Evaluate the limit, lim x -> 0 (x * cos theta - theta * cos x)/(x - theta)​

The area of the interior above the polar axis is -0.858 square units

The area bounded by a polar curve between θ = θ₁ and θ = θ₂ is given by

[tex]A = \int\limits^{\theta_{2} }_{\theta_{1} } {\frac{1}{2}r^{2} } \, d\theta[/tex]

Now, since we have the curve r = 1 - sinθ and we want to find the area of the interior above the polar axis, we integrate from θ = 0 to θ = π, since this is the region above the polar axis.

So, [tex]A = \int\limits^{\theta_{2} }_{\theta_{1} } {\frac{1}{2}r^{2} } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}(1 - sin\theta)^{2} } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}[1 - 2sin\theta + (sin\theta)^{2}] } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}\, d\theta - \int\limits^{\pi}_{0} 2sin\theta \, d\theta+ \int\limits^{\pi}_{0} (sin\theta)^{2} } \, d\theta\\[/tex]

[tex]A = \int\limits^{\pi}_{0} {\frac{1}{2}\, d\theta - 2\int\limits^{\pi}_{0} sin\theta \, d\theta+ \int\limits^{\pi}_{0} \frac{(1 - cos2\theta)}{2} } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}\, d\theta - 2\int\limits^{\pi}_{0} sin\theta \, d\theta+ \int\limits^{\pi}_{0} \frac{1}{2} } \, d\theta -\int\limits^{\pi}_{0} \frac{cos2\theta}{2} } \, d\theta\\[/tex]

[tex]A = \int\limits^{\pi}_{0} \, d\theta - 2\int\limits^{\pi}_{0} sin\theta \, d\theta -\int\limits^{\pi}_{0} \frac{cos2\theta}{2} } \, d\theta\\= [\theta]_{0}^{\pi} - 2[-cos\theta]^{\pi} _{0} - [\frac{sin2\theta}{4}]_{0}^{\pi} \\= [\theta]_{0}^{\pi} + 2[cos\theta]^{\pi} _{0} - [\frac{sin2\theta}{4}]_{0}^{\pi}\\= [\pi - 0] + 2[cos\pi - cos0] - \frac{ [sin2\pi - sin0]}{4}\\= \pi + 2[-1 - 1] - \frac{ [0 - 0]}{4}\\= \pi + 2[-2] - \frac{ [0]}{4}\\= \pi - 4 - 0\\= \pi - 4\\= 3.142 - 4\\= -0.858[/tex]

So, the area of the interior above the polar axis is -0.858 square units

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In order to have a greater fertilizer coverage, you need to increase flow rate and: A. increase speed to approximately 7.1 mph so that you cover the field more quickly.

A flow rate is an average rate and it can be defined as the number of flow units that are created per unit time such as fertilizer coverage on a farm.

In this context, we can infer and logically deduce that you have to increase flow rate and your increase speed to approximately 7.1 mph, so that you can cover the field more quickly and have a greater fertilizer coverage.

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Complete Question:

The soil samples for the next field indicate that fertilizer coverage needs to be greater. To achieve this, you need to increase flow rate. How would you achieve this?

A. Increase speed to approximately 7.1 mph so that you cover the field more quickly

B. Increase the engine speed to approximately 2,000 rpm

C. Decrease speed to approximately 6.0 mph so that you cover the field more slowly

D. Shift to second gear so that the engine speed slows

Answer:

[tex]\bf m = - 1[/tex]

Step-by-step explanation:

[tex]\bf m= −(4+m)+2[/tex]

First of all, let's Rearrange the given terms :-

[tex]\bf m = - ( m +4) + 2[/tex]

Now, let's distribute:-

[tex]\bf m = - m - 4 + 2[/tex]

Add numbers -4 + 2 = -2

[tex]\bf m = - m - 2[/tex]

Add m to both sides, and combine like terms :-

[tex]\bf 2m = - 2[/tex]

Divide both sides by 2 :-

[tex]\bf m = - 1[/tex]

The value of x is -23 and the values of AB, BC, and AC are -33, 12 and -21

The equations are given as

AB = 13 + 2x

BC = 12

AC = x + 2

Point B is between A and C on Line AC.

So, we have

AC = AB + BC

This gives

x + 2 = 13 + 2x + 12

Evaluate the like terms

x - 2x = 13 + 12 - 2

This gives

-x = 23

Divide by -1

x =-23

Substitute x =-23 in AB = 13 + 2x and AC = x + 2

AB = 13 - 2 * 23 = -33

AC = -23 + 2 = -21

Hence, the value of x is -23 and the values of AB, BC, and AC are -33, 12 and -21

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Answer:

3

Step-by-step explanation:

Numbers of beads used in making triangles:

3, 6, 10, 15, 21, 28, 36, ...

We need to pick a few numbers from those above that add to 41.

3 + 10 + 28 = 41

Answer: 3 triangles

Answer:

A = -5

B = 1

Step-by-step explanation:

Hello!

You simply plug in the given x-value into the equation to find the missing y-value.

Plug in -1 for x into y = 2x - 3.

Plug in 2 for x into y = 2x - 3.

Step-by-step explanation:

you need to use the ruler to measure the lengths of it, like place the ruler from A to B and look at the measurement of it

Answer:

8.1

Step-by-step explanation:

[tex]\sqrt{(-1-0)^2 + (5-(-3))^2}=\sqrt{65} \approx 8.1[/tex]

1. Consult the linked question. [tex]n(A\cap B\cap W) = \boxed{3}[/tex].

2. We have

[tex]n(A\cap B) = n(A\cap B \cap W) + n(A\cap B \cap W') \\\\ \implies 11 = 3 + n(A\cap B \cap W') \\\\ \implies n(A\cap B\cap W') = 8[/tex]

Similarly we can find

[tex]n(A\cap B' \cap W) = 3[/tex]

[tex]n(A'\cap B\cap W) = 6[/tex]

[tex]n(A\cap B'\cap W') = 16[/tex]

[tex]n(A'\cap B\cap W') = 8[/tex]

[tex]n(A'\cap B'\cap W) = 12[/tex]

Then the total number of students that applied to at least two of the universities is

[tex]\underbrace{n(A\cap B\cap W') + n(A\cap B'\cap W) + n(A'\cap B\cap W)}_{\text{only 2}} + \underbrace{n(A\cap B\cap W)}_{\text{all 3}} = \boxed{20}[/tex]

3. There's a small ambiguity here. Are we interested in students that applied to zero universities? If so, there are

[tex]\underbrace{n(U\setminus(A\cup B\cup W))}_{\text{none}} + \underbrace{n(A\cap B'\cap W') + n(A'\cap B\cap W') + n(A'\cap B'\cap W)}_{\text{only 1}} = \boxed{40}[/tex]

If we want students that applied to at least 1 school, we omit the first term and get a total of 36.

4. Split this set of students into those that also applied to Wachemo and those that did not.

[tex]n(A \cap B') = n(A\cap B' \cap W) + n(A\cap B'\cap W') = \boxed{19}[/tex]

The inverse of the given function [tex]f(x)=\sqrt[5]{x^3}[/tex] is [tex]f^{-1}(x)=x^{\frac{5}{3}}[/tex]

The given function is:

[tex]f(x)=\sqrt[5]{x^3}[/tex]

This function can be written in exponent form as:

[tex]f(x)=x^\frac{3}{5}[/tex]

Make x the subject of the formula:

[tex]x=[f(x)]^\frac{5}{3}[/tex]

Let x be replaced by [tex]f^{-1}(x)[/tex] and f(x) be replaced by x

The inverse function therefore becomes:

[tex]f^{-1}(x)=x^{\frac{5}{3}}\\\\ f^{-1}(x)=\sqrt[3]{5}[/tex]

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Applying the angles of intersecting secants theorem, the measure of arc XY is: 32°.

According to the angles of intersecting secants theorem, m∠XZY = 1/2(arc VW - arc XY).

Plug in the values:

39 = 1/2(110 - arc XY)

2(39) = 110 - arc XY

78 = 110 - arc XY

78 - 110 = - arc XY

-32 = - arc XY

Arc XY = 32°

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It would Julia 0.60 years more to double the initial investment.

What is future value?

Future value means the initial investment multiplied by 2 since the future value is meant to double.

The formula for future value of a continuously interest rate is provided below:

FV=PV*e^(rt)

FV=future value=$4,200*2=$8,400

PV=initial investment=$4,200

e=exponential constant=2.7182818

r=interest rate=9.5%

t=number of years it takes for the investment to double=unknown

$8,400=$4,200*2.7182818^(9.5%*t)

$8,400/$4,200=2.7182818^(0.095t)

2=2.7182818^0.095t

take log of both sides

ln(2)=0.095t* ln(2.7182818)

0.095t=ln(2)/ln(2.7182818)

0.095t=0.69314718781684800

t=0.69314718781684800/0.095

t=7.30 years

The future value when interest is compounded quarterly is shown thus:

FV=PV*(1+r/4)^(N*4)

FV=$8,400

PV=$4,200

r=8 7/8%

r=8.875%

N=the number of years it would take for the initial investment to double=unknown

$8,400=$4,200*(1+8.875%/4)^(N4)

$8,400=$4,200*(1+0.0221875)^(N4)

$8,400/$4,200=(1+0.0221875)^(N4)

2=(1+0.0221875)^(N4)

2=(1.0221875)^(N4)

take log of both sides

ln(2)=N4*ln(1.0221875)

N4=ln(2)/ln(1.0221875)

N4=31.5857423180125

N=31.5857423180125/4

N=7.90

Difference in years=7.90-7.30

difference in years=0.60 years

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Answer:

244 ft

Step-by-step explanation:

Scale: 1/2 in = 16 ft

Scale length = 7 5/8 inches

Real length = ?

We can use a proportion.

1/2 in : 16 ft = 7 5/8 in. : x

(1/2)/16 = (7 5/8)/x

(1/2)x = 16 × 7 5/8

(1/2)x = 122

x = 244

Answer: C. 244 ft  

Answer:

Rafael earns $375

Andrew earns $340

Step-by-step explanation:

Here, we will set up a system of equations where r represents Rafael and a represents andrew.

r = a + 35

r + a = 715

Next, we substitute a + 35 for r in the second equation.

(a + 35) + a = 715 (combine like terms)

2a + 35 = 715 (subtract 35 from both sides)

2a = 680 (divide both sides by 2)

a = 340

To solve for r, we can plug 340 for a into either equation.

r = (340) + 35

r = 375

We can double check by plugging both values into the other equation.

375 + 340 = 715 (yes, it works!)

Brainliest, please :)

The simplest form of the expression given in the task content is; 5.5x +7.42.

The expression given in the task content is;

5.5(x+6 1/2)-(x+9 1/3) - (19-x)

Hence, the expression can be written in its simplest form as follows;

5.5x + 35.75 -x -9.33 -19 +x

= 5.5x +7.42.

Remarks: The complete task requires that the expression be written in its simplest form.

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The function that will have a horizontal asymptotes are function B, C and D

A function is said to have a horizontal asymptotes of the degree of the numerator is less or equal to that of the denominator.

From the rational expression given, the function A and E wil have no horizontal asymptotes since the power of the numerator is greater abd equal to the denominator,

The function that will have a horizontal asymptotes are function B, C and D

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Using the binomial distribution, the probabilities are given as follows:

a) 0.37 = 37%.

b) 0.5065 = 50.65%.

c) 0.3260 = 32.60%.

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

For this problem, the fixed parameter is:

p = 0.37.

Item a:

The probability is P(X = 1) when n = 1, hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 1) = C_{1,1}.(0.37)^{1}.(0.63)^{0} = 0.37[/tex]

Item b:

The probability is P(X = 3) when n = 3, hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 3) = C_{3,3}.(0.37)^{3}.(0.63)^{0} = 0.5065[/tex]

Item c:

The probability is P(X = 2) when n = 4, hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 2) = C_{4,2}.(0.37)^{2}.(0.63)^{2} = 0.3260[/tex]

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The graphed piecewise function is the one in option B.

On the graph, we can see that on the lowest part we have a closed dot at x = 1

And the above part has a open dot at x = 1.

Then the piecewise function is of the form:

Such that the above part seems to be quadratic, and the part below seems to have a larger degree.

With that, we can conclude that the correct option is:

Which is option b.

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He pays $0.24 less when using his loyalty card.

We know that, without the card, he would pay $1.679 dollars per liter of gasoline.

With his card, he would pay $1.439

To find the relation between the two prices, we just need to take the difference between the two values (a difference is a subtraction).

d = $1.679 - $1.439 = $0.24

He pays $0.24 less when using his loyalty card.

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Answer:

12 years old

Step-by-step explanation:

Set up the following equation:

x + 2 = 2(x - 5) (distribute the 2)

x + 2 = 2x - 10 (subtract 2 from both sides)

x = 2x - 12 (subtract 2x from both sides)

-x = -12 (multiply both sides by negative 1)

x = 12

Brainliest, please :)

Answer: 12

Step-by-step explanation:

Let's algebraically represent this situation. Say his current age is x. We can rephrase the sentence to say that in two years, his age is the same as twice his age 5 years ago.

The phrase "in two years" means when his age is 2 more than his current age, or x + 2.

The phrase "same as" represents equality. The expressions on both sides have the same value.

The phrase "twice his age 5 years ago" is his age minus 5, multiplied by 2. We can write this as 2*(x-5).

Now that we have "translated" the equation from words to numbers, let's put it all in one equation and solve:

[tex]x+2=2(x-5)[/tex]

[tex]x+2=2x-10[/tex] [Distributive Property]

[tex]x+2-2x-2=2x-10-2x-2[/tex] [Adding -2x-2 to both sides]

[tex]-x=-12[/tex] [Combining like terms and simplifying]

[tex]x=12[/tex] [Multiplying both sides by -1 to remove the negative]

Hence, Billy's age is 12.

Answer:

0

Step-by-step explanation:

sin²t + cos²t = 1, so since cos²t = 1, sin²t = 0, and thus sint = y = 0.

Answer: [tex]x=\frac{\log (3/5)}{\log 5}[/tex]

Step-by-step explanation:

[tex]3=5^{x+1}\\\\\log_{5} 3=x+1\\\\x=\log_{5} 3-1\\\\x=\log_{5} 3-\log_{5} 5\\\\x=\log_{5} \left(\frac{3}{5} \right)\\\\x=\frac{\log (3/5)}{\log 5}[/tex]

Numbers whose decimals form goes on forever are best represented as fractions because they are non-terminating decimals

The decimal numbers are represented as:

A and B

Where

A > B

Numbers whose decimals form goes on forever are non-terminating numbers, and the complete form cannot be represented using decimals except with the sign .......

Because non-terminating decimals have no end, it is best to represent them as fractions

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The person which is more vulnerable to identity theft include:

This involves using someone's personal information to commit fraud and other forms of crime.

The set of people mentioned above are usually naive and innocent thereby making them more vulnerable to identity theft.

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Answer:

Step-by-step explanation:

hello : here is a solution

One tree is 31.044 m taller than the other one in height.

Given Information and Formula Used:

The distance between the trees, BC (from the figure) = 120 m

Elevation of angles to the top of the trees,

∠AMB = 52°

∠DMC = 36°

In a right angled triangle,

tan x = Perpendicular/Height

Here, x is the angle opposite to the Perpendicular.

Calculating the Height Difference:

Let's compute the height of the taller tree first.

In ΔAMB,

tan 52° = AB / BM

Now, since M is the point halfway between the trees,

BM = CM = BC/2

BM = CM = 60 m

⇒ 1.2799 = AB / 60

AB = 1.2799 × 60

Thus, the height of the taller tree, AB =  76.794 m

Now, we will compute the height of the smaller tree.

In ΔDCM,

tan 36° = DC / CM

⇒ 0.7625 = DC / 60

DC = 0.7625 × 60

Thus, the height of the smaller tree, DC = 45.75 m

The difference in the heights of the trees, AP = AB - DC

AP = (76.794 - 45.75)m

AP = 31.044m

Hence one tree is 31.044m taller in height than the other.

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Answer:

Step-by-step explanation:

[tex] \frac{x {}^{2} }{3y} \div \frac{3x}{2y} \\ \frac{x {}^{2} }{3y} \times \frac{2y}{3x} \\ \frac{x {}^{2} }{3} \times \frac{2}{3x} \\ \frac{x}{3} \times \frac{2}{3} = \frac{2x}{9} [/tex]