Please answer thankssss

Answer:

The vehicle should start with [tex](100\, e^{ke} - 100)\; {\rm kg}[/tex] of fuel and drive at a speed of [tex](1/k)\; {\rm km \cdot hr^{-1}}[/tex].

Step-by-step explanation:

Let [tex]t\; {\rm hr}[/tex] denote the number of hours after the vehicle started. As in the question, let the amount of fuel currently on this vehicle be [tex]x\; {\rm kg}[/tex]. The question states that the vehicle consumes fuel at a rate ([tex]dx/dt[/tex]) of [tex]((100 + x) / 100)\, e^{kv}[/tex]. In other words:

[tex]\displaystyle \frac{dx}{dt} = -\frac{100 + x}{100}\, e^{k\, v}[/tex].

Note the minus sign in front of the right-hand side. The amount of fuel on this vehicle decreases over time. Hence, the rate of change in [tex]x[/tex] should be negative.

This equation is a separable ordinary differential equation. The variables are [tex]x[/tex] and [tex]t[/tex]. Solve this ODE to find an expression of [tex]x\![/tex] (fuel in the vehicle) in terms of [tex]t\![/tex] (time.) Follow these steps:

Rearrange this equation such that all [tex]x[/tex] and [tex]dx[/tex] are are on the same side of the equation, while [tex]t[/tex] and [tex]dt[/tex] on all on the other side.

[tex]\displaystyle \frac{dx}{100 + x} = -\frac{e^{k\, v}\, dt}{100}[/tex].

Integrate both sides, and the equality should still hold. Note that [tex]k[/tex] and [tex]v[/tex] are considered as constants. Be sure to include the constant of integration [tex]C[/tex] on one side of the equation.

[tex]\displaystyle \int \frac{dx}{100 + x} = -\frac{e^{k\, v}}{100}\int dt[/tex].

[tex]\displaystyle \ln | 100 + x | = -\frac{(e^{k\, v})\, t}{100} + C[/tex].

Let [tex]x_{0}[/tex] denote the initial amount of fuel on this vehicle (i.e., the value of [tex]x[/tex] when [tex]t = 0[/tex]). The constant of integration [tex]C[/tex] should ensure that [tex]x = x_{0}[/tex] when [tex]t = 0[/tex]. Thus:

[tex]\displaystyle \ln | 100 + x_{0} | = C[/tex].

Hence, the value of the constant of integration should be [tex]\ln | 100 + x_{0} |[/tex]. Therefore:

[tex]\displaystyle \ln | 100 + x | = -\frac{(e^{k\, v})\, t}{100} + \ln | 100 + x_{0}|[/tex].

Since the speed of the vehicle is constant at [tex]v\; {\rm km\cdot hr^{-1}}[/tex], the time required to travel [tex]100\; {\rm km}[/tex] would be [tex](100 / v)\; {\rm hr}[/tex].

For optimal use of the fuel, the vehicle should have exactly [tex]x = 0[/tex] fuel when the destination is reached. Therefore, [tex]x = 0[/tex] at [tex]t = 100 / v[/tex]. Hence:

[tex]\displaystyle \ln | 100 | = -\frac{(e^{k\, v})\, (100 / v)}{100} + \ln | 100 + x_{0}|[/tex].

[tex]\displaystyle \ln | 100 | = -\frac{e^{k\, v}}{v} + \ln | 100 + x_{0}|[/tex].

Notice that [tex]\ln|100 + x_{0}|[/tex] is monotone increasing with respect to [tex]x_{0}[/tex] as long as [tex]100 + x_{0} > 0[/tex]. Thus, given that [tex]x_{0} > 0[/tex], [tex]x_{0}\![/tex] would be minimized if and only if the surrogate [tex]\ln|100 + x_{0}|\![/tex] is minimized.

While the goal is to find the [tex]v[/tex] that minimize [tex]x_{0}\![/tex], finding the [tex]v\![/tex] that minimizes [tex]\ln|100 + x_{0}|[/tex] would achieve the same purpose.

[tex]\displaystyle \ln | 100 + x_{0}| = \ln | 100 | + \frac{e^{k\, v}}{v}}[/tex].

The [tex]\texttt{RHS}[/tex] of this equation is indeed convex with respect to [tex]v[/tex] ([tex]v > 0[/tex].) Thus, the [tex]\texttt{RHS}\![/tex] could be minimized by setting the first derivative with respect to [tex]v[/tex] to [tex]0[/tex].

Differentiate the right hand side with respect to [tex]v[/tex]:

[tex]\begin{aligned} & \frac{d}{dv}\left[\frac{e^{k\, v}}{v}\right] \\ =\; & \frac{k\, e^{k\, v}}{v} - \frac{e^{k\, v}}{v^{2}}\\ =\; & \frac{(k\, v - 1)\, e^{k\, v}}{v^{2}}\end{aligned}[/tex].

Setting this first derivative to [tex]0[/tex] and solving for [tex]v[/tex] gives:

[tex]k\,v - 1 = 0[/tex].

[tex]v = (1/k)[/tex].

Therefore, the amount of fuel required for this trip is minimized when [tex]v = (1/k)\; {\rm km \cdot hr^{-1}}[/tex].

Substitute [tex]v[/tex] back and solve for [tex]x_{0}[/tex] (initial amount of fuel on the vehicle.)

[tex]\displaystyle \ln | 100 + x_{0}| = \ln | 100 | + \frac{e^{k\, (1/k)}}{(1/k)}}[/tex].

[tex]\displaystyle \ln | 100 + x_{0}| = \ln | 100 | + k\, e[/tex].

[tex]e^{\ln| 100 + x_{0}|} = e^{\ln|100| + k\, e}[/tex].

[tex]e^{\ln| 100 + x_{0}|} = e^{\ln|100|}\, e^{k\, e}[/tex].

[tex]100 + x_{0} = 100\, e^{k\, e}[/tex].

[tex]x_{0} = 100\, e^{k\, e} - 100[/tex].

In other words, the initial amount of fuel on the vehicle should be [tex](100\, e^{k\, e} - 100)\; {\rm kg}[/tex].

Answer:

31 and 1

Step-by-step explanation:

let the first number be x

let the second number be y

x + y = 32 ---------- eq 1

x - y = 30 ---------- eq 2

eq 1 - eq 2

(x - x) + (y -(-y)) = 32 - 30

y + y = 2

2y = 2

y = 1

Substitute the value of y in eq 1

x + y = 32

x + 1 = 32

x = 32 - 1

x = 31

Therefore the first number is 31 and the second number is 1

x - y = 0  is the standard form of the polar equation θ = -135

[tex]\sf \theta \ =-135[/tex]

[tex]\rightarrow \sf tan(\theta) \ =tan(-135)[/tex]

[tex]\rightarrow \sf tan(\theta) \ =1[/tex]

[tex]\rightarrow \sf \dfrac{y}{x} \ = 1[/tex]

[tex]\rightarrow \sf y =x[/tex]

[tex]\rightarrow \sf x -y = 0[/tex]

[tex]\\ \rm\Rrightarrow \theta=-135[/tex]

[tex]\\ \rm\Rrightarrow tan\theta=tan(-135)[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{Perpendicular}{Base}=tan(-\dfrac{3\pi}{4})[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{y}{x}=1[/tex]

[tex]\\ \rm\Rrightarrow y=x[/tex]

[tex]\\ \rm\Rrightarrow x-y=0[/tex]

Answer:

x = 4

Step-by-step explanation:

4/4 = 1      1 + 8 = 9      9x2 = 18 give me brainliest pls

Answer:

its my alt ima give u

Step-by-step explanation:

[tex] = (x^2 \times x^3)^3 \\ = x {}^{6} \times {x}^{9} \\ = (x) {}^{6 + 9} \\ = {x}^{15} [/tex]

Answer:

A = 11.7

Step-by-step explanation:

Just substitute the known values into the equation:

[tex]A=\frac{1}{2}(4.5)(5.2)\\A=11.7[/tex]

Answer:

The building is 16 m tall.

Step-by-step explanation:

x/2 = 80/10

10x=160

10x/10 = 160/10

x=16

Answer:

this is a ratio question. The answer is 16 m

Step-by-step explanation:

The persons height to shadow length ratio is 2 : 10

We want to know what the height of the building is when the shadows length is 80.

Let H represent height:

Persons ratio:

2 : 10

Buildings ratio:

H : 80

In other words for every 2 meters height a 10 m shadow is casted so we want to know how many of this ratio can fit into the building:

Let’s find the equivalent ratio:

80/10=8

That means 8 of the persons ratio makes up the building. Let’s find the height of building by 2*8=16.

therefore the building is 16m tall.

The complete factor of the polynomial 36x² - 25 is (6x + 5)(6x − 5). Then the correct option is D.

Polynomial is an algebraic expression that consists of variables and coefficients. Variable are called unknown. We can apply arithmetic operations such as addition, subtraction, etc. But not divisible by variable.

Factor completely 36x² - 25.

We know that

[tex]\rm a^2 - b^2 = (a+b)(a-b)[/tex]

(6x)² − 5²

(6x + 5)(6x − 5)

The complete factor of the polynomial 36x² - 25 is (6x + 5)(6x − 5). Then the correct option is D.

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The factors of the equation are [tex]\rm (x+12)(x-4)[/tex]

Factorization of an algebraic expression means writing the given expression as a product of its factors.

These factors can be numbers, variables, or an algebraic expression.

Given

The equation is

[tex]\rm x^2+8x-48[/tex]

The factors of the equation are given by;

[tex]\rm x^2+8x-48\\\\x^2+12x-4x-48\\\\x(x+12)-4(x+12)\\\\(x+12)(x-4)[/tex]

Hence, the factors of the equation are [tex]\rm (x+12)(x-4)[/tex]/

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The package that contains more seed is brand b's package.

In order to determine which package contains more seeds, the volume of each package has to be determined.

The volume of brand A's package = 1/3(nr^2h)

1/3 x 2.5^2 x 10 x 22/7 = 65.48

The volume of brand b's package = 2.5 x 5 x 5  = 62.5

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Volume of an object is the measure of the space that object occupies. The correct comparison of volume of considered objects is: Volume of A = Volume of B

Suppose that the right rectangular prism in consideration be having its dimensions as 'a' units, 'b' units, and 'c' units, then its volume is given as:

[tex]V = a\times b \times c \: \rm unit^3[/tex]

It can be obtained by multiplying the cross sectional triangle's area to the height of the considered triangular prism.

Thus, for the given case, we get:

Its dimensions are 1.5 units by 1 units by 1.81 units

Thus, [tex]V_A = 1.5 \times 1 \times 1.81 = 1.5 \times 1.81 = 2.715 \: \rm unit^3[/tex]

Its cross section triangle has base of 2 units and height of 1.5 unit. The height of the prism is equal to the height of the considered rectangular prism = 1.81 units,

Thus,

[tex]V_B = \text{Area of triangular cross-section} \times 1.81 = \dfrac{1}{2} \times 2\times 1.5 \times 1.81 \: \rm unit^3\\V_B = 2.715\: \rm unit^3\\\\[/tex]

Thus, we get [tex]V_A = V_B[/tex]((Volume of A = Volume of B)

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Answer: A

Step-by-step explanation:

The total area would have to be 1, so (1/2)(5)(height)=1

2.5height = 1

height = 1/2.5 = 2/10 = 1/5

The height for the density curve will be 1/5. The correct option is A.

The area under the curve represents 100% of all probabilities. Because we normally use decimals in probabilities, we can also say that the area is equal to 1 (because 100% is a decimal). The density curve shown above depicts how body weights are distributed.

The image of the triangle is showing the base 5 units. For the density curve, the area under the curve must be 1. The height of the curve will be calculated as,

Area = 1

1/2 x B x H = 1

1/2 x 5 x H = 1

2.5 x H = 1

H = 1 / 2.5

H = 1 / 5

Therefore, the height will be 1 / 5.

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Answer:

(x+4) (x+6)

Step-by-step explanation:

use backwards foil

two numbers that add up to 10 and multiply to 24 are 6 and 4

both signs are positive so you will be adding

Answer:

(x+4) (x+6)

Step-by-step explanation:

Find 2 numbers that added =  10  and multiplied =24

(6 &4)     6+4 = 10                  6x4 = 24

=x²+10x+24

=(x²+6x)+(4x+24)

=x(x+6)+4(x+6)

=(x+4)(x+6)

By using a trigonometric table, we will see that:

cotangent(3pi/4) = -1

We want to find:

cotangent(3pi/4)

This is equal to:

[tex]\frac{cos(3pi/4)}{sin(3pi/4)}[/tex]

We can use a table to evaluate these two, we know that:

cos(3pi/4) = -√2/2

sin(3pi/4) = √2/2

Then:

[tex]\frac{cos(3pi/4)}{sin(3pi/4)} = \frac{-\sqrt{2}*2 }{\sqrt{2}*2 } = -1[/tex]

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Answer:

1st/ 4th options

Step-by-step explanation:

the work column shows the rule is

- 8 + 4 ( term number - 1 ) , that is with term n

= - 8 + 4(n - 1) → 1st option

distribute parenthesis and simplify

= - 8 + 4n - 4

= 4n - 12 ← 4th option

If Ted has $55. 00 in his pocket then out of  two books for $29. 95 each headphones for $29. 99 and a portable speaker set for $30. 95 an MP3 player for $35. 99 and a $20. 00 gift card for downloading music three books for $16. 99 each he can purchase  three books for $16. 99 each.

To calculate total money spent we have to add all the money we spent on all things.

Given that Ted has $55. 00 in his pocket.

(i)

two books for $29. 95 each

Money needed for 2 books = [tex]2\times 29.95=59.9[/tex]

So Ted is not able to purchase 2 books

(ii) headphones for $29. 99 and a portable speaker set for $30. 95

Money needed for headphones and portable speaker = 29.99+30.95=60.94

So Ted is not able to purchase  headphones and portable speaker

(iii)  an MP3 player for $35. 99 and a $20. 00 gift card for downloading music

Money needed for MP3 player and gift card= 35.99+20.00=55.99

So Ted is not able to purchase  MP3 player and gift card

(iv) three books for $16. 99 each.

Money needed for 3 books =[tex]3\times16.99[/tex]=50.97

So ted is able to purchase these three books .

If Ted has $55. 00 in his pocket then out of  two books for $29. 95 each headphones for $29. 99 and a portable speaker set for $30. 95 an MP3 player for $35. 99 and a $20. 00 gift card for downloading music three books for $16. 99 each he can purchase  three books for $16. 99 each.

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Answer:

d)three books for $16. 99 each.

Step-by-step explanation:

Answer:

7/16

Step-by-step explanation:

[tex]\dfrac78 \times \dfrac24=\dfrac{7 \times 2}{8 \times 4}=\dfrac{14}{32}=\dfrac{7}{16}[/tex]

Answer:

Step-by-step explanation:

I'll give you the false statements, the rest are true:

Answer:

are you sure?

Step-by-step explanation:

hind ko alm yan eh HAHAHAHA

bro I thought I had the right answer but I didn't so I'm now contemplating my life's decisions

Answer:

-3x + 6

Step-by-step explanation:

Distribute the 3 across the parentheses, like you are multiplying. 3 x (-x) = -3x and 3 x 2 = 6. Then write as an expression.

Answer:

This question is incomplete

where is the length width and breath

Step-by-step explanation:

meanwhile

[tex]23 √149 \\ 23 \times 12.20655561573370 \\ 280.7507791618751 \\ approx \\ 280.8[/tex]

Answer:

5(5x-t)

=13×5

25x - 5t

=65

25x -5t + 4x +5t

=65 + 22

29x

= 87

x

= 87÷29

x =3

substituting x=3 in 5x-t=13

15- t

=13

15 - 13

= t

t = 2

Step-by-step explanation:

Answer:

let us suppose the number of boys be x

now,

10=4+2x

or, 10-4/2=x

or,x=3

therefore, 3 boys are there

Sarah scored 82 points on her first test of the year. On the second test,

82%

On the Second Test, She scored 77 points

77%

82

-77

=5

The Percent of Change is 5%

or -5% Because The second test Is lower than her first test

Answer:

5%

Step-by-step explanation:

82% - 77%=5%

82%

77%

so there fore the answer is 5%

Probabilities are used to determine the chances of events

There are 4 C2 cards in the first deck out of a total of 12 cards, and there are 3 C6 cards in the second deck out of a total of 14 cards.

So, the probability that the player will choose a C2 card from the first deck or a C6 card from the second deck is:

[tex]P = \frac{4}{12} + \frac{3}{14}[/tex]

Simplify

[tex]P = \frac{1}{3} + \frac{3}{14}[/tex]

Take the LCM

[tex]P = \frac{14 + 3}{42}[/tex]

[tex]P = \frac{17}{42}[/tex]

Also;

There are 3 C3 cards in the first deck out of a total of 12 cards, and there are 4 C5 cards in the second deck out of a total of 14 cards.

So, the probability that the player will choose a C3 card from the first deck or a C5 card from the second deck is:

[tex]P = \frac{3}{12} + \frac{4}{14}[/tex]

Simplify

[tex]P = \frac{1}{4} + \frac{2}{7}[/tex]

Take the LCM

[tex]P = \frac{4 + 14}{28}[/tex]

[tex]P = \frac{18}{28}[/tex]

[tex]P = \frac{9}{14}[/tex]

Hence, the probability that the player will choose a C3 card from the first deck or a C5 card from the second deck is 9/14

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