please explain it step by step​

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:

divide 99 by 5

99/5= 19.8

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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Answer:

[tex]P'(x)=2.2-0.12x[/tex]

Step-by-step explanation:

start by finding [tex]P(x) = R(x)-C(x)[/tex]

[tex]P(x)=3x-0.06x^2-286-0.8x[/tex]

to find P'(x), you take the derivative of P(x)

[tex]P'(x)=\frac{d}{dx} (2.2x-286-0.06x^2)[/tex]

[tex]\frac{d}{dx} (2.2x-286-0.06x^2)=2.2-0.12x[/tex]

so [tex]P'(x)=2.2-0.12x[/tex]

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.

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

Answer:

40°

Step-by-step explanation:

x +2x+50=180 [sum of interior angles of a triangle]

or,3x=180-50

or,x=130/3

x=130/3

2x=2×130/3

Answer:

83

Step-by-step explanation:

1∇2= (2+2^1)

=2+2=4

(4)∇3= (2+3^4)

=2+81

=83

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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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:

-2 would be right next to -3 because its negative and -1 would be right next to -2, 2 would be two points away from 0 bc its a whole number

Answer: 5 and 6

Step-by-step explanation:

X^2 - 11 = (X-1)^2

X^2 - 11 = X^2-2X+1

X^2 - X^2 + 2X = 11+1

2X = 12

X = 6

The preceding number is 5

(6)(6)=36 and (5)(5)=25

36-25=11

The number required is 6

If the number is multiplied by itself, it becomes x²

If the result is 11 greater than the preceding number when it is multiplied by itself is expressed as:

x² - 11  = (x - 1)²

x² - 11 = x² - 2x + 1

2x = 11 + 1

2x = 12

x = 6

Hence the number required is 6

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Answer:

Step-by-step explanation:

6x=1/2(2x +7)            Multiply both sides by 2

2*6x = 1/2(2x + 7)*2

12x = 2x + 7              Subtract 2x from sides

12x-2x =2x-2x+7

10x = 7                      Divide by 10

x = 7/10

x = 0.7

Let's check it

6(0.7) = 4.2

1/2 (2*0.7 + 7)

1/2 (1.4 + 7)

1/2 ( 8.4)

4.2

Both sides check. The answer must be x = 0.7

Answer:

x=12

Step-by-step explanation:

3x=2x+12

3x-2x=12

x=12

Answer:

(5*sqrt(2), 5pi/4)

Step-by-step explanation:

In Polar coordinates, tan(theta)=y/x and r=sqrt(x^2+y^2)

tan(theta)=-5/5=-1. Theta=5pi/4

r=sqrt(5^2+5^2)=5*sqrt(2)

Hence the Polar coordinate is (5*sqrt(2), 5pi/4)

The polar coordinate is [tex](5\sqrt{2},\frac{-\pi }{4} )[/tex].

The polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction.

To convert rectangular coordinate (x, y) to polar coordinate(r, θ) by using some formula

tanθ = y/x and [tex]r =\sqrt{x^{2} +y^{2} }[/tex]

According to the given question

We have

A rectangular coordinate (5, -5).

⇒ x = 5 and y = -5

Therefore,

[tex]r=\sqrt{(5)^{2} +(-5)^{2} } =\sqrt{25+25} =\sqrt{50} =5\sqrt{2}[/tex]

and

tanθ = [tex]\frac{-5}{5} =-1[/tex]

⇒ θ = [tex]tan^{-1} (-1)[/tex] = [tex]-\frac{\pi }{4}[/tex]

Therefore, the polar coordinate is [tex](5\sqrt{2},\frac{-\pi }{4} )[/tex].

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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 <===

Using the normal distribution, there is a 0.7967 = 79.67% probability that at most 12 adults in a sample of 100 adults pass this fitness test.

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The parameters of the binomial distribution are given by:

p = 0.1, n = 100.

Hence the mean and the standard deviation for the approximation are given by:

The probability at most 12 adults in a sample of 100 adults pass this fitness test, using continuity correction, is P(X < 12.5), which is the p-value of Z when X = 12.5, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{12.5 - 10}{3}[/tex]

Z = 0.83

Z = 0.83 has a p-value of 0.7967.

0.7967 = 79.67% probability that at most 12 adults in a sample of 100 adults pass this fitness test.

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Answer:

Spherical-Like Shape

Step-by-step explanation:

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

Answer:

I think cubic polynomial cause degree is 3

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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Answer:

12

Step-by-step explanation:

f(-2) = -3*(-2^2)

f(-2) = -3*-4

f(-2) = 12

Answer:

i doesn,t understand this language

Answer:

C: 70 Mins

Step-by-step explanation:

1, 20ft*15ft=300ft^2

2, 30ft*25ft=750ft^2

3, 750ft-350ft=450ft^2

4, 450 ft^2 = 30 mins

5, 350ft=750ft=1050ft^2

6, 1050/450=2.3333

7, 30*2.3333=70

8, 70 mins

At the same rate per square foot , Bert will take 80 minutes to mow the both lawns.

Rate is the ratio between two related quantities in different units.

Area of the rectangular lawn = lw

where

area of the lawn1  = 30 × 25 = 750 ft²

area of the lawn2 = 20 × 15 = 300 ft²

Therefore,

He mow the firts lawn 30 minutes longer than the second lawn. Therefore,

let

x = time to mow the second lawn

x + 30 = time to mow the first lawn

rate for the first lawn = 30 + x / 750

rate for the second lawn = x / 300

Hence,

30 + x / 750 = x / 300

cross multiply

9000 + 300x = 750x

9000 = 750x - 300x

9000 =  450x

x = 9000 / 450

x = 20

it will take him 30 + 20 + 20 = 80 minutes to mow the both lawns.

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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.

Graph 1

Part (a)

The function is increasing when x > 0. The function is decreasing when x < 0.

The function is never constant

An increasing portion is when the graph goes uphill when moving left to right. A decreasing portion goes in the opposite direction: it goes downhill when moving left to right.

The reason why the function is never constant is because there aren't any flat horizontal sections. Such sections are when x changes but y does not. No such sections occur.

------------------------

Graph 1

Part (b)

Domain = set of all real numbers

Range = set of y values such that [tex]y \ge 0[/tex]

The domain is the set of all real numbers because we can plug in any value for x without any restriction. There are no division by zero errors to worry about, or square roots of negative numbers to worry about either.

The range is the set of nonnegative numbers as the graph indicates. The lowest y gets is y = 0.

------------------------

Graph 1

Part (c)

The function is even

The function f(x) = 1.6x^12 is an even function due to the even number exponent. For any polynomial, as long as the exponents are all even, then the function itself is even. If all the exponents were odd, then the function would be odd. This applies to polynomials only. A power function is a specific type of polynomial.

Note in the graph, we have y axis symmetry. The mirror line is vertical and placed along the y axis. This is a visual trait of any even function.

We could use algebra to show that f(-x) = f(x) like so

f(x) = 1.6x^12

f(-x) = 1.6(-x)^12

f(-x) = 1.6x^12

The third step is possible because (-x)^12 = x^12 for all real numbers x. It's similar to how (-x)^2 = x^2. You could think of it like (-1)^2 = (1)^2

============================================================

Graph 2

Part (a)

The function is decreasing when x < 0 and when x > 0

The function is never increasing

The function is never constant

In other words, the function is decreasing over the entire domain (see part b). The only time it's not decreasing is when x = 0.

The function is decreasing because the curve is going downhill when moving to the right. You can think of it like a roller coaster of sorts.

At no point of this curve goes uphill when moving to the right. Therefore, it is never increasing. The same idea applies to flat horizontal sections, so there are no constant intervals either.

------------------------

Graph 2

Part (b)

Domain: x is any real number but [tex]x \ne 0[/tex]

Range: y is any real number but [tex]y \ne 0[/tex]

Explanation: If we tried plugging x = 0 into the function, we get a division by zero error. This doesn't happen with any other number. Therefore, the set of allowed inputs is any number but 0.

The range is a similar story. There's no way to get y = 0 as an output.

If we plugged y = 0 into the equation, then we'd get this

y = 17x^(-3)

0 = 17/(x^3)

There's no way to have the right hand side turn into 0. The numerator is 17 and won't change. Only the denominator changes. We can't have the denominator be 0.  

------------------------

Graph 2

Part (c)

The function is odd

We can prove this by showing that f(-x) = -f(x)

f(x) = 17x^(-3)

f(-x) = 17(-x)^(-3)

f(-x) = 17* ( -(x)^(-3) )

f(-x) = -17x^(-3)

f(-x) = -f(x)

This is true for nearly all real numbers x, except we can't have x = 0.

Graphic 1:

(A) If f(x) = 1.6x ¹², then f '(x) = 19.2x ¹¹. Both f '(x) and x have the same sign, which means

• for -∞ < x < 0, we have f '(x) < 0, so that f(x) is decreasing on this interval

• for 0 < x < ∞, we have f '(x) > 0, so f(x) is increasing on this interval

f(x) is not constant anywhere on its domain.

(B) Speaking of domain, since f(x) is a polynomial (albeit only one term), it has

• a domain of all real numbers

• a range of {y ∈ ℝ : y = f(x) and y ≥ 0} (in other words, all real numbers y such that y = 1.6x ¹² and y is non-negative)

(C) This function is even, since

f(-x) = 1.6 (-x)¹² = (-1)¹² × 1.6x ¹² = 1.6x ¹² = f(x)

Graphic 2:

(A) Now if f(x) = 17/x ³, then f '(x) = -51/x ⁴. Because x ⁴ ≥ 0 for all x, this means f '(x) < 0 everywhere, except at x = 0. So f(x) is decreasing for (-∞ < x < 0) U (0 < x < ∞).

(B) f(x) has

• a domain of {x ∈ ℝ : x ≠ 0} (or all non-zero real numbers)

• a range of {y ∈ ℝ : y = f(x) and y ≠ 0} (also all non-zero reals)

(C) This function is odd:

f(-x) = 17/(-x)³ = 1/(-1)³ × 17/x ³ = -17/x ³ = -f(x)

Answer:

100 different shoes

Step-by-step explanation:

4 styles * 5 sizes * 5 colors

4*5*5 = 100

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]