What is the value of x?
90 degrees
170 degrees
X
Z

Answer:

20 L volume and 12.5 cm depth.

Step-by-step explanation:

See attached image.

We have

[tex]\displaystyle \int_{-2}^2 f(x) \, dx - \int_{-2}^{-1} f(x) \, dx = \int_{-1}^2 f(x) \, dx[/tex]

so that

[tex]\displaystyle \int_{-2}^2 f(x) \, dx + \int_2^5 f(x) \, dx - \int_{-2}^{-1} f(x) \, dx \\\\ ~~~~~~~~~~~~ =  \int_{-1}^2 f(x) \, dx + \int_2^5 f(x) \, dx \\\\ ~~~~~~~~~~~~ = \boxed{\int_{-1}^5 f(x) \, dx}[/tex]

Answer:

The slope of the line would be y=x-1

Step-by-step explanation:

y=x-1

Subsite

3=4-1

3=3  True,

Thus y=x-1

Answer:

Step-by-step explanation:

I gotcha my dude.......

x-5 and 2x+3

The smallest possible value of m according to the task is; 1/1540.

Since it follows from the task content that the product of 1540 and m is a square number and the smallest possible small number is; 1.

The equation which holds true is; 1540 × m = 1

Consequently, m = 1/1540.

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Answer:

Graph b

Step-by-step explanation:

y = |x| is a "v"at x = 0

y = |x-1| is a "v" shifted to the right by 1 unit

It takes a machine at a seafood company 20 seconds to clean 3 pounds of shrimp. The rate of the machine is 0.15 pound per second

An equation is an expression that shows the relationship between two numbers and variables.

An independent variable is a variable that does not depend on any other variable for its value whereas a dependent variable is a variable that depend on any other variable for its value.

It takes a machine at a seafood company 20 seconds to clean 3 pounds of shrimp. Hence:

Rate of the machine = 3 pounds / 20 seconds = 0.15 pound per second

It takes a machine at a seafood company 20 seconds to clean 3 pounds of shrimp. The rate of the machine is 0.15 pound per second

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Answer: [tex]\Large\boxed{f(-9)=-189}[/tex]

Step-by-step explanation:

f(x) = -3x² - 6x

Requirements of the question

Find the value of f(-9)

Substitute values into the given function

f(x) = -3x² - 6x

f(9) = -3 (-9)² - 6 (-9)

Simplify the exponent

f(9) = -3 (81) - 6 (-9)

Simplify by multiplication

f(-9) = (-243) - (-54)

Simplify by subtraction

[tex]\Large\boxed{f(-9)=-189}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

The rule of the translation that maps the rocket from position I to position 2 is 4 units right and 4 units up

The translation is added as an attachment

From the attached figure, we have the following corresponding coordinates:

Figure 1 = (0, 0)

Figure 2 = (4, 4)

The rule of translation is calculated as:

(x, y) = T<Figure 2 - Figure 1>

This gives

(x, y) = T<4 - 0, 4 - 0>

Evaluate

(x, y) = T<4, 4>

Hence, the rule of the translation that maps the rocket from position I to position 2 is 4 units right and 4 units up

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Complete question

In an animated film, a simple scene can be created by translating a figure against a still background. Write a rule for the translation that maps the rocket from position I to position 2.

This is a telescoping sum. The K-th partial sum is

[tex]S_K = \displaystyle \sum_{k=1}^K \left(\frac1{\sqrt{k+1}} - \frac1{\sqrt{k+3}}\right) \\\\ ~~~= \left(\frac1{\sqrt2} - \frac1{\sqrt4}\right) + \left(\frac1{\sqrt3} - \frac1{\sqrt5}\right) + \left(\frac1{\sqrt4} - \frac1{\sqrt6}\right) + \left(\frac1{\sqrt5} - \frac1{\sqrt7}\right) + \cdots \\\\ ~~~~~~~~+ \left(\frac1{\sqrt{K-1}} - \frac1{\sqrt{K+1}}\right) \\\\ ~~~~~~~~+ \left(\frac1{\sqrt K} - \frac1{\sqrt{K+2}}\right) + \left(\frac1{\sqrt{K+1}} - \frac1{\sqrt{K+3}}\right)[/tex]

[tex]\displaystyle = \frac1{\sqrt2} + \frac1{\sqrt3} - \frac1{\sqrt{K+2}} - \frac1{\sqrt{K+3}}[/tex]

As [tex]K\to\infty[/tex], the two trailing terms will converge to 0, and the overall infinite sum will converge to

[tex]\displaystyle \sum_{k=1}^\infty \left(\frac1{\sqrt{k+1}} - \frac1{\sqrt{k+3}}\right) = \lim_{k\to\infty} S_k = \boxed{\frac1{\sqrt2} + \frac1{\sqrt3}}[/tex]

By the limit comparison test, the expression √[1 / (1 + 1 / k)] - √[1 / (1 + 3 / k)] has a limit, then the expression [1 / √(k + 1)] / [1 /√k] -  [1 / √(k + 3)] / [1 /√k] has a limit and the series ∑ [1 / √(k + 1)] - ∑ [1 / √(k + 3)] is convergent.

Herein we have a series that involves radical components. First, we simplify the expression given:

∑ [1 / √(k + 1) - 1 / √(k + 3)] = ∑ [1 / √(k + 1)] - ∑ [1 / √(k + 3)]

The convergence of the series can be proved by the limit comparison test, where each component of the subtraction of the series is compared with a series that is convergent. We notice that both 1 / √(k + 1) and 1 / √(k + 3) resembles the expresion 1 /√k. Then, we have the following subtraction of ratios:

[1 / √(k + 1)] / [1 /√k] - [1 / √(k + 3)] / [1 /√k]

√k / √(k + 1) - √k / √(k + 3)

√[k / (k + 1)] - √[k / (k + 3)]

Then, by using the limit property for rational functions we find the following result for n → + ∞:

√[1 / (1 + 0)] - √[1 / (1 + 0)]

√1 - √1

1 - 1

0

By the limit comparison test, the expression √[1 / (1 + 1 / k)] - √[1 / (1 + 3 / k)] has a limit, then the expression [1 / √(k + 1)] / [1 /√k] -  [1 / √(k + 3)] / [1 /√k] has a limit and the series ∑ [1 / √(k + 1)] - ∑ [1 / √(k + 3)] is convergent.

The statement is incomplete and complete form cannot be found, therefore, we decided to determine if the series is convergent or not.

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Answer:

3024

Step-by-step explanation:

Each chosen member (out of 9 members) will occupy a different position

out of these four (president , vice-president , secretary , treasurer).

So, here we have to calculate the Number of Permutations of 9 members Taken 4 at a Time :

= 9P4

= 9 × 8 × 7 × 6

= 3024

∠ADB + ∠BDC = ∠ADC

39° + (3x - 4) = 8x + 5

3x - 4 = 8x + 5 - 39

3x - 4 = 8x - 34

34 - 4 = 8x - 3x

5x = 30

x = 6

∠ADC = 39° + (3(6) - 4) = 39 + 14 = 53°

Hope it helps!

The distance on map exists 32 cm and actual area exists 37.5 km².

Given: Scale of the map exists at 1:250000.

(a) Distance between two cities = 80 km

= 80000 m

= 8000000 cm

Distance on map = 8000000 [tex]*[/tex] 1/ 250000

= 32 cm

(b) Area of map = 6 cm²

Actual area = [tex]6(250000)^2[/tex] cm²

[tex]= 6 * 625 * 10^8[/tex] cm²

[tex]= 3750 * 10^8/ 10^{10}[/tex]

= 37.5 km²

Therefore, the distance on map exists 32 cm and actual area exists 37.5 km².

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Using decimal multipliers, the correct statement is given by:

(iii) S1 = S2.

Increases of a% or decreases of a% re represented by decimal values, as follows:

Hence, in the context of this problem:

The options are given as follows:

They are equal, hence option (iii) is correct.

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The correct option is (A) P(X) × P(Y) = P(X ∩ Y)

What is probability and example?

 We are given to consider a situation in which X and Y are two events such that

P(X) = 4/5, P(Y) = 1/4, P(X ∩ Y) = 1/5

We are to select the statement that best describes the events X and Y

We know that

any two events A and B are said to be independent if

P(A) × P(B) = P (A ∩ B)

We have, for events X and Y,

P(X) × P(Y) = 4/5 × 1/4 = 1/5 = P (X ∩ Y)

P(X) × P(Y) = P(X ∩ Y)

Thus, X and Y are independent because P(X) × P(Y) = P(X ∩ Y)

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The complete question is -

Consider a situation in which P(X) = 4/5 and P(Y) = 1/4. If P(X and Y) is = 1/5, which best describes the events?

They are independent because P(X) x P(Y) = P(X and Y).

They are independent because P(X) + P(Y) = P(X and Y).

They are dependent because P(X) x P(Y) = P(X and Y).

They are dependent because P(X) + P(Y) = P(X and Y).

Answer: a

Step-by-step explanation:

just took the test

The value of k if (x+1) is a factor of f(x) is -4

The polynomial function is given as:

f(x)=x^3-kx^2+x+6

(x+1) is a factor of f(x)

So, we start by setting x + 1 to 0

x + 1 = 0

Solve for x

x = -1

Substitute x = -1 in f(x)=x^3-kx^2+x+6 and set the equation to 0

(-1)^3-k(-1)^2+(-1)+6 = 0

Evaluate the exponents

-1 - k - 1 + 6 = 0

Evaluate the like terms

k + 4 = 0

Solve for k

k = -4

Hence, the value of k if (x+1) is a factor of f(x) is -4

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Answer:

I think it's A (parallel)

Answer:

A. its parallel

Step-by-step explanation:

all angles are equal... two sides are perpendicular and two are parallel

Answer:

when x=-7

Step-by-step explanation:

if u are solving a function like this, u must not have a division by zero so if we put x=-7 in the function we hav a division by zero

Answer: Look into step by step

Step-by-step explanation:

1. Multiply the second equation by 2; 6x - 2y = -2 then add to first equation

7x = -7 so x = -1, substitute x = -1 into the first equation -1 + 2y = -5 so y = -2

2. Multiply the first equation by 3; 3x + 6y = 6 then subtract it by second equation

y = 2, substitute y = 2 into first equation, x + 4 = 2, x = -2

3. Multiply the first equation by 2; 2x + 6y = 14 then subtract it by second equation

2y = 3 so y = 1.5, substitute y = 1.5 into the first equation x + 4.5 = 7, x = 2.5

I chose this method as it is easy

Answer:

[tex]\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1[/tex]

Step-by-step explanation:

Given equation:

[tex]4x^2-9y^2-8x+36y-68=0[/tex]

This is an equation for a horizontal hyperbola.

To complete the square for a hyperbola

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

[tex]\implies 4x^2-8x-9y^2+36y=68[/tex]

Factor out the coefficient of the x² term and the y² term.

[tex]\implies 4(x^2-2x)-9(y^2-4y)=68[/tex]

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

[tex]\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2[/tex]

[tex]\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36[/tex]

Factor the two perfect trinomials on the left side:

[tex]\implies 4(x-1)^2-9(y-2)^2=36[/tex]

Divide both sides by the number of the right side so the right side equals 1:

[tex]\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}[/tex]

Simplify:

[tex]\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1[/tex]

Therefore, this is the standard equation for a horizontal hyperbola with:

Answer:

Step-by-step explanation:

0= x2 – 2x – 3

a = 1, b = -2 and c = -3.

Answer:

A on edge

Step-by-step explanation:

Give me the brainiest please

Equation (A) C= 1.46n shows the relationship between the total cost of gasoline, c, and the number of gallons purchased, n.

To find the right equation:

Therefore, equation (A) C= 1.46n shows the relationship between the total cost of gasoline, c, and the number of gallons purchased, n.

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The complete question is given below:
The total cost of gasoline varies directly with the number of gallons purchased. Kathy pays $23.36 for 16 gallons of gasoline. Which equation

shows the relationship between the total cost of gasoline, c, and the number of gallons purchased, n?

A.C= 1.46n

B. n = 1.46c

C.C = 23.361

D. n = 23.36c

Answer:

[tex]\displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x=-\dfrac{2}{x+1}+\text{C}[/tex]

Step-by-step explanation:

Fundamental Theorem of Calculus

[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given integral:

[tex]\displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x[/tex]

Factor the denominator:

[tex]\begin{aligned}\implies x^2+2x+1 & = x^2+x+x+1\\& = x(x+1)+1(x+1)\\& = (x+1)(x+1)\\& = (x+1)^2\end{aligned}[/tex]

[tex]\implies \displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x=\int \dfrac{2}{(x+1)^2}\:\:\text{d}x[/tex]

[tex]\textsf{Apply exponent rule} \quad \dfrac{1}{a^n}=a^{-n}[/tex]

[tex]\implies \displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x=\int 2(x+1)^{-2}\:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{4 cm}\underline{Integrating $ax^n$}\\\\$\displaystyle \int ax^n\:\text{d}x=\dfrac{ax^{n+1}}{n+1}+\text{C}$\end{minipage}}[/tex]

Use Integration by Substitution:

[tex]\textsf{Let }u=(x+1) \implies \dfrac{\text{d}u}{\text{d}x}=1 \implies \text{d}x=\text{d}u}[/tex]

Therefore:

[tex]\begin{aligned}\displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x & = \int 2(x+1)^{-2}\:\:\text{d}x\\\\& = \int 2u^{-2}\:\:\text{d}u\\\\& = \dfrac{2}{-1}u^{-2+1}+\text{C}\\\\& = -2u^{-1}+\text{C}\\\\& = -\dfrac{2}{u}+\text{C}\\\\& = -\dfrac{2}{x+1}+\text{C}\end{aligned}[/tex]

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Answer:

  5x^4 -37x^3 -6x^2 +41x -6

Step-by-step explanation:

We simplify this expression by removing parentheses and combining like terms. Parentheses are removed using the distributive property.

The product of the final pair of polynomials in parentheses is ...

  (-4x^3 +5x -1)(2x -7) = (-4x^3 +5x -1)(2x) +(-4x^3 +5x -1)(-7)

  = -8x^4 +10x^2 -2x +28x^3 -35x +7

  = -8x^4 +28x^3 +10x^2 -37x +7

  = (5x^4 -9x^3 +7x -1) + (-8x^4 +4x^2 -3x +2) - (-8x^4 +28x^3 +10x^2 -37x +7)

  = (5 -8 -(-8))x^4 +(-9 -28)x^3 +(4 -10)x^2 +(7 -3 -(-37))x +(-1 +2 -7)

  = 5x^4 -37x^3 -6x^2 +41x -6

Answer:

$0.71/cup

Step-by-step explanation:

1 gallon = 16 cups

2 gallons = 2 × 1 gallon = 2 × 16 cups = 32 cups

$22.72/(2 gal) = $22.72/(32 cups) = $0.71/cup

Answer:

3/4

Step-by-step explanation:

multiply each fraction by 100 or you could find the LCM

Let [tex]A(t)[/tex] = amount of salt (in pounds) in the tank at time [tex]t[/tex] (in minutes). Then [tex]A(0) = 11[/tex].

Salt flows in at a rate

[tex]\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}[/tex]

and flows out at a rate

[tex]\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}[/tex]

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

[tex]\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}[/tex]

which I'll solve with the integrating factor method.

[tex]\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95[/tex]

[tex]-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}[/tex]

[tex]\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}[/tex]

Integrate both sides. By the fundamental theorem of calculus,

[tex]\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}} [/tex]

[tex]\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right) [/tex]

[tex]\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}[/tex]

[tex]\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}[/tex]

[tex]\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}[/tex]

After 1 hour = 60 minutes, the tank will contain

[tex]A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131[/tex]

pounds of salt.