Answer:
y = - [tex]\frac{4}{3}[/tex] x
Step-by-step explanation:
1). [tex]y_{1}[/tex] = [tex]m_{1}[/tex] [tex]x_{1}[/tex] + [tex]b_{1}[/tex]
[tex]y_{2}[/tex] = [tex]m_{2}[/tex] [tex]x_{2}[/tex] + [tex]b_{2}[/tex]
[tex]y_{1}[/tex] ⊥ [tex]y_{2}[/tex] if [tex]m_{2}[/tex] = - [tex]\frac{1}{m_{1} }[/tex]
2). ( [tex]x_{3}[/tex] , [tex]y_{3}[/tex] )
y - [tex]y_{3}[/tex] = m( x - [tex]x_{3}[/tex] )
~~~~~~~~~~~~~~~~~~
y = [tex]\frac{3}{4}[/tex] x + 7
[tex]m_{1}[/tex] = [tex]\frac{3}{4}[/tex]
[tex]m_{2}[/tex] = - [tex]\frac{4}{3}[/tex]
( 3, - 4 )
y - ( - 4) = - [tex]\frac{4}{3}[/tex] ( x - 3 )
y + 4 = - [tex]\frac{4}{3}[/tex] x + 4
y = - [tex]\frac{4}{3}[/tex] x
What is the value of X? HELP
with no further informations, just go by looking at it.
it's 90°, all other options are too far off
Write in a shorter form:7m -7 +7m +7
Answer:
14m
Step-by-step explanation:
[tex]7m-7+7m-7\\[/tex]
First, we need to eliminate the like term and collect the like term.
[tex]-7+-7=0[/tex]
Now, we have 7m +7m, sum them up and you will get the answer.[tex]7m+7m=14m[/tex]
So, the answer is 14m.
Answer:
14m
Step-by-step explanation:
7m -7 +7m +7
7m + 7m - 7 + 7
14m
tentukan himpinan penyelasaian dari plsv berikut. A:3x+=2x+12
Answer:
x=12
Step-by-step explanation:
3x=2x+12
3x-2x=12
x=12
Define the operation a∇b = 2+b^a What is the value of (1∇2)∇3?
Answer:
83
Step-by-step explanation:
1∇2= (2+2^1)
=2+2=4
(4)∇3= (2+3^4)
=2+81
=83
Abigail plans to repaint some classroom bookcases. She has 6/25
gallons of paint. All of the bookcases are the same size and each requires 2/3
gallon of paint. How many bookcases will the custodian be able to repaint with that amount of paint?
Answer:
Step-by-step explanation: Hello! Do
I have know idea how to do this problem and teach my child
k(t)=13t-2
Answer:
K(t)=37
Step-by-step explanation:
k(t) = 13t - 2
k(3) = 13(3) - 2
k(3) = 39 - 2
k(3) = 37 <===
What is the shape of a sorbital
Answer:
Spherical-Like Shape
Step-by-step explanation:
An s-orbital is spherical with the nucleus at its center.
Solve the formula for t
V = 4(3.14)ct + 6(3.14)c^2
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Answer:
t = (V -6(3.14)c^2)/(4(3.14)c)
Step-by-step explanation:
Isolate the term containing t, then divide by the coefficient of t
[tex]V=4(3.14)ct+6(3.14)c^2\\\\V-6(3.14)c^2=t(4(3.14)c)\\\\\boxed{t=\dfrac{V-6(3.14)c^2}{4(3.14)c}}[/tex]
What is the length of the line?
The percent of data between z=0.23 and z = 1.27 is
(Round to two decimal places as needed.)
Answer:
0.40905 - 0.10204 = .30701 = 30.7 %
Step-by-step explanation:
0.23 0.40905
1.27 0.10204
Given that V=4/3πrcube , make r the subject of formula
Answer:
[tex]V = \frac{4}{3} \pi {r}^{3} \\ \\ 3V = 4\pi {r}^{3} \\ {r}^{3} = \frac{3V}{4\pi} \\ \\ r = \sqrt[3]{ \frac{3V}{4\pi} } [/tex]
.......... is a factor of every even number.
Answer:
2 is the factor of every even number hope this help you
Find the differential coefficient of
[tex]e^{2x}(1+Lnx)[/tex]
Answer:
[tex] \rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} [/tex]
Step-by-step explanation:
we would like to figure out the differential coefficient of [tex]e^{2x}(1+\ln(x))[/tex]
remember that,
the differential coefficient of a function y is what is now called its derivative y', therefore let,
[tex] \displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )[/tex]
to do so distribute:
[tex] \displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x} [/tex]
take derivative in both sides which yields:
[tex] \displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )[/tex]
by sum derivation rule we acquire:
[tex] \rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x} [/tex]
Part-A: differentiating $e^{2x}$
[tex] \displaystyle \frac{d}{dx} {e}^{2x} [/tex]
the rule of composite function derivation is given by:
[tex] \rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)[/tex]
so let g(x) [2x] be u and transform it:
[tex] \displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x[/tex]
differentiate:
[tex] \displaystyle {e}^{u} \cdot 2[/tex]
substitute back:
[tex] \displaystyle \boxed{2{e}^{2x} }[/tex]
Part-B: differentiating ln(x)•e^2x
Product rule of differentiating is given by:
[tex] \displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)[/tex]
let
[tex]f(x) \implies \ln(x) [/tex][tex]g(x) \implies {e}^{2x} [/tex]substitute
[tex] \rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x} [/tex]
differentiate:
[tex] \rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }[/tex]
Final part:
substitute what we got:
[tex] \rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }[/tex]
and we're done!
Answer:
Product Rule for Differentiation
[tex]\textsf{If }y=uv[/tex]
[tex]\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}[/tex]
Given equation:
[tex]y=e^{2x}(1+\ln x)[/tex]
Define the variables:
[tex]\textsf{Let }u=e^{2x} \implies \dfrac{du}{dx}=2e^{2x}[/tex]
[tex]\textsf{Let }v=1+\ln x \implies \dfrac{dv}{dx}=\dfrac{1}{x}[/tex]
Therefore:
[tex]\begin{aligned}\dfrac{dy}{dx} & =u\dfrac{dv}{dx}+v\dfrac{du}{dx}\\\\\implies \dfrac{dy}{dx} & =e^{2x} \cdot \dfrac{1}{x}+(1+\ln x) \cdot 2e^{2x}\\\\& = \dfrac{e^{2x}}{x}+2e^{2x}(1+\ln x)\\\\ & = \dfrac{e^{2x}}{x}+2e^{2x}+2e^{2x} \ln x\\\\& = e^{2x}\left(\dfrac{1}{x}+2+2 \ln x \right)\end{aligned}[/tex]
Geometry Identify the sides or angles that need to be congruent in order to make the given triangles congruent by AAS. Please help me!!!!!!!!!
Answer:
A. Sides AC and DF
B. Angles BAC and EDF
Answer:
Step-by-step explanation:
8 meters for every 2inches what is the area of 144 meters squared
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Answer:
9 square inches
Step-by-step explanation:
The area is proportional to the square of the linear scale factor. We can use this to write the proportion ...
A/(144 m²) = ((2 in)/(8 m))²
A = (144·4/64) in² = 9 in²
The area representing 144 square meters is 9 square inches.
You found the prime factorization of the number 73 explain how you can check your answer
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Explanation:
You can check your answer by making sure that each of the primes you found is actually a prime. (Compare to a list of known primes, for example.) After you have determined your factors are primes, multiply them together to see if the result is 73. If so, you have found the correct prime factorization.
__
Additional comment
73 is prime, so its prime factor is 73.
73 = 73
They are 10 ice cream flavors, 5 different toppings and it could be either in cup or in cone, how many 2-scoop combinations are possible?
Using the fundamental counting principle, it is found that: 50 2-scoop combinations are possible.
----------------------------------
The flavors and the toppings are independent, which means that the fundamental counting principle is used to solve this question, which states that if there are p ways to do a thing, and q ways to do another thing, and these two things are independent, there are p*q ways to do both things.
----------------------------------
In this question:
10 ice cream flavors.5 toppings.So,
[tex]10 \times 5 = 50[/tex]
50 2-scoop combinations are possible.
A similar question is found at https://brainly.com/question/24067651
Use the power series method to solve the given initial-value problem. (Format your final answer as an elementary function.)
(x − 1)y'' − xy' + y = 0, y(0) = −7, y'(0) = 3
You're looking for a solution of the form
[tex]\displaystyle y = \sum_{n=0}^\infty a_n x^n[/tex]
Differentiating twice yields
[tex]\displaystyle y' = \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n[/tex]
[tex]\displaystyle y'' = \sum_{n=0}^\infty n(n-1) a_n x^{n-2} = \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n[/tex]
Substitute these series into the DE:
[tex]\displaystyle (x-1) \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n - x \sum_{n=0}^\infty (n+1) a_{n+1} x^n + \sum_{n=0}^\infty a_n x^n = 0[/tex]
[tex]\displaystyle \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^{n+1} - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=0}^\infty (n+1) a_{n+1} x^{n+1} + \sum_{n=0}^\infty a_n x^n = 0[/tex]
[tex]\displaystyle \sum_{n=1}^\infty n(n+1) a_{n+1} x^n - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=1}^\infty n a_n x^n + \sum_{n=0}^\infty a_n x^n = 0[/tex]
Two of these series start with a linear term, while the other two start with a constant. Remove the constant terms of the latter two series, then condense the remaining series into one:
[tex]\displaystyle a_0-2a_2 + \sum_{n=1}^\infty \bigg(n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}-na_n+a_n\bigg) x^n = 0[/tex]
which indicates that the coefficients in the series solution are governed by the recurrence,
[tex]\begin{cases}y(0)=a_0 = -7\\y'(0)=a_1 = 3\\(n+1)(n+2)a_{n+2}-n(n+1)a_{n+1}+(n-1)a_n=0&\text{for }n\ge0\end{cases}[/tex]
Use the recurrence to get the first few coefficients:
[tex]\{a_n\}_{n\ge0} = \left\{-7,3,-\dfrac72,-\dfrac76,-\dfrac7{24},-\dfrac7{120},\ldots\right\}[/tex]
You might recognize that each coefficient in the n-th position of the list (starting at n = 0) involving a factor of -7 has a denominator resembling a factorial. Indeed,
-7 = -7/0!
-7/2 = -7/2!
-7/6 = -7/3!
and so on, with only the coefficient in the n = 1 position being the odd one out. So we have
[tex]\displaystyle y = \sum_{n=0}^\infty a_n x^n \\\\ y = -\frac7{0!} + 3x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots[/tex]
which looks a lot like the power series expansion for -7eˣ.
Fortunately, we can rewrite the linear term as
3x = 10x - 7x = 10x - 7/1! x
and in doing so, we can condense this solution to
[tex]\displaystyle y = 10x -\frac7{0!} - \frac7{1!}x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots \\\\ \boxed{y = 10x - 7e^x}[/tex]
Just to confirm this solution is valid: we have
y = 10x - 7eˣ ==> y (0) = 0 - 7 = -7
y' = 10 - 7eˣ ==> y' (0) = 10 - 7 = 3
y'' = -7eˣ
and substituting into the DE gives
-7eˣ (x - 1) - x (10 - 7eˣ ) + (10x - 7eˣ ) = 0
as required.
Find the area of the shaded regions:
QUICKLY!!!!!
Answer:
[tex]41.89\ cm^2[/tex]
Step-by-step explanation:
[tex]We\ are\ given:\\In\ two\ concentric\ circles,\\OD=3\ cm\\BC=4\ cm\\\angle DOB=\angle AOC=120\\Now,\\We\ know\ that:\\Area\ of\ a\ sector\ with\ a\ central\ angle\ \theta\ and\ a\ radius\ r\ is:\\A=\frac{\theta}{360}* \pi r^2\\Here,\\Area\ between\ the\ sectors=Area\ of\ Larger\ Sector - Area\ of\ smaller\ sector=\frac{\theta}{360}*\pi(R^2-r^2),\ where\ R\ and\ r\ are\ radii\ of\ the\ respective\ circles\ and\\ \theta\ is\ the\ common\ central\ angle.\\Here,\\R=4+3=7\ cm\\r=3\ cm\\ \theta=120\\ Hence,[/tex]
[tex]Area\ of\ the\ shaded\ region=\frac{120}{360}*\pi(7^2-3^2)=\frac{1}{3}*\pi(49-9)=\frac{1}{3}*\pi(40) \approx 41.89\ cm^2[/tex]
A triangle has side lengths of (4.6x-4.4y)(4.6x−4.4y) centimeters, (7.5x-8.8z)(7.5x−8.8z) centimeters, and (7.7z-9.2y)(7.7z−9.2y) centimeters. Which expression represents the perimeter, in centimeters, of the triangle?
The expression that represents the perimeter, in centimeters, of the triangle is [tex]77.41x^{2}+104y^{2}+ 136.73z^{2} -40.48xy - 132xz - 141.64yz[/tex] cm OR 77.41x²+104y²+136.73z²-40.48xy-132xz-141.64yz centimeters
From the question, the side lengths are
(4.6x-4.4y)(4.6x−4.4y) cm, (7.5x-8.8z)(7.5x−8.8z) cm, and (7.7z-9.2y)(7.7z−9.2y) cm.
First, we will clear the brackets one after the other
For (4.6x-4.4y)(4.6x−4.4y) cm
[tex]4.6x(4.6x-4.4y) -4.4y(4.6x-4.4y)[/tex]
[tex]21.16x^{2} -20.24xy -20.24xy+19.36y^{2}[/tex]
[tex]21.16x^{2} -40.48xy+19.36y^{2}[/tex]
∴ (4.6x-4.4y)(4.6x−4.4y) cm = [tex]21.16x^{2} -40.48xy+19.36y^{2}[/tex] cm
For (7.5x-8.8z)(7.5x−8.8z) cm
[tex]7.5x(7.5x-8.8z) -8.8z(7.5x-8.8z)[/tex]
[tex]56.25x^{2} - 66xz -66xz + 77.44z^{2}[/tex]
[tex]56.25x^{2} - 132xz + 77.44z^{2}[/tex]
∴ (7.5x-8.8z)(7.5x−8.8z) cm = [tex]56.25x^{2} - 132xz + 77.44z^{2}[/tex] cm
For (7.7z-9.2y)(7.7z−9.2y) cm
[tex]7.7z(7.7z-9.2y)-9.2y(7.7z-9.2y)[/tex]
[tex]59.29z^{2} - 70.84yz-70.84yz+84.64y^{2}[/tex]
[tex]59.29z^{2} - 141.64yz+84.64y^{2}[/tex]
∴ (7.7z-9.2y)(7.7z−9.2y) cm = [tex]59.29z^{2} - 141.64yz+84.64y^{2}[/tex] cm
Now, for the expression that represents the perimeter of the triangle,
Perimeter of a triangle can be calculated by determining the sum of all its sides
That is,
Perimeter of the triangle = [tex]21.16x^{2} -40.48xy+19.36y^{2}[/tex] cm + [tex]56.25x^{2} - 132xz + 77.44z^{2}[/tex] cm + [tex]59.29z^{2} - 141.64yz+84.64y^{2}[/tex] cm
[tex]=21.16x^{2} -40.48xy+19.36y^{2} + 56.25x^{2} - 132xz + 77.44z^{2} + 59.29z^{2} - 141.64yz+84.64y^{2}[/tex]
Collect like terms
[tex]= 21.16x^{2}+ 56.25x^{2} -40.48xy+19.36y^{2}+84.64y^{2} - 132xz + 77.44z^{2} + 59.29z^{2} - 141.64yz[/tex]
[tex]= 77.41x^{2} -40.48xy+104y^{2} - 132xz + 136.73z^{2} - 141.64yz[/tex]
[tex]= 77.41x^{2}+104y^{2}+ 136.73z^{2} -40.48xy - 132xz - 141.64yz[/tex] cm
Hence, the expression that represents the perimeter, in centimeters, of the triangle is [tex]77.41x^{2}+104y^{2}+ 136.73z^{2} -40.48xy - 132xz - 141.64yz[/tex] cm OR 77.41x²+104y²+136.73z²-40.48xy-132xz-141.64yz centimeters
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A vector has a magnitude of 43
meters at an angle (0). If the
Y-component of the vector is 30
meters, what is the X-component?
[?]m
30.8 m
Step-by-step explanation:
Given: [tex]V = 43\:\text{m}[/tex], [tex]V_y = 30\:\text{m}[/tex]
The x-component of vector [tex]\vec{\text{V}}[/tex] is
[tex]V_x = \sqrt{V^2 - V_y^2} = \sqrt{(43)^2 - (30)^2} = 30.8\:\text{m}[/tex]
The x- component of the vector is 30.8meters.
What is the magnitude and direction of vector?If [tex]v = < a. b >[/tex] be a position vector then the magnitude of vector v is found by [tex]|v| =\sqrt{a^{2}+b^{2} } }[/tex] , where a and b are the x and y component respectively.
And the direction is equals to the angle formed x- axis or y axis.
According to the given question
We have
Magnitude of the vector, |v| = 43meters
Y- component of the vector, b = 30meters
Since, we know that
[tex]|v| =\sqrt{a^{2} +b^{2} }[/tex]
Substitute the value of |v| = 43 and b = 30 in the above formula of magnitude.
⇒ [tex]43 = \sqrt{a^{2}+30^{2} }[/tex]
⇒ [tex]43 = \sqrt{a^{2}+900 }[/tex]
⇒ [tex]43^{2} =a^{2} + 900[/tex]
⇒ [tex]1849 = a^{2} + 900[/tex]
⇒ [tex]1849-900=a^{2}[/tex]
⇒ [tex]949=a^{2}[/tex]
⇒ [tex]a =\sqrt{949}[/tex]
⇒ [tex]a = 30.8[/tex]
Hence, the x- component of the vector is 30.8meters.
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Determine x&y
(2+i) (x+yi) = -7+3i
Answer:
x = -11/5 or -2.2
y = 13/5 or 2.6
Step-by-step explanation:
well, start by doing the multiplication. then we will see better.
2x + 2yi + xi + yii = -7 + 3i
2x + 2yi + xi - y = -7 + 3i
this is because, remember, i = sqrt(-1), and ii = -1.
now we group the i-factors and the terms without i and compare it to the corresponding parts on the right side.
2x - y = -7
2yi + xi = 3i
=> 2y + x = 3
x = 3 - 2y
and that we use ihr the first equation again
2×(3-2y) - y = -7
6 - 4y - y = -7
-5y = -13
y = 13/5
x = 3 - 2×13/5 = 3 - 26/5 = 15/5 - 26/5 = -11/5
What is the value of x that makes l1||l2
A. 35
B. 25
C. 37
D. 18
Answer:
B
Step-by-step explanation:
For l1 and l2 to be parallel, these two angles need to be equal. 3x-15=2x+10, x=25
Complete the function table.
Input (n) Output (n-2)
Answer: Choice C
This is because the input n = 2 leads to the output n-2 = 2-2 = 0
As another example: the input n = 4 leads to the output n-2 = 4-2 = 2
Whatever the input is, subtract 2 from it to get the output.
A 100.0 m long polymer cable of uniform circular cross section and of diameter 0.4cm has a mass of 1885.0 gm. What is the mass density, of the polymer in kg/m3?
The mass density in kg/m3 will be - 15 x [tex]10^{-2}[/tex] Kg/m3.
We have a 100.0 m long polymer cable of uniform circular cross section and of diameter 0.4cm has a mass of 1885.0 gm.
We have to determine its mass density in kg/m3.
What is Mass density ?The amount of mass per unit volume present in the body is called its mass density.
According to question, we have -
Length of polymer cable = 100.0 m
diameter of polymer cable = 0.4 cm = 0.004 m
Therefore, its radius = 0.002 m
The mass density of the wire will be -
[tex]\rho =\frac{m}{\pi r^{2} l}[/tex]
[tex]\rho[/tex] = [tex]\frac{1885}{3.14 \times0.002 \times 0.002 \times 100 }[/tex]
[tex]\rho = \frac{1885}{0.001256}[/tex] = 1500796.1 g/m3
1 Kg = 1000g
1g = 1/1000kg
1500796.1g = 1500.7 Kg = 15 x [tex]10^{-2}[/tex] Kg
Therefore, mass density = 15 x [tex]10^{-2}[/tex] Kg/m3
Hence, the mass density in kg/m3 will be - 15 x [tex]10^{-2}[/tex] Kg/m3.
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A shoe store carries one brand of shoe in 4
different styles, 5 sizes, and 5 colors. How many
different shoes are available of this one brand?
Answer:
100 different shoes
Step-by-step explanation:
4 styles * 5 sizes * 5 colors
4*5*5 = 100
if the area of the triangle is 5cm^2. find the angle.
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Answer:
the angle is 30°
Step-by-step explanation:
The area of a triangle given two sides and the included angle is ...
A = 1/2ab·sin(C)
5 cm² = 1/2(4 cm)(5 cm)sin(θ)
0.5 = sin(θ) . . . . . . divide by 10 cm²
θ = arcsin(.5) ≈ 30°
_____
Additional comment
An obtuse angle of 150° in that location will give a triangle with the same area.
a bag contains 7 red chips and 11 blue chips. two chips are selected randomly without replacement from the bag. what is the probability that the two chips are NOT the same coler
Answer:
77/306 or around 25.2%
Step-by-step explanation:
[tex]\frac{7}{18} *\frac{11}{17}[/tex] section 1/total * section 2/(total-1) since there is no replacement
just solve and you get 77/306
Suppose y varies inversely with x, and y = 32 when x = 4. What is the value of y when x = 8?
a. 1/8
b. 64
c. 16
d. 8
NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW!!!
Answer:
16
Step-by-step explanation:
Inverse variation is of the form
xy = k where k is a constant
x=4 and y = 32
4*32 = k
128 = k
xy = 128
Let x = 8
8y = 128
Divide each side by 8
8y/8 = 128/8
y =16
convert the following to decimal fractions 99 by 5
Answer:
divide 99 by 5
99/5= 19.8