The third side of the triangle falls between (50, 60).
How to find the third side of a triangle?Inequality triangle theorem states that the sum of any two sides of a triangle is greater than or equal to the third side.
Therefore, the other two sides are 5 cm and 55 cm. The range of the third side x can be computed using inequality triangle theorem.
Hence,
x < 5 + 55
x < 60
And,
x > 55 - 5
x > 50
Therefore, 50 < x < 60.
Hence, the third side of the triangle falls between (50, 60).
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Paula finished the race at 2:14 p.m Beatrice finished the race 22 minutes earlier what time did Beatrice finish the race a 1:54 p.m b 1:48 p.m. c 1:58 p.m. d 1:52 p.m. e none of these f I don't know yet
Answer: d: 1:52pm
Step-by-step explanation: Since Beatrice finished 22 minutes earlier, we subtract 22 minutes from 2:14. 2:14 - 14 is 2:00. 22-14 is 8. 2:00 - 8 is 1:52.
find the value of x,y in the given figure with reasons.
Answer:
answer
x= 40°
Step-by-step explanation:
x= 40° [ base angle of isocles triangle]
In politics, marketing, etc. we often want to estimate a percentage or proportion p. One calculation in statistical polling is the margin of error - the largest (reasonble) error that the poll could have. For example, a poll result of 72% with a margin of error of 4% indicates that p is most likely to be between 68% and 76% (72% minus 4% to 72% plus 4%). In a (made-up) poll, the proportion of people who like dark chocolate more than milk chocolate was 32% with a margin of error of 2.5%. Describe the conclusion about p using an absolute value inequality. The answer field below uses the symbolic entry option in Mobius. That lets you type in a vertical bar | to represent absolute values. Also, when you type in < and then =, the symbolic entry option will automatically convert that to ≤ . In the same way, if you type in > and then =, the symbolic entry option will automatically convert that to ≥. Be sure to use decimal numbers in your answer (such as using 0.40 for 40%). __________
Answer:
yes yes yes yes yes Yes Yes you are cute
Could someone show me a step by step process on how to do this problem? Calculus 2
The arc length is given by the definite integral
[tex]\displaystyle \int_1^3 \sqrt{1 + \left(y'\right)^2} \, dx = \int_1^3 \sqrt{1+9x} \, dx[/tex]
since by the power rule for differentiation,
[tex]y = 2x^{3/2} \implies y' = \dfrac32 \cdot 2x^{3/2-1} = 3x^{1/2} \implies \left(y'\right)^2 = 9x[/tex]
To compute the integral, substitute
[tex]u = 1+9x \implies du = 9\,dx[/tex]
so that by the power rule for integration and the fundamental theorem of calculus,
[tex]\displaystyle \int_{x=1}^{x=3} \sqrt{1+9x} \, dx = \frac19 \int_{u=10}^{u=28} u^{1/2} \, du = \frac19\times\frac23 u^{1/2+1} \bigg|_{10}^{28} = \boxed{\frac2{27}\left(28^{3/2} - 10^{3/2}\right)}[/tex]
Using two six-sided number cubes, each labeled with the numbers 1 through 6, event A is rolling a sum less than 6. Which of the following shows the sample space of event A?
{(1, 1), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3)}
{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
{(1, 1), (1, 2), (1, 3), (1, 5), (2, 1), (2, 2), (2, 3), (3, 1), (3, 3), (4, 1)}
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (3, 3), (4, 1), (4, 2)}
the sample space is:
{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1), (2, 2), (2, 3), (3, 2)}
Which of the following shows the sample space of event A?Event A is rolling a sum less than 6.
Let's define the possible elements in this experiment as:
(outcome of dice 1, outcome of dice 2)
The outcomes where the sum is less than 6 are:
dice 1 dice 2 sum
1 1 2
1 2 3
1 3 4
1 4 5
2 1 3
3 1 4
4 1 5
2 2 4
3 2 5
2 3 5
So there are 10 outcomes, then the sample space is:
{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1), (2, 2), (2, 3), (3, 2)}
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Answer:
B) {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
Step-by-step explanation:
If the question said less than 6 meaning you have to find all possible solution that are 5 or lower.
However, if the problem said equal or less than 6 then you have to find all possible solution that are 6 or lower.
B option is only option that don't have sum of 6. Therefore, option B is correct.
Write down the answer for Q 7 and 8
The answers to the given addition operations are
7) 6 Hundredths add to 4 tenths add to 6 ones is equal to 6.46
8) 82 Hundredths add to 9 tenths add to 4 tens is equal to 41.72
Addition operationFrom the question, we are to add the given numbers
7. 6 Hundredths add to 4 tenths add to 6 ones is equal to
6 Hundredths = 0.06
4 tenths = 0.4
6 ones = 6
Thus, we get
0.06 + 0.4 + 6 = 6.46
8. 82 Hundredths add to 9 tenths add to 4 tens is equal to
82 Hundredths = 0.82
9 tenths = 0.9
4 tens = 40
Thus, we get
0.82 + 0.9 + 40 = 41.72
Hence, the answers to the given addition operations are
7) 6 Hundredths add to 4 tenths add to 6 ones is equal to 6.46
8) 82 Hundredths add to 9 tenths add to 4 tens is equal to 41.72
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NO LINKS!!! Please help me with this problem
Answer:
9
Step-by-step explanation:
To find the rate of change we use the formula
f(x2) - f(x1)
------------------
x2 -x1
f(x) = 9x
x2 = 9 and x1 = 0
f(x2) = 9( 8) = 72
f(x1) = 9(0) =0
The rate of change is
72 - 0
------------
8-0
72
----
8
9
The rate of change is 9
Answer:
9
Step-by-step explanation:
The average rate of change of function f(x) over the interval a ≤ x ≤ b is given by:
[tex]\dfrac{f(b)-f(a)}{b-a}[/tex]
Given:
f(x) = 9xinterval: 0 ≤ x ≤ 8Therefore:
a = 0b = 8Substitute the given values into the average rate of change formula:
[tex]\begin{aligned}\implies \dfrac{f(8)-f(0)}{8-0} & = \dfrac{9(8)-9(0)}{8-0}\\\\& = \dfrac{72-0}{8-0}\\\\& = \dfrac{72}{8}\\\\& = 9\end{aligned}[/tex]
Therefore, the average rate of change of the function f(x) over the interval 0 ≤ x ≤ 8 is 9.
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The function f(x) is shown in the graph
f(a)
Which type of function describes ((x)?
© Exponential
O Logarithmic
O Rational
O Polynomial
Answer:
the function is an exponential funtion.
Step-by-step explanation:
learned it
What is the solution of the inequality shown
below?
c+3>8
Answer:
c > 5
Step-by-step explanation:
The properties of equality can be used to solve inequalities. Attention needs to be paid to ordering.
ApplicationWe can subtract 3 from both sides to solve this.
c +3 > 8 . . . . . . given
c +3 -3 > 8 -3 . . . . subtract 3 from both sides
c > 5 . . . . . . . . . simplify
The solution is c > 5.
__
Additional comment
Adding or subtracting a value to a number on a number line is equivalent to shifting it right or left. It does not change ordering.
Multiplying or dividing by a positive number is equivalent to expanding or compressing the distance from zero. It does not change ordering.
Multiplying or dividing by a negative number reflects the value across the origin, in addition to expanding or compressing the distance from zero. This reflection reverses the left-right ordering. For example, -2 < -1, but 2 > 1. (Both numbers multiplied by -1.)
As long as you're aware of the effect on ordering, you can use any of the properties of equality to solve inequalities.
Sketch the graphic y=|x+1|
Answer:
Consider the table for y= |x+1| :
x | y
---------
0 1
1 2
2 3
-1 2
-2 3
This would give us the parent function of y=|x| but translated up one unit. It should look like a v starting at (0, 1)
what is the greatest number that can divide 13,17 and 21 and have one as a remainder
Answer:
4
Step-by-step explanation:
this is the same question as what number can divide
13-1 = 12, 17-1 = 16 and 21-1 = 20 and has 0 remainder ?
the greatest number that can do that is 4.
we can easily see that, but formally, let's do prime factorization :
12 ÷ 2 = 6
6 ÷ 2 = 3
3 ÷ 2 no
3 ÷ 3 = 1 finished
12 = 2×2×3
16 ÷ 2 = 8
8 ÷ 2 = 4
4 ÷ 2 = 2
2 ÷ 2 = 1 finished
16 = 2×2×2×2
20 ÷ 2 = 10
10 ÷ 2 = 5
5 ÷ 2 no
5 ÷ 3 no
5 ÷ 5 = 1 finished
20 = 2×2×5
so, the largest common factor is the combination of the longest streaks per factor they have in common.
they only have 2s in common.
and the longest common streak is 2×2 = 4.
hence the answer
Throughout this course, you have examined how real-world scenarios can be modelled using quadratic functions, exponential functions, trigonometric ratios sinusoidal functions, and sequences and series. Part A:- In this task, you will be creating unique real-world problems that can be modelled using the functions that we have learned. You may use real-world scenarios that we have examined throughout the course, but your problem should be created by you and have a unique description. Choose three (3) of the five (5) topics below and create a real-world scenario related to each of the three. 1. Exploring Quadratic Functions to Find Zeros or the Vertex; 2. Exponential Growth or Decay; 3. Using Trigonometric Ratios to Solve Three Dimensional Problems; 4. Representing Periodic Behaviour with Sinusoidal Functions: 5. Solving Financial Problems using Sequences & Series.
PLEASE SOLVE WITHOUT USING RADINAS
The exponential function is illustrated below.
How to illustrate the example?An exponential function has a growth factor or 3.76. What is the percentage growth rate?
The growth factor (b) is given as:
b = 3.76
So, the percentage growth rate (r) is calculated as:
r = b - 1
Substitute known values
r = 3.76 - 1
Evaluate the difference
r = 276%
The way to solve Financial Problems using Sequences & Series will be:
The first salary that Mr James earn is 10000 and there is a yearly increase of 2000. Find his salary in the 5th year. This will be:
= a + (n - 1)d
= 1000 + (5 - 1)2000
= 10000 + (4 × 2000)
= 10000 + 8000.
= 18000
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A restaurant manager has the option of a 30-year loan of $417,000 at an annual interest rate of 3.85% or the same interest rate but on a loan for 15 years.
(a)
Calculate the monthly payment for each loan. (Round your answers to the nearest cent.)
30-year $
15-year $
(b)
Calculate the savings in interest by using the 15-year loan. (Round your answer to the nearest cent.)
$
(c)
The term of the 15-year loan is one-half the term of the 30-year loan. Is the monthly payment for the 15-year loan twice that of the 30-year loan?
Yes
No
(d)
Is the interest savings for the 15-year loan more or less than one-half of the interest paid on the 30-year loan?
more
less
a) The monthly payment for each loan is as follows:
30-year $1,954.93
15-year $3,053.25
b) The savings in interest by using the 15-year loan is $154,189,20 ($286,774.20 - $132,585).
c) No. the monthly payment for the 15-year loan is not twice that of the 30-year loan as the loan term.
d) The interest savings for the 15-year loan are more than one-half of the interest paid on the 30-year loan.
How are the calculations for periodic payments done?The calculations for the monthly payments, including interests can be carried out using an online finance calculator, as follows:
30-year Loan:N (# of periods) = 360 months (12 x 30 years)
I/Y (Interest per year) = 3.85%
PV (Present Value) = $417000
FV (Future Value) = $0
Results:
PMT = $1,954.93
Sum of all periodic payments = $703,774.80 ($1,954.93 x 360)
Total Interest = $286,774.20 ($703,774.80 - $417,000)
15-year Loan:N (# of periods) = 180 months (12 x 15 years)
I/Y (Interest per year) = 3.85%
PV (Present Value) = $417000
FV (Future Value) = $0
Results:
PMT = $3,053.25
Sum of all periodic payments = $549,585 ($3,053.25 x 180)
Total Interest = $132,585 ($549,585 - $417,000)
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For the equation 2x - y = 1, if x = 0, then y = ?
Answer:
y= -1
Step-by-step explanation:
(2) (0) − y = 1
0 + − y = 1
(−y) + (0) = 1
−y = 1
Step 2: Divide both sides by -1
Y = −1
A line contains the point (4, 5) and has a slope of -2.
Which point is also on the line?
(5,7)
(6,2)
(5,3)
(4.1)
Answer: (5,3)
Step-by-step explanation:
Substituting into point-slope form, the equation of the line is
[tex]y-5=-2(x-4)[/tex]
Which rearranges as follows:
[tex]y-5=-2x+8\\\\y=-2x+13[/tex]
To determine if a point lies on a line, you can see if its coordinates satisfy the equation.
Of all the options, only (5,3) works.
What is the standard form equation of an ellipse that has vertices (−2,−18) and (−2,8) and foci (−2,−14) and (−2,4)?
Answer:
Hello,
Step-by-step explanation:
All is in the picture.
B=(-2,8), O=(-2,-5)
b=BO=8+5=13
F_1=(-2,4) O=(-2,5) Focus distance=4+5=9
Horizontal half axis=√(b²-f²)=√88
please help!!
maths functions
Answer:
The Co-ordinate of C is (4/3, -1/2)
Step-by-step explanation:
We have two equation one is of straight line equation which is:
y=2x-3 (i)
Other equation is of quadratic function which is:
y=-3x^2+5 (ii)
Put the value of y from equation (i) in equation (ii)
So, we have:
2x-3=-3x^2+5
3x^2+2x-8=0
By factorization:
3x^2+6x-4x-8=0
3x(x+2)-4x(x+2)=0
(x+2)(3x-4)=0
x+2=0 ; 3x-4=0
x=-2 ; x=4/3
Put first x=-2 in equation (i)
y=2(-2)-3
y=-4-3
y=-7
Now Put x=4/3 in equation (i)
y=2(4/3)-3
y=8/3-3
y=-1/2
So, we have two Order pair One is (-2 , -7) and Second one is (4/3 , -1/2)
Hence the Co-ordinate of C is:
C=(4/3 , -1/2)
Answer:
Point C: (3, 3)
Point D: (3, -22)
Step-by-step explanation:
If the distance between points C and D is 25 units, the y-value of point D will be 25 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-3[/tex]
[tex]\textsf{Equation 2}: \quad y-25=-3x^2+5[/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-3-25=-3x^2+5[/tex]
[tex]\implies 3x^2+2x-33=0[/tex]
[tex]\implies 3x^2-9x+11x-33=0[/tex]
[tex]\implies 3x(x-3)+11(x-3)=0[/tex]
[tex]\implies (x-3)(3x+11)=0[/tex]
[tex]\implies x=3, -\dfrac{11}{3}[/tex]
From inspection of the given graph, the x-value of points C and D is positive, therefore x = 3.
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-3 & =y\\2(3)-3 & =3\\ \implies & (3, 3)\end{aligned}[/tex]
[tex]\begin{aligned} \textsf{Point D}: \quad -3x^2+5 & = y \\ -3(3)^2+5 & =-22\\ \implies & (3, -22)\end{aligned}[/tex]
Therefore, point C is (3, 3) and point D is (3, -22).
If the equation below is solved by graphing, which statement is true? log (6 x + 10) = log 1/2 x
The solution to the given expression is x = -20/11
Logarithmic function are inverse of exponential functions. Given the equation below;
log (6 x + 10) = log 1/2 x
In order to determine the solution to the given logarithmic equation, we will first have to cancel the logarithm on both sides to have
6x + 10 = 1/2x
Collect the like terms
6x - 1/2x = 0 - 10
Find the LCD
12x-x/2 = -10
11x/2 = -10
Cross multiply
11x = -2 * 10
11x = -20
Divide both sides by 11
11x/11 = -20/11
x = -20/11
Hence the solution to the given expression is x = -20/11
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What is the range of the exponential function f(x) = 2*+25? Check all that
apply.
A. (2,+00)
B. (25,+00)
C. f(x) 22
D. f(x) > 25
A normal population has a mean u = 31 and standard deviation = 10. What proportion of the population is less than 30?
The proportion of the population exists less than 30 then
(x< 30) = 0.986.
How to estimate the proportion of the population that exists less than 30?To estimate the z-score using the formula, z = (x - µ)/σ
Where, x be the randomly chosen values = 30
µ be the mean = 31
σ be the standard deviation = 10
Proportion of the population that exists less than 18 = P(x < 30)
Plug in the values into z = (x - µ)/σ, to get z-score.
Substitute the values in the above equation, we get
z = (30 - 31)/10
z = -1/10 = -0.1
Therefore, the value of z = - 0.1.
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Which of the following options have the same value as 5\%5%5, percent of 353535?
Answer:
[tex]\frac{5}{100}\times35[/tex]
[tex]0.05\times35[/tex]
Step-by-step explanation:
Given:
Which of the following options have the same value as 5% 35, percent of 35?
Following Options:
[tex]5 \times 35[/tex]
[tex]\frac{5}{100}\times35[/tex]
[tex]0.5\times0.35[/tex]
[tex]0.05\times35[/tex]
[tex]\frac{5}{10}\times35[/tex]
Solve:
[tex]5[/tex] % [tex]= 0.05=\frac{5}{100}[/tex]
Thus the following options:
[tex]5 \times 35[/tex] [ False x ]
5 does not equal 5%
[tex]\frac{5}{100}\times35[/tex] [True √ ]
[tex]5[/tex]% [tex]=\frac{5}{100}[/tex]
[tex]0.5\times0.35[/tex] [ False x ]
[tex]0.5 = 0.50[/tex]
[tex]0.05\times35[/tex] [True √ ]
[tex]5[/tex]% [tex]= 0.05=\frac{5}{100}[/tex]
[tex]\frac{5}{10}\times35[/tex] [ False x ]
[tex]\frac{5}{10}=0.50[/tex]
Therefore, the options [B] [tex]\frac{5}{100}\times35[/tex] and [D] [tex]0.05\times35[/tex] is True.
Kavinsky
Find d²y/dx² for implicitly in terms of x and y
xy-1=2x+y²
The second derivative of the implicit function x · y - 1 = 2 · x + y² is equal to y'' = [2 / (2 · y - x)] · [(2 - y) / (x - 2 · y)] · [1 - [(2 - y) / (x - 2 · y)]].
What is the second derivative of an implicit equation?
In this problem we have a function in implicit form, that is, an expression of the form: f(x, y, c) = 0, where c is a constant. Then, we should apply implicit differentiation twice to determine the second derivative of the function:
Original expression
x · y - 1 = 2 · x + y²
First derivative
y + x · y' = 2 + 2 · y · y'
(x - 2 · y) · y' = 2 - y
y' = (2 - y) / (x - 2 · y)
Second derivative
y' + y' + x · y'' = 2 · (y')² + 2 · y · y''
2 · y' - 2 · (y')² = (2 · y - x) · y''
y'' = 2 · [y' - (y')²] / (2 · y - x)
y'' = [2 / (2 · y - x)] · [(2 - y) / (x - 2 · y)] · [1 - [(2 - y) / (x - 2 · y)]]
The second derivative of the implicit function x · y - 1 = 2 · x + y² is equal to y'' = [2 / (2 · y - x)] · [(2 - y) / (x - 2 · y)] · [1 - [(2 - y) / (x - 2 · y)]].
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NO LINKS! Help me with this problem
[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]
Let's solve ~
Equation of directrix is : y = 1, so we can say that it's a parabola of form : -
[tex]\qquad \sf \dashrightarrow \: (x - h) {}^{2} = 4a(y - k)[/tex]
h = x - coordinate of focus = -4k = y - coordinate of focus = 5a = half the perpendicular distance between directrix and focus = 1/2(5 - 1) = 1/2(4) = 2and since the focus is above the directrix, it's a parabola with upward opening.
[tex]\qquad \sf \dashrightarrow \: (x - ( - 4)) {}^{2} = 4(2)(y - 5)[/tex]
[tex]\qquad \sf \dashrightarrow \: (x + 4) {}^{2} = 8(y - 5)[/tex]
[tex]\qquad \sf \dashrightarrow \: {x}^{2} + 8x + 16 = 8y - 40[/tex]
[tex]\qquad \sf \dashrightarrow \: 8y = {x}^{2} + 8x + 56[/tex]
[tex]\qquad \sf \dashrightarrow \: y = \cfrac{1}{8} {x}^{2} + x + 7[/tex]
Directrix
y=1Focus
(h,k)=(-4,5)Focus lies in Q3 and above y=1
Parabola is opening upwardsThen
Perpendicular distance
(5-1)=4Find a for the equation
a=4/2=2Now the equation is
[tex]\\ \rm\dashrightarrow 4a(y-k)=(x-h)^2[/tex]
[tex]\\ \rm\dashrightarrow 4(2)(y-5)=(x+4)^2[/tex]
[tex]\\ \rm\dashrightarrow 8(y-5)=x^2+8x+16[/tex]
[tex]\\ \rm\dashrightarrow 8y-40=x^2+8x+16[/tex]
[tex]\\ \rm\dashrightarrow 8y=x^2+8x+16+40[/tex]
[tex]\\ \rm\dashrightarrow 8y=x^2+8x+56[/tex]
[tex]\\ \rm\dashrightarrow y=\dfrac{x^2}{8}+x+7[/tex]
If t1 = 4, s1 = 5, and s2 = 2, determine the value of t2.
Answer:
t2=8/5
Step-by-step explanation:
using this formula
t1/s1 =t2/s2
4/5=t2/2
cross multiply
5t2=8
t2=8/5
The correct answer for the value of t₂ is [tex]1.6[/tex].
Given:
Time t₁ = 4,
Distance s₂ =2
Distance s₁ = 5.
To find value of t₂ , use the concept of proportion:
[tex]\dfrac{t_1}{s_1} = \dfrac{t_2}{s_2}[/tex]
Put value of [tex]t_1 ,s_1 ,s_2[/tex]:
[tex]\dfrac{t_2}{2} =\dfrac{4}{5}\\\\t_2 =\dfrac{8}{5}\\\\ t_2 = 1.6[/tex]
The correct value of [tex]t_2[/tex] is [tex]1.6[/tex].
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What is the measure of
angle x?
Enter your answer in the box.
X =
Answer:
x = 48°
Step-by-step explanation:
Complementary angles
Angles that sum to 90°.
Vertical Angle Theorem
When two straight lines intersect, the vertical angles are congruent (equal).
Therefore, angle x is equal to the angle that is complementary to 42°.
To find x, subtract 42° from 90°:
⇒ x = 90° - 42°
⇒ x = 48°
Is 25x²-40xy+16y²a perfect square number? why?
Answer:
yes
Step-by-step explanation:
25x² - 40xy + 16y² can be factored as
(5x - 4y)² ← a perfect square
Evaluate the following expression at x = 3 and y = -4. 7x - 3y + 2.
provide your answer below:
Answer:
35
Step-by-step explanation:
first, you look at 7x, from the previous equation, you know that x=3, so you take 7x3=21 then you evaluate -3y. as you did with x on the last one you will look at the equation for y and see that it's -4. A negative times a negative is a positive, so -4x(-3)= 12. Then you add them all together, since 12 is a positive, the equation would now look like 21+12+2. After adding all three numbers together, you get 12.
PLEASE HELP ASAP !!!!
Which functions have a range of {y e R-00 < y < ∞0}?
O f(x) = -4x + 11
Of(x) = x - 8
O
f(x) = 2x+3
O f(x) = -(x + 1)² - 4
O
f(x)
f(x) = x²
x² + 7x9
Answer:
Options 1 and 2
Step-by-step explanation:
Correct. The range of all linear functions without a restricted domain is the set of all real numbers.Correct. The range of all linear functions without a restricted domain is the set of all real numbers.Wrong. The exponential cannot take negative values.Wrong. Quadratics do not have a range that is the set of real numbers.Wrong. Quadratics do not have a range that is the set of real numbers.If [tex]\mathrm {y = (x + \sqrt{1+x^{2}})^{m}}[/tex], then prove that [tex]\mathrm {(x^{2} +1)y_{2} +x y_{1} - m^{2}y = 0}[/tex].
Note : y₁ and y₂ refer to the first and second derivatives.
Answer:
See below for proof.
Step-by-step explanation:
Given:
[tex]y=\left(x+\sqrt{1+x^2}\right)^m[/tex]
First derivative
[tex]\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}[/tex]
[tex]\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}[/tex]
Second derivative
[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}[/tex]
[tex]\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}[/tex]
[tex]\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}[/tex]
[tex]\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m[/tex]
[tex]\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m[/tex]
[tex]\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}[/tex]
[tex]= \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)[/tex]
Proof
[tex](x^2+1)y_2+xy_1-m^2y[/tex]
[tex]= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m[/tex]
[tex]= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m[/tex]
[tex]= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right][/tex]
[tex]= \left(x+\sqrt{1+x^2}\right)^m\left[0][/tex]
[tex]= 0[/tex]
Which of the triangles in the diagram are congruent?
Triangle 1, triangle 3 and triangle 4 are congruent triangles bases on side-side-side and side-angle-side congruency.
What are congruent triangles?Triangle is a polygon that has three sides and three angles. Types of triangles are isosceles, equilateral and scalene triangle.
Two triangles are said to be congruent if they have the same shape and their corresponding sides are congruent to each other. Also, their corresponding angles are congruent.
Triangle 1, triangle 3 and triangle 4 are congruent triangles bases on side-side-side and side-angle-side congruency.
Find out more on congruent triangles at: https://brainly.com/question/2938476
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