in 3⅓ hoursif it maintains a constant speed of 80¼ km/h?
600 g of 3 kg as a percentage
show method
Answer:
20%
Step-by-step explanation:
600÷3 of 100 = 20%
600/3 of 100 = 20%
estion list For the given equation, state the value of the discriminant and the number of real sol 36x^(2)-24x+4=0
The value of the discriminant is 0 and the number of real solutions is 1
For the given equation, 36x^(2)-24x+4=0, we can find the value of the discriminant and the number of real solutions using the quadratic formula.
The quadratic formula is x = (-b ± √(b^(2)-4ac))/(2a), where a, b, and c are the coefficients of the equation.
The discriminant is the part of the formula under the square root, b^(2)-4ac.
In this equation, a = 36, b = -24, and c = 4. Plugging these values into the discriminant, we get:
Discriminant = (-24)^(2)-4(36)(4) = 576-576 = 0
Since the discriminant is equal to 0, there is only one real solution to this equation.
Therefore, the value of the discriminant is 0 and the number of real solutions is 1.
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Assume that the sales of automobiles in Brandon follow a Poisson distribution with a mean of 3 per day.
a. What is the probability that none is sold on a particular day?
b. What is the probability that at least 2 automobiles are sold on a particular day?
c. What is probability that for five consecutive days at least one automobile is sold?
Therefore, the answers are:
a. 0.0498
b. 0.8008
c. 0.7745
The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event.
a. The probability that none is sold on a particular day can be calculated using the formula:
P(X = 0) = (e^-λ)(λ^0) / 0!
Where λ is the mean and e is the base of the natural logarithm.
Substituting the values, we get:
P(X = 0) = (e^-3)(3^0) / 0!
P(X = 0) = (0.0498)(1) / 1
P(X = 0) = 0.0498
b. The probability that at least 2 automobiles are sold on a particular day can be calculated using the formula:
P(X ≥ 2) = 1 - P(X < 2)
P(X ≥ 2) = 1 - (P(X = 0) + P(X = 1))
P(X ≥ 2) = 1 - ((e^-3)(3^0) / 0! + (e^-3)(3^1) / 1!)
P(X ≥ 2) = 1 - (0.0498 + 0.1494)
P(X ≥ 2) = 1 - 0.1992
P(X ≥ 2) = 0.8008
c. The probability that for five consecutive days at least one automobile is sold can be calculated using the formula:
P(X ≥ 1) = 1 - P(X = 0)
P(X ≥ 1) = 1 - (e^-3)(3^0) / 0!
P(X ≥ 1) = 1 - 0.0498
P(X ≥ 1) = 0.9502
The probability that for five consecutive days at least one automobile is sold is:
P(X ≥ 1)^5 = 0.9502^5 = 0.7745
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Teri has 12 paper weights in her collection. She has twice as many glass paperweights as metal, and three are wood. How many of each paper weight does she have?
Let's say that the number of metal paperweights is "x", then the number of glass paperweights is "2x" since she has twice as many glass paperweights as metal.
We know that she has a total of 12 paperweights, so we can set up an equation:
x + 2x + 3 = 12
Simplifying the equation, we get:
3x + 3 = 12
Subtracting 3 from both sides, we get:
3x = 9
Dividing both sides by 3, we get:-
x = 3
So Teri has 3 metal paperweights, and since she has twice as many glass paperweights as metal, she has:
2x = 2(3) = 6 glass paperweights.
Therefore, Teri has 3 metal paperweights, 6 glass paperweights, and 3 wood paperweights in her collection.
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Let X1, ... , Xn be an i.i.d. sample from the Pareto distribution with the density function f(x) = θx^θ x^(-θ-1) x>x0. where xo >0 and θ > 0. Assume that xo is given. Let Yi = log(Xi/xo), i = 1,..., n. 1. 1. Find θ3, the method of moments estimate of 0 based on Y1,..., Yn. 2. Find the distribution of Y. 3. Find the mean and variance of θ3. You may assume that n > 3 and use the following facts: (a) r(a +1) = ar (a) for a > 0. (b) If U follows a gamma distribution, then E(U") = r (a+r)/ [λ'T(a)] for r > -a.
Var(θ3) = (a-1)(a-2) / λ^2 - θ^2.
First, let's find the method of moments estimate of θ based on Y1,...,Yn. We know that E(Y) = E(log(X/x0)) = E(log(X)) - log(x0) = θ^-1 - log(x0). Therefore, θ^-1 = E(Y) + log(x0) and θ3 = 1 / (E(Y) + log(x0)).
Next, let's find the distribution of Y. Since Y = log(X/x0), we can use the change of variables formula to find the density function of Y. Let g(y) = x0 * exp(y), then the Jacobian is |g'(y)| = x0 * exp(y). The density function of Y is fY(y) = fX(g(y)) * |g'(y)| = θ * (x0 * exp(y))^θ * (x0 * exp(y))^(-θ-1) * x0 * exp(y) = θ * x0^θ * exp(-θy).
Finally, let's find the mean and variance of θ3. We know that E(θ3) = E(1 / (E(Y) + log(x0))) = 1 / (E(Y) + log(x0)) = θ. To find the variance, we can use the fact that Var(θ3) = E(θ3^2) - E(θ3)^2. We can use the fact that if U follows a gamma distribution, then E(U^r) = Γ(a+r) / [λ^r * Γ(a)] to find E(θ3^2). Let U = E(Y) + log(x0), then E(θ3^2) = E(U^-2) = Γ(a-2) / [λ^2 * Γ(a)] = (a-1)(a-2) / λ^2. Therefore, Var(θ3) = (a-1)(a-2) / λ^2 - θ^2.
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The restrictions on x, when (x+4)/(5x-1) is divided by (3x+12)/(6x), can be written in the form x!
The restrictions on x are x = 1/5 and x = 0.
The restrictions on x, when (x+4)/(5x-1) is divided by (3x+12)/(6x), can be found by looking at the denominators of each fraction. The restrictions are values of x that would make the denominator equal to zero, which would make the fraction undefined.
For the first fraction, (x+4)/(5x-1), the restriction is when 5x - 1 = 0. Solving for x, we get:
5x = 1
x = 1/5
For the second fraction, (3x+12)/(6x), the restriction is when 6x = 0. Solving for x, we get:
x = 0
Therefore, the restrictions on x are x = 1/5 and x = 0. These values of x cannot be used in the original expression because they would make the denominator equal to zero and the expression undefined.
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For f(x)= 11x and g(x)= 1/11x, (fog)(x) and (gof)(x).Then determine whether (fog)(x) = (gof)(x). What is (fog)(x)?
(fog)(x)=____
The answer to the question "What is (fog)(x)?" is:
(fog)(x) = x
The first step in solving this problem is to find the composition of f and g, or (fog)(x). This is done by substituting the expression for g(x) into the expression for f(x):
(fog)(x) = f(g(x)) = f(1/11x) = 11(1/11x) = x
Similarly, we can find the composition of g and f, or (gof)(x), by substituting the expression for f(x) into the expression for g(x):
(gof)(x) = g(f(x)) = g(11x) = 1/11(11x) = x
Now, we can determine whether (fog)(x) = (gof)(x) by comparing the two expressions. Since both (fog)(x) and (gof)(x) are equal to x, we can conclude that (fog)(x) = (gof)(x).
Therefore, the answer to the question "What is (fog)(x)?" is:
(fog)(x) = x
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What is 1/1 of a full rotation?
Answer:
Try out
Step-by-step explanation:
1/4
1/2
1
1 1/2
Answer: 360 degrees
So basically when you spin around you make 360 degrees lol.
A number is equal to the sum of half a sebond number and 3. The first number is also equal to the sum of one-quarter of the second number and 5. The situation can be represented by using the graph below, where x represents the second number.
A patient weighing 21.1 pounds presents with a bacterial infection and is prescribed a course of Amoxicillin. An adult dose for the same type and severity of infection would be 771 mg given every 12 hours.
Choose the most appropriate formula given the information you have about the patient to calculate the dose to administer every 12 hours. Round your answer to the nearest tenth of a milligram as necessary.
Please help me by showing work . Thank you!
The most appropriate formula to use in this situation (patient with a bacterial infection and prescribed with Amoxicillin) is Clark's rule, and the dose to administer every 12 hours would be 108.4 mg.
The best formula to use in this situation is the Clark's rule, which is used to calculate the dose of medication for a child based on their weight and the adult dose. The formula is as follows:
In this case, the weight of the child is 21.1 pounds and the adult dose is 771 mg every 12 hours. Plugging these values into the formula, we get:
Child dose = (21.1 / 150) x 771Child dose = 0.1406666666666667 x 771Child dose = 108.384 mg every 12 hoursRounding to the nearest tenth of a milligram, the child dose would be 108.4 mg and to administer every 12 hours.
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