Answer: Logarithmic
Explanation:
This curve is a reflection of the exponential curve over the line y = x, to show that it is the inverse of exponentials. We use logs to help isolate the exponent among other useful properties.
HOW MANY DIFFERENT ARRANGEMENTS CAN BE MADE WITH THE NUMBERS
28535852
Using the arrangements formula, it is found that 1680 arrangements can be made with these numbers.
What is the arrangements formula?The number of possible arrangements of n elements is given by the factorial of n, that is:
[tex]A_n = n![/tex]
When there are repeated elements, repeating [tex]n_1, n_2, \cdots, n_n[/tex] times, the number of arrangements is given by:
[tex]A_n^{n_1, n_2, \cdots, n_n} = \frac{n!}{n_1!n_2! \cdots n_n!}[/tex]
For the number 28535852, we have that:
There are 8 numbers.5 repeats 3 times.2 repeats two times.8 repeats two times.Hence the number of arrangements is:
[tex]A_8^{3,2,2} = \frac{8!}{3!2!2!} = 1680[/tex]
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A company's sales increased 20% this year, to $6740. What were their sales last year?
The sales last year of the company is 5617
How to determine the sales last year?The given parameters are:
Company sales = 6740
Percentage increase = 20%
The sales of the company last year (x) to date is calculated as:
Company sales = (1 + Proportion) * Last year sales
Substitute the known values in the above equation
6740 = (1 + 20%) * x
Evaluate the sum
6740 = 1.2 * x
Divide both sides by 1.2
x = 5617
Hence, the sales last year is 5617
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PLEASE HELP ME WITH THIS QUESTION I NEED HELP
Answer:
True
Step-by-step explanation:
This is true in an equilateral triangle.
PLEASE HELP URGENT
Which equation could be solved using this application of the quadratic formula?
Answer:
Step-by-step explanation:
D because the quadratic formula is basically [tex]\frac{-b+\sqrt{b^{2}-4ac } }{2a}[/tex]±
so if we place the equation D into the formula
a = 1
b = +3
c = +24
we can clearly say that D eq is correctly used and can be.
one of the acute angles of a right triangle is 36° find the other acute angles
Step-by-step explanation:
An acute angle is an angle less or equal to 90 degrees.
let the other angle be x.
x + 36 = 90
grouping like terms
X = 90 - 36
X = 54 degrees
Answer:
54
Step-by-step explanation:
= ∠A + ∠B + ∠C = 180o … [∵sum of the angles of a triangle is 180]
or 180 = 36 + 90 + ∠C
C=54
Jake goes to the grocery store and buys 3 apples, 2 cans of soup, and 1 box of cereal. The apples cost $0.89 each; the soup costs $2.98 per can; and the box of cereal costs $4.99. Write an equation that represents the total cost c of Jake’s purchases.
The equation that represents the total cost, c, of the purchases made by Jake is: c = 3(0.89) + 2(2.98) + 4.99.
How to Write the Equation of Total Cost?To write the equation that represents the total cost of a given scenario like the one given, you can use variables to represent each component that makes up the total cost in the given situation.
Thus, let:
a = apple = 3a
s = can of soup = 2s
b = box of cereal = b
c = total cost
We know that unit price of each items are:
Apple = $0.89
Can of soup = $2.98
Box of cereal = $4.99
Equation would be:
c = 3a + 2s + b
Plug in the values
c = 3(0.89) + 2(2.98) + 4.99
Therefore, the equation that represents the total cost, c, of the purchases made by Jake is: c = 3(0.89) + 2(2.98) + 4.99.
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Does this set of ordered pairs represent a function? {(–2, 3), (–1, 3), (0, 2), (1, 4), (5, 5)} A. The relation is a function. Each input value is paired with more than one output value. B. The relation is a function. Each input value is paired with one output value. C. The relation is not a function. Each input value is paired with only one output value. D. The relation is not a function. Each input value is paired with more than one output value.
The correct option regarding whether the relation is a function is:
B. The relation is a function. Each input value is paired with one output value.
When does a relation represent a function?A relation represent a function if each value of the input is paired with one value of the output.
In this problem, when the input - output mappings are given by:
{(–2, 3), (–1, 3), (0, 2), (1, 4), (5, 5)}.
Which means that yes, each input value is paired with one output value, hence the relation is a function and option B is correct.
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What is the domain of the relation graphed below?
The domain of the relation shown in the graph is -4 <= x <= 4
How to determine the domain of the relation shown in the graph?The relation on the graph is an ellipse function.
The domain is the set of x values the function can take
From the graph, we have the following x values
This means that the domain of x in the graph is -4 <= x <= 4
Hence, the domain of the relation shown in the graph is -4 <= x <= 4
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Find the area of the sector formed by the 60 degree central angle.
503π in2503π in2
103π in2103π in2
100π in2100π in2
None of the Above
The area of the sector of the circle is: A. 50/3π in.².
What is the Area of a Sector of a Circle?The area of a sector that is bounded by two radii of a circle is calculated using the formula, ∅/360 × πr², where we have the following parameters:
r = radius of the circle∅ = central angle formed by the sector.Thus, we are given the following regarding the sector of the circle:
Central angle (∅) = 60 degrees
Radius (r) = 10 inches.
Plug in the values into ∅/360 × πr²:
Area of sector = 60/360 × π(10²)
Area of sector = 1/6 × π(100)
Area of sector = 100/6 × π
Area of sector = 50/3 × π
Area of sector = 50/3π in.²
Thus, the area of the sector of the circle is: A. 50/3π in.².
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s−3(s+6)= please help
Answer:
-2s-18 is the simplified answer
Step-by-step explanation:
s-3(s+6)
distributive property
s-3s-18
-2s-18
a dust mite is 400 pm long under a microscope, it looks 100,000 pm long. what magnification scale was used
Answer: 250:1
Step-by-step explanation:
Divide the measurement of appearance by the actual measurement.
[tex]100,000/400=250[/tex]
The magnification scale is 250:1. The scale is used by a 250X magnifying lense.
Evaluate the limits
[tex]x > \ln(x)[/tex] for all [tex]x[/tex], so
[tex]\displaystyle \lim_{x\to\infty} (\ln(x) - x) = - \lim_{x\to\infty} x = \boxed{-\infty}[/tex]
Similarly, [tex]\displaystyle \lim_{x\to\infty} (x-e^x) = - \lim_{x\to\infty} e^x = \boxed{-\infty}[/tex]
We can of course see the limits are identical by replacing [tex]x\mapsto e^x[/tex], so that
[tex]\displaystyle \lim_{x\to\infty} (\ln(x) - x) = \lim_{x\to\infty} (\ln(e^x) - e^x) = \lim_{x\to\infty} (x - e^x)[/tex]
You can also rewrite the limands to accommodate the application of l'Hôpital's rule. For instance,
[tex]\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\exp\left(\lim_{x\to\infty} (x - e^x)\right)\right) = \ln\left(\lim_{x\to\infty} e^{x-e^x}\right) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right)[/tex]
Using the rule, the limit here is
[tex]\displaystyle \lim_{x\to\infty} \frac{(e^x)'}{\left(e^{e^x}\right)'} = \lim_{x\to\infty} \frac{e^x}{e^x e^{e^x}} = \lim_{x\to\infty} \frac1{e^{e^x}} = 0[/tex]
so the overall limit is
[tex]\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right) = \ln(0) = \lim_{x\to0^+} \ln(x) = -\infty[/tex]
Suppose you know that the distribution of sample proportions of fifth grade students in a large school district who read below grade level in samples of 100 students is normal with a mean of 0.30 and a standard deviation of 0.12. You select a sample of 100 fifth grade students from this district and find that the proportion who read below grade level in the sample is 0.54. This sample proportion lies 2.0 standard deviations above the mean of the sampling distribution. What is the probability that a second sample would be selected with a proportion greater than 0.54 ?
Based on the mean of the sample and the proportion who read below grade level, the probability that a second sample would have a proportion greater than 0.54 is 0.9772.
What is the probability of the second sample being greater than 0.54?The probability that the second sample would be selected with a proportion greater than 0.54 can be found as:
P (x > 0.54) = P ( z > (0.54 - 0.30) / 0.12))
Solving gives:
P (x > 0.54) = P (z > 2)
P (x > 0.54) = 0.9772
In conclusion, the probability that the second sample would be selected with a proportion greater than 0.54 is 0.9772.
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5. Write the following inequality in slope-intercept form. −8x + 4y ≥ −52 y ≤ 2x − 13 y ≥ 2x − 13 y ≤ 2x + 13 y ≥ 2x + 13
Answer:35443
Step-by-step explanation:
Solve the following system of equations. 4x + 3y = -5
- 3x + 7y=13
Answer: 13
Step-by-step explanation:
4x+3y=-5 solve x :
4x = -3y + -5 | -3y
1x = -0.75y + -1.25 | : 4
-3x + 7y = 13 solve x :
-3x + 7y = 13 | -7y
-3x = -7y = 13 | : (-3)
Equalization Method Solution: -0.75y+-1.25=2.333y+-4.333
-0, 75y - 1,25 = 2,333y - 4, 333 solve y:
-0, 75y - 1,25 = 2,333y -4, 333 | -2,333y
-3, 083y - 1,25 = -4,333 | + 1, 25
-3. 083y = -3,088 | : (-3, 083)
y = 1
Plug y = 1 into the equation 4x + 3y = -5 :
4x + 3 · 1 | Multiply 3 with 1
4x + 3 = -5 | -3
4x = -8 | : 4
x = -2
So the solution is:
y = 1, x = -2
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6n−8(n+4)= what? Please help me
Answer:
-2n -32
Step-by-step explanation:
Acc. to BODMAS
we multiply first
⇒-6n - -8 x [(+4) (n)]
⇒ 6n - 8n x -32
[tex]\huge\text{Hey there!}[/tex]
[tex]\huge\textbf{Equation:}[/tex]
[tex]\mathsf{6n - 8(n + 4)}[/tex]
[tex]\huge\textbf{Solve: }[/tex]
[tex]\mathsf{6n - 8(n + 4)}[/tex]
[tex]\huge\textbf{Distribute \boxed{\bf -8} within the parenthesis:}[/tex]
[tex]\mathsf{= 6n - 8(n) - 8(4)}[/tex]
[tex]\mathsf{= 6n - 8n - 32}[/tex]
[tex]\huge\textbf{Combine the like terms: }[/tex]
[tex]\mathsf{= (6n - 8n) - (32)}[/tex]
[tex]\mathsf{= 6n - 8n - 32}[/tex]
[tex]\mathsf{= -2n - 32}[/tex]
[tex]\huge\textbf{Therefore, your answer should be:}[/tex]
[tex]\huge\boxed{\frak{{ -2n - 32}}}\huge\checkmark[/tex]
[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]
~[tex]\frak{Amphitrite1040:)}}[/tex]
Instructions: Find the missing side of the triangle.
X= Answer
Answer: 15
Step-by-step explanation: Using the pythagorean theorem A^2 + B^2 = C^2 where C is the hypotenuse of the triangle and any A and B are any sides of the triangle that are not the hypotenuse; we can rewrite this as A^2 = C^2 - B^2 to find out the missing side. I picked A to be the unknown side but you could also choose B. We know the hypotenuse is 39, and side B is 36. 39^2 - 36^2, or 1521 - 1296, is 225.
Now we have A^2 = 225. To get rid of the square, we have to take the square root of both sides. The A^2 cancels to be A, and sqrt(225) is 15. Thus, side A, or the missing side is 15.
Hope this helped!
15. Find the young modulus of a brass rod of diameter 25mm and of length 250mm which is
subjected to a tensile load of 50KN when the extension of the rod is equal to 0.3mm.
The Young Modulus of the brass rod ≈ 8.5 × 10¹⁰ Pa or N/m².
How to find the young modulus of a brass rod?
Given: Physical Properties of the Brass rod exist
Length, L = 250 mm
Radius, R = Diameter / 2 = 25/2 mm
Load of 50 kN hanging on the brass rod which results in elongation of the rod to 0.3 mm.
To estimate the Young Modulus of the brass rod we will require Stress on the rod, Ratio of elongation to the total length.
To find the stress on the rod,
Stress(Pressure) = Force / Area
⇒ Stress = 50,000 / (Area of cross-section)
The rod exists shaped like a cylinder then the area of its cross-section will be πR² ( R = radius)
⇒ Stress = 50000 / (π (25/2 mm)² )
The force exists given in newton, then we must convert the unit of the area into m² to obtain the solution in pascal.
1 mm = 0.001 m
⇒ Stress = 50,000 / 22/7 ( 25/2 × 10⁻³ m)²
take, π = 22/7
⇒ Stress = 50000 / ( 22/7 × 625 / 4 × 10⁻⁶ )
⇒ Stress = 50000 / ( 11/14 × 625 × 10⁻6 )
⇒ Stress = 50000 × 14 × 10⁶ / 625×11
⇒ Stress = 80 × 14 × 10⁶ / 11
⇒ Stress = 1120/11 × 10⁶ Pa
Now, that we maintain Stressed, to estimate Strain (ratio of elongation to original length),
⇒ Strain = (Elongation) / (Original length)
⇒ Strain = 0.3 / 250
⇒ Strain = 3 / 2500
Therefore, Young Modulus, Y = Stress / Strain
⇒ Y = 1120/11 × 10⁶ / 3/2500
⇒ Y = 1120/11 × 2500/3 × 10⁶
⇒ Y = 101.8 × 833.3 × 10⁶
⇒ Y = 84829.94 × 10⁶
⇒ Y ≈ 8.5 × 10¹⁰ Pa
Therefore, the Young Modulus of the brass rod ≈ 8.5 × 10¹⁰ Pa or N/m².
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Help me asappp w this question
[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]
The straight angles are angles that form a straight line, and their measure = 180°
In the given figure, the Straight angles is :
GEB[tex] \qquad \large \sf {Conclusion} : [/tex]
Correct choice is DHOW MANY DISTINCT ARRANGEMENTS CAN BE MADE WITH THE LETTERS IN THE
WORD CONNECTION?
There are 151200 distinct ways to arrange the letters of the word CONNECTION
How to determine the number of arrangements?The word is given as:
CONNECTION
In the above word, we have the following parameters:
Total number of characters, n = 10
The repeated letters are:
C's = 2
O's = 2
N's = 3
The number of arrangements of the letters is then calculated as:
Arrangements = n!/(C! * O! * N!)
Substitute the known values in the above equation
Arrangements = 10!/(2! * 2! * 3!)
Evaluate the expression
Arrangements = 151200
Hence, there are 151200 distinct ways to arrange the letters of the word CONNECTION
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consider the function y=-2-cos(x-pi). What effect does pi have on the basic graph?
Using translation concepts, it is found that pi is the phase shift of the graph, and since it is negative, the graph is shifted right pi units.
What is a translation?A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction either in it’s definition or in it’s domain. Examples are shift left/right or bottom/up, vertical or horizontal stretching or compression, and reflections over the x-axis or the y-axis.
In this problem, the change is given as follows:
x -> x - pi
It means that the change is in the domain, in which pi is the phase shift of the graph, and since it is negative, the graph is shifted right pi units.
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Find a potential function for the vector field
(a) We want to find a scalar function [tex]f(x,y,z)[/tex] such that [tex]\mathbf F = \nabla f[/tex]. This means
[tex]\dfrac{\partial f}{\partial x} = 2xy + 24[/tex]
[tex]\dfrac{\partial f}{\partial y} = x^2 + 16[/tex]
Looking at the first equation, integrating both sides with respect to [tex]x[/tex] gives
[tex]f(x,y) = x^2y + 24x + g(y)[/tex]
Differentiating both sides of this with respect to [tex]y[/tex] gives
[tex]\dfrac{\partial f}{\partial y} = x^2 + 16 = x^2 + \dfrac{dg}{dy} \implies \dfrac{dg}{dy} = 16 \implies g(y) = 16y + C[/tex]
Then the potential function is
[tex]f(x,y) = \boxed{x^2y + 24x + 16y + C}[/tex]
(b) By the FTCoLI, we have
[tex]\displaystyle \int_{(1,1)}^{(-1,2)} \mathbf F \cdot d\mathbf r = f(-1,2) - f(1,1) = 10-41 = \boxed{-31}[/tex]
[tex]\displaystyle \int_{(-1,2)}^{(0,4)} \mathbf F \cdot d\mathbf r = f(0,4) - f(-1,2) = 64 - 41 = \boxed{23}[/tex]
[tex]\displaystyle \int_{(0,4)}^{(2,3)} \mathbf F \cdot d\mathbf r = f(2,3) - f(0,4) = 108 - 64 = \boxed{44}[/tex]
Identify the most reasonable unit to measure the time spent at school on an average school day. Seconds, minutes, or hours?
The most reasonable unit to measure the time spent at school on an average school day is in hours.
How to find the most reasonable unit for a measure?
In anything we are going to measure, we have to pay special attention to the unit. When doing this, we want to keep the absolute value of the measure, that is, the number small, hence the do this conversion of units may be used.
For example, it is easier and more practical to say one day instead of 87,840 seconds. The same logic is applied to this problem, in which the most reasonable unit to measure the time spent at school on an average school day is in hours, as it keeps the numerical measure smaller than it would be in minutes or seconds.
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please help!!!
maths functions
Answer:
Point A: (2, 10)
Point B: (-3, 0)
Point C: (-5, -4)
Point D: (-5, -32)
Step-by-step explanation:
Part (a)Points A and B are the points of intersection between the two graphs.
Therefore, to find the x-values of the points of intersection, substitute one equation into the other and solve for x:
[tex]\implies 2x+6=-2x^2+18[/tex]
[tex]\implies 2x^2+2x-12=0[/tex]
[tex]\implies 2(x^2+x-6)=0[/tex]
[tex]\implies x^2+x-6=0[/tex]
[tex]\implies x^2+3x-2x-6=0[/tex]
[tex]\implies x(x+3)-2(x+3)=0[/tex]
[tex]\implies (x-2)(x+3)=0[/tex]
[tex]\implies x=2, -3[/tex]
From inspection of the graph:
The x-value of point A is positive ⇒ x = 2The x-value of point B is negative ⇒ x = -3To find the y-values, substitute the found x-values into either of the equations:
[tex]\begin{aligned} \textsf{Point A}: \quad 2x+6 & =y\\2(2)+6 & =10\\ \implies & (2, 10)\end{aligned}[/tex]
[tex]\begin{aligned} \textsf{Point B}: \quad -2x^2+18 & =y\\-2(-3)^2+18 & =0\\ \implies & (-3,0)\end{aligned}[/tex]
Therefore, point A is (2, 10) and point B is (-3, 0).
Part (b)If the distance between points C and D is 28 units, the y-value of point D will be 28 less than the y-value of point C. The x-values of the two points are the same.
Therefore:
[tex]\textsf{Equation 1}: \quad y=2x+6[/tex]
[tex]\textsf{Equation 2}: \quad y-28=-2x^2+18[/tex]
As the x-values are the same, substitute the first equation into the second equation and solve for x to find the x-value of points C and D:
[tex]\implies 2x+6-28=-2x^2+18[/tex]
[tex]\implies 2x^2+2x-40=0[/tex]
[tex]\implies 2(x^2+x-20)=0[/tex]
[tex]\implies x^2+x-20=0[/tex]
[tex]\implies x^2+5x-4x-20=0[/tex]
[tex]\implies x(x+5)-4(x+5)=0[/tex]
[tex]\implies (x-4)(x+5)=0[/tex]
[tex]\implies x=4,-5[/tex]
From inspection of the given graph, the x-value of points C and D is negative, therefore x = -5.
To find the y-value of points C and D, substitute the found value of x into the two original equations of the lines:
[tex]\begin{aligned} \textsf{Point C}: \quad 2x+6 & =y\\2(-5)+6 & =-4\\ \implies & (-5,-4)\end{aligned}[/tex]
[tex]\begin{aligned} \textsf{Point D}: \quad -2x^2+18 & = y \\ -2(-5)^2+18 & =-32\\ \implies & (-5, -32)\end{aligned}[/tex]
Therefore, point C is (-5, -4) and point D is (-5, -32).
Answer:
a) A = (2, 10) and B = (-3, 0)
b) C = (-5, -4) and D = (-5, -32)
Explanation:
a) To determine the coordinates of A and B, find the intersection points of the line "y = 2x + 6" and curve "y = -2x² + 18".
Solve the equation's simultaneously:
y = y
⇒ 2x + 6 = -2x² + 18
⇒ 2x² + 2x + 6 - 18 = 0
⇒ 2x² + 2x - 12= 0
⇒ 2x² + 6x - 4x - 12 = 0
⇒ 2x(x + 3) - 4(x + 3) = 0
⇒ (2x - 4)(x + 3) = 0
⇒ 2x - 4 = 0, x + 3 = 0
⇒ x = 2, x = -3
Then find value of y at this x points,
at x = 2, y = 2(2) + 6 = 10
at x = -3, y = 2(-3) + 6 = 0
Intersection points: A(2, 10) and B(-3, 0)
b) Given that CD = 28 units. Also stated parallel to y axis so x coordinates for both will be same but differ in y coordinate.
[tex]2x + 6 = -2x^2 + 18 + 28[/tex]
[tex]-2x^2 + 18 + 28-2x - 6 = 0[/tex]
[tex]-2x^2-2x+40=0[/tex]
[tex]-2x^2-10x+8x+40=0[/tex]
[tex]-2(x+5)+8(x+5)=0[/tex]
[tex](-2x+8)(x+5)=0[/tex]
[tex]x = -5, 4[/tex]
[tex]\leftrightarrow \sf C(-5, y_2), \ D(-5, y_2)[/tex]
Find y value for Point C : 2x + 6 = 2(-5) + 6 = -4
Find y value for Point D : -2x² + 18 = -2(-5)² + 18 = -32
[tex]\sf \rightarrow Point \ C = (-5, -4)\\ \\\rightarrow Point \ D = (-5, -32)[/tex]
(4x³-12x +11) + (2x - 2)
the figure at the right shows the dimensions of the garden in Marissa's back yard. What is the area of the garden?
The area of the garden is 74 square feet
How to determine the area of the garden?The complete question is added as an attachment
From the attached figure, we have the following shapes and dimensions:
Rectangle: 10 by 5 feetTrapezoid: Bases = 10 and 6; Height = 3The rectangular area is
A1 = 10 * 5
Evaluate
A1 = 50
The area of the trapezoid is
A2 = 0.5 * Sum of parallel bases * height
This gives
A2 = 0.5 * (10 + 6) * 3
Evaluate
A2 = 24
The total area is
Total = A1 + A2
This gives
Total = 50 + 24
Evaluate
Total = 74
Hence, the area of the garden is 74 square feet
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TIME REMAINING
44:36
Which point is an x-intercept of the quadratic function f(x) = (x + 6)(x – 3)?
(0,6)
(0,–6)
(6,0)
(–6,0)
Answer:
d. (-6, 0)
Step-by-step explanation:
no explanation bc u need this quick
HOW MANY DIFFERENT COMMITTEES CAN BE FORMED FROM 4 TEACHERS AND 20
STUDENTS IF THE COMMITTEE IS TO CONSIST OF TWO TEACHERS AND THREE
STUDENTS?
Answer:
6840
Step-by-step explanation:
multiply combinations
n!/(r!(n − r)!)
n choose r
n (objects) = |n
r (sample) = r
(4 teachers, CHOOSE 2)*(20 students, CHOOSE 3)
C(4,2) times C(20,3) =
C(n,r)=?
C(n,r) = C(4,2)
4! / (2!(4-2)!)
4! /2! x 2!
= 6
C(n,r)=?
C(n,r) = C(20,3)
20! / (3!(20-3)!)
20! / 3! x 17!
= 1140
6 * 1140 = 6840
alegbracom Edwin McCravy
Consider the ordinary differential equation (answer questions in picture)
a. Given the 2nd order ODE
[tex]y''(x) = 4y(x) + 4[/tex]
if we substitute [tex]z(x)=y'(x)+2y(x)[/tex] and its derivative, [tex]z'(x)=y''(x)+2y'(x)[/tex], we can eliminate [tex]y(x)[/tex] and [tex]y''(x)[/tex] to end up with the ODE
[tex]z'(x) - 2y'(x) = 4\left(\dfrac{z(x)-y'(x)}2\right) + 4[/tex]
[tex]z'(x) - 2y'(x) = 2z(x) - 2y'(x) + 4[/tex]
[tex]\boxed{z'(x) = 2z(x) + 4}[/tex]
and since [tex]y(0)=y'(0)=1[/tex], it follows that [tex]z(0)=y'(0)+2y(0)=3[/tex].
b. I'll solve with an integrating factor.
[tex]z'(x) = 2z(x) + 4[/tex]
[tex]z'(x) - 2z(x) = 4[/tex]
[tex]e^{-2x} z'(x) - 2 e^{-2x} z(x) = 4e^{-2x}[/tex]
[tex]\left(e^{-2x} z(x)\right)' = 4e^{-2x}[/tex]
[tex]e^{-2x} z(x) = -2e^{-2x} + C[/tex]
[tex]z(x) = -2 + Ce^{2x}[/tex]
Since [tex]z(0)=3[/tex], we find
[tex]3 = -2 + Ce^0 \implies C=5[/tex]
so the particular solution for [tex]z(x)[/tex] is
[tex]\boxed{z(x) = 5e^{-2x} - 2}[/tex]
c. Knowing [tex]z(x)[/tex], we recover a 1st order ODE for [tex]y(x)[/tex],
[tex]z(x) = y'(x) + 2y(x) \implies y'(x) + 2y(x) = 5e^{-2x} - 2[/tex]
Using an integrating factor again, we have
[tex]e^{2x} y'(x) + 2e^{2x} y(x) = 5 - 2e^{2x}[/tex]
[tex]\left(e^{2x} y(x)\right)' = 5 - 2e^{2x}[/tex]
[tex]e^{2x} y(x) = 5x - e^{2x} + C[/tex]
[tex]y(x) = 5xe^{-2x} - 1 + Ce^{-2x}[/tex]
Since [tex]y(0)=1[/tex], we find
[tex]1 = 0 - 1 + Ce^0 \implies C=2[/tex]
so that
[tex]\boxed{y(x) = (5x+2)e^{-2x} - 1}[/tex]
6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.
6.b) The value of z(x) is [tex]z(x) = 5e^{2x} - 2[/tex].
6.c) The value of y(x) is [tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex].
The given ordinary differential equation is y''(x) = 4y(x) + 4, y(0) = y'(0) = 1 ... (d).
We are also given a substitution function, z(x) = y'(x) + 2y(x) ... (z).
Putting x = 0, we get:
z(0) = y'(0) + 2y(0),
or, z(0) = 1 + 2*1 = 3.
Rearranging (z), we can write it as:
z(x) = y'(x) + 2y(x),
or, y'(x) = z(x) - 2y(x) ... (i).
Differentiating (z) with respect to x, we get:
z'(x) = y''(x) + 2y'(x),
or, y''(x) = z'(x) - 2y'(x) ... (ii).
Substituting the value of y''(x) from (ii) in (d) we get:
y''(x) = 4y(x) + 4,
or, z'(x) - 2y'(x) = 4y(x) + 4.
Substituting the value of y'(x) from (i) we get:
z'(x) - 2y'(x) = 4y(x) + 4,
or, z'(x) - 2(z(x) - 2y(x)) = 4y(x) + 4,
or, z'(x) - 2z(x) + 4y(x) = 4y(x) + 4,
or, z'(x) = 2z(x) + 4y(x) - 4y(x) + 4,
or, z'(x) = 2z(x) + 4.
The initial value of z(0) was calculated to be 3.
6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.
Transforming z(x) = dz/dx, and z = z(x), we get:
dz/dx = 2z + 4,
or, dz/(2z + 4) = dx.
Integrating both sides, we get:
∫dz/(2z + 4) = ∫dx,
or, {ln (z + 2)}/2 = x + C,
or, [tex]\sqrt{z+2} = e^{x + C}[/tex],
or, [tex]z =Ce^{2x}-2[/tex] ... (iii).
Substituting z = 3, and x = 0, we get:
[tex]3 = Ce^{2*0} - 2\\\Rightarrow C - 2 = 3\\\Rightarrow C = 5.[/tex]
Substituting C = 5, in (iii), we get:
[tex]z = 5e^{2x} - 2[/tex].
6.b) The value of z(x) is [tex]z(x) = 5e^{2x} - 2[/tex].
Substituting the value of z(x) in (z), we get:
z(x) = y'(x) + 2y(x),
or, 5e²ˣ - 2 = y'(x) + 2y(x),
which gives us:
[tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex] for the initial condition y(x) = 0.
6.c) The value of y(x) is [tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex].
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The width of a rectangle measures
(7h+3) centimeters, and its length measures
(8h−4) centimeters. Which expression represents the perimeter, in centimeters, of the rectangle?
The expression that represents the perimeter of the rectangle is 30h - 2
How to determine the perimeter expression?The dimensions of the rectangle are given as:
Length = 7h + 3
Width = 8h - 4
The perimeter of the rectangle is given as:
P = 2 * (Length + Width)
Substitute the known values in the above equation
P =2 * (7h + 3 + 8h - 4)
Evaluate the like terms
P = 2 * (15h - 1)
Evaluate the product
P = 30h - 2
Hence, the expression that represents the perimeter of the rectangle is 30h - 2
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