In the picture below, which lines are lines of symmetry for the figure?



A. 1 and 3
B. only 1
C. only 3
D. 2 and 4

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

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

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]

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

with no further informations, just go by looking at it.

it's 90°, all other options are too far off

Answer:

2 is the factor of every even number hope this help you

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

Answer:

x=12

Step-by-step explanation:

3x=2x+12

3x-2x=12

x=12

Answer:

divide 99 by 5

99/5= 19.8

Answer:

I would rather use:

(b) 10,000 batches of 10 observations each.

Step-by-step explanation:

It is easier to have 10,000 batches of 10 observations each than to have 50 batches of 2,000 observations.  Human errors are reduced with fewer observations. For example, Hadoop, a framework used for storing and processing big data, relies on batch processing.  Using batch processing that divides the 100,000 stationary waiting times into 10 observations with 10,000 batches each is more efficient than having 2,000 observations with 50 batches each.

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.

Answer:

0.40905 - 0.10204 = .30701 = 30.7 %

Step-by-step explanation:

0.23 0.40905

1.27 0.10204

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:

So,

[tex]10 \times 5 = 50[/tex]

50 2-scoop combinations are possible.

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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.

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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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.

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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9514 1404 393

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

9514 1404 393

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.

9514 1404 393

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.

Answer:

100 different shoes

Step-by-step explanation:

4 styles * 5 sizes * 5 colors

4*5*5 = 100

Answer:

A. Sides AC and DF

B. Angles BAC and EDF

Answer:

Step-by-step explanation:

Answer:

K(t)=37

Step-by-step explanation:

k(t) = 13t - 2

k(3) = 13(3) - 2

k(3) = 39 - 2

k(3) = 37 <===

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]

Answer:

Step-by-step explanation: Hello! Do

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

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]

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

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.

Answer:

83

Step-by-step explanation:

1∇2= (2+2^1)

=2+2=4

(4)∇3= (2+3^4)

=2+81

=83

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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Answer:

Spherical-Like Shape

Step-by-step explanation:

An s-orbital is spherical with the nucleus at its center.