3 pipes take 60 minutes to water the field. How much time will it take to water the field with 6 pipes? Is this direct proportion or inverse proportion and why? Full working out and equation please.​

Answer:

Option C and option D

Step-by-step explanation:

we have that

[tex]y=x^{2}-4x+5[/tex]

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert the equation into vertex form

Complete the square

[tex]y-5=x^{2}-4x[/tex]

[tex]y-5+4=x^{2}-4x+4[/tex]

[tex]y-1=x^{2}-4x+4[/tex]

[tex]y-1=(x-2)^{2}[/tex]

[tex]y=(x-2)^{2}+1[/tex] ----> equation of the parabola in vertex form

The vertex is the point (2,1)

therefore

when the bird is 2 feet away from the first fence post, it reaches its minimum height of 1 foot

Answer: C and D

Step-by-step explanation:

Answer:all real number greater than 0

Step-by-step explanation:

Firstly if you input any number equal to 0 or less than 0 you will not find the defined range...

You cant use o or any negetive number as domain in the term of log or ln type math..

But if u put any value more than 0 you can find all real number as range

Such as, log(0.001)=-3

log(1)=0

log(120)=2.07

So the domain is all real number above o...but the range is all real number including 0 and negetive number..

The domain of y=logx is all real numbers greater than 0.

Hence, option b is the correct answer.

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Answer:

2(d - 3) is the equation. You cannot solve for d. You can only simplify it

The rectangular wall that is painted in 15 minute. 6.4 square feet area is painted per minute.

A rectangle is a part of a quadrilateral, whose sides are parallel to each other and equal.

Given that,

The length of the rectangular wall = 12 ft.

The width of the rectangular wall = 8 ft.

The area of rectangle = 12 x 8 = 96 square feet.

Since, in 15 minutes, the rectangular wall is painted.

To find the area that painted in one minute,

Use ratio property,

15 minutes = 96 square feet painted,

1 minute = 96 / 15 square feet painted.

1 minute = 6.4 square feet.

6.4 square feet painted in 1 minute.

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Step-by-step explanation:

(1)

Let the numerator be x and denominator be y. A/Q x + y = 4 + 2x → - x + y = 4

(2)

multiplying each term by 2, 2x-2y= -8

(3)

Also, (x+3) / (y+3) = 2 / 3 → 3x - 2y = -3

Subtracting (2) from (3) → x = 5 and by putting this in (1) we can get y=9. Hence, the fraction is 5 / 9

Answer:

[tex]\frac{5}{9}[/tex]

Step-by-step explanation:

let the fraction be [tex]\frac{x}{y}[/tex], then

x + y = 2x + 4 ( subtract x from both sides )

y = x + 4 → (1)

If 3 is added to numerator and denominator, then

[tex]\frac{x+3}{y+3}[/tex] = [tex]\frac{2}{3}[/tex] ( cross- multiply )

3(x + 3) = 2(y + 3) ← distribute both sides

3x + 9 = 2y + 6 ← substitute y = x + 4

3x + 9 = 2(x + 4) + 6

3x + 9 = 2x + 8 + 6 = 2x + 14 ( subtract 2x from both sides )

x + 9 = 14 ( subtract 9 from both sides )

x = 5

Substitute x = 5 into (1)

y = 5 + 4 = 9

Hence the original fraction is [tex]\frac{5}{9}[/tex]

[tex]\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{5}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{5-2}{1-(-3)}\implies \cfrac{3}{1+3}\implies \cfrac{3}{4}[/tex]

[tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=\cfrac{3}{4}[x-(-3)] \implies y-2=\cfrac{3}{4}(x+3) \\\\\\ y-2=\cfrac{3}{4}x+\cfrac{9}{4}\implies y=\cfrac{3}{4}x+\cfrac{9}{4}+2\implies y=\cfrac{3}{4}x+\cfrac{17}{4}[/tex]

Answer:b

Step-by-step explanation: you graph each answer choice and see which one looks like the graph

Answer:  The correct option is

(B) [tex]y=\left(\dfrac{1}{2}\right)^{x-3}+1.[/tex]

Step-by-step explanation:  We are given to select the function that is shown in the graph.

From the graph, we know that

if the function is represented by y = f(x), then f(0) = 9. That is, the value of y at x = 0 is 9.

Option (A) :  Here, the given function is

[tex]y=\left(\dfrac{1}{2}\right)^{x+3}-1.[/tex]

So, at x = 0, the value of y is given by

[tex]y(x=0)=\left(\dfrac{1}{2}\right)^{0+3}-1=\dfrac{1}{8}-1=-\dfrac{7}{8}\neq 9.[/tex]

So, this option is not correct.

Option (B) :  Here, the given function is

[tex]y=\left(\dfrac{1}{2}\right)^{x-3}+1.[/tex]

So, at x = 0, the value of y is given by

[tex]y(x=0)=\left(\dfrac{1}{2}\right)^{0-3}+1=\left(\dfrac{1}{8}\right)^{-1}+1=8+1=9.[/tex]

So, this option is CORRECT.

Option (C) :  Here, the given function is

[tex]y=\left(\dfrac{1}{2}\right)^{x-1}+3.[/tex]

So, at x = 0, the value of y is given by

[tex]y(x=0)=\left(\dfrac{1}{2}\right)^{0-1}+3=2+3=5\neq 9.[/tex]

So, this option is not correct.

Option (D) :  Here, the given function is

[tex]y=\left(\dfrac{1}{2}\right)^{x+1}-3.[/tex]

So, at x = 0, the value of y is given by

[tex]y(x=0)=\left(\dfrac{1}{2}\right)^{0+1}-3=\dfrac{1}{2}-3=-\dfrac{5}{2}\neq 9.[/tex]

So, this option is not correct.

Thus, (B) is the correct option.

Answer:

Answer is 11+6i

Step-by-step explanation:

You just have to add imaginary part together and the real part. The answer will be 11+6i

Answer: 11+6i

Step-by-step explanation:

Answer:

[tex]\large\boxed{(f-g)(x)=2x-3}[/tex]

Step-by-step explanation:

[tex](f-g)(x)=f(x)-g(x)\\\\f(x)=3x-1,\ g(x)=x+2\\\\\text{substitute:}\\\\(f-g)(x)=(3x-1)-(x+2)\\\\=3x-1-x-2\qquad\text{combine like terms}\\\\=(3x-x)+(-1-2)\\\\=2x-3[/tex]

Answer: 64π

Step-by-step explanation: A = (d)^2 π

64 x π

64π

Answer:

y = -7

Step-by-step explanation:

The easisest way to find the slope of this line is to use slope-intercept form.

Slope-intercept form:

y = mx + b

Where m = slope and b = y -intercept

In this graph, the y-intercept is -7. However, the line doesn't have a slope since its a straight horizontal line.

So, the mx part of the equation isn't a part of this new equation.

So, your equation would just y = -7

Answer:

The answer in the procedure

Step-by-step explanation:

we know that

The cross  section formed by the slice is a square with the same dimensions of the base of rectangular prism

The length side of the square is 6 cm

The area of the cross section is equal to

[tex]A=b^{2}[/tex]

[tex]b=6\ cm[/tex]

substitute

[tex]A=6^{2}[/tex]

[tex]A=36\ cm^{2}[/tex]

Answer:

square and 36

Step-by-step explanation:

I took the test

Answer:

[tex]y=25,700(0.85)^{x}[/tex]

Step-by-step explanation:

we know that

In this problem we have a exponential function of the form

[tex]y=a(b)^{x}[/tex]

where

y ----> represent the pool’s loss of water

x ----> the number of days

a is the initial value

a=25,700 gal

b ----> is the base

r=15%=15/100=0.15

b=(1-r)=1-0.15=0.85

The function is equal to

[tex]y=25,700(0.85)^{x}[/tex]

Answer:

B) x +20 is greater than or less to 45

Step-by-step explanation:

Answer:

x + 20 ≥  45

Step-by-step explanation:

You currently have : $20

You will earn : $x

Tomorrow you will end up with (20 + x) dollars

the pair of jeans cost $45, in order to afford the jeans, the amount of money that you will need must be equal or more than $45

Hence,

Money you will have tomorrow ≥ 45

or

x + 20 ≥  45 (Answer)

Answer:

C. 6x - 7y = 21

Step-by-step explanation:

y=2/3 x - 3, y-intercept = -3

A. 2/3 x + 3y = -3

   3y = -2/3 x - 3

     y = -2x - 1; y-intercept = -1

B.-2/3 x + 3y = 6

   3y = 2/3 x + 6

     y = 2x + 2; y-intercept = 2

C. 6x-7y=21

    7y = 6x - 21

     y = 6/7 x - 3; y-intercept = -3

D. x+4y = 12

   4y = -x +12

     y = -1/4 + 3; y-intercept = 3

Answer is C. 6x - 7y = 21

Answer:

46°

Step-by-step explanation:

Alternate angles from 46° and alternate angles are equal

Answer:

After these reflections, the coordinates of P′ will be (4 , -4)

Step-by-step explanation:

* Lets revise some transformation

- If point (x , y) reflected across the x-axis

∴ Its image is (x , -y)

- If point (x , y) reflected across the y-axis

∴ Its image is (-x , y)

* Lets solve the problem

- The triangle PQR with coordinates P(-4, -4), Q(-1, -3), and R(-3, -1)

- The triangle is reflected across the x-axis

∵ Δ PQR is reflected across the x-axis

∴ All y-coordinates of the vertices P, Q , R reversed their signs

∴ The new points will be (-4 , 4) , (-1 , 3) , (-3 , 1)

- The new vertices will reflected across the y-axis

∴ All x-coordinates of the new vertices reversed their signs

∴ The new points will be (4 , 4) , (1 , 3) , (3 , 1)

- The new vertices will reflected across the x-axis to form Δ P'Q'R'

∴ All y-coordinates of the new vertices reversed their signs

∴ P' = (4 , -4) , Q' = (1 , -3) , R' = (3 , -1)

* After these reflections, the coordinates of P′ will be (4 , -4)

Final answer:

After reflecting triangle PQR across the x-axis, then the y-axis, and the x-axis again, point P' will have coordinates (4, -4).

Explanation:

Reflecting a triangle across an axis involves flipping the triangle over that axis. Each reflection inverses the corresponding coordinate (x or y) of each vertex of the triangle, while the other coordinate remains the same. Starting with the first reflection across the x-axis, the y-coordinate of each point negates, but the x-coordinate remains unchanged. The second reflection is across the y-axis, which negates the x-coordinate and keeps the y-coordinate (already negated from the first reflection) the same. The third reflection across the x-axis negates the y-coordinate again, effectively returning it to its original value before the first reflection. So for point P(-4, -4), after these reflections, the new coordinates for P' will be (4, -4).

Answer:

a) y = sin x

b) y = csc x

c) y = cot x

Step-by-step explanation:

only d is even

Answer:

The roots are {-3, -2, 2, 3}.

Step-by-step explanation:

The trick here is to represent y² by some other letter, such as x.  If we do that, then y^4-13y^2+36=0 becomes x² - 13x + 36 = 0.

Recognize that the factors -4 and -9 of 36 sum up to 13.  Thus,

x² - 13x + 36 = 0 is equivalent to (x - 4)(x - 9) = 0, and x = 4 or x = 9.

Recall that x = y².

When x = 4:  y² = 4, and so y = ±2.

When x = 9, y² = 9, and so y = ±3.

The roots are {-3, -2, 2, 3}.

The roots of the bi-quadratic equation y⁴ - 13y² + 36 = 0 are,

- 2, 2, -3, 3.

A polynomial is an algebraic expression.

A polynomial of degree n in variable x can be written as,

a₀xⁿ + a₁xⁿ⁻¹ + a₂xⁿ⁻² +...+ aₙ.

Given, y⁴ - 13y² + 36 = 0.

This a bi-quadratic polynomial.

Let, x = y².

Therefore,

x² - 13x + 36 = 0.

x² - 4x - 9x + 36 = 0.

x(x - 4) - 9(x - 4) = 0.

(x - 4)(x - 9) = 0.

x = 4 Or x = 9.

Now,

For x = 4 ⇒ y² = 4 ⇒ y = ± 2.

For x = 9 ⇒ y² = 9 ⇒ y = ± 3.

So, The roots of the biquadratic equation are - 2, 2, -3, 3.

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Answer:

= [tex]3ab^{\frac{2}{3} }[/tex]

Step-by-step explanation:

∛(27 a³ [tex]b^{7}[/tex])

= ∛27 · ∛a³ · ∛b³ · ∛b³ · ∛b

= 3 · a · b · b ∛b

= 3ab² [tex]b^{\frac{1}{3} }[/tex]

or

= [tex]3ab^{\frac{2}{3} }[/tex]

Answer:

The maximum speed of Hal's airplane in still air is:

[tex]v= 250\ miles/h[/tex]

The speed of the wind

[tex]c = 50\ miles/h[/tex]

Step-by-step explanation:

Remember that the velocity v equals the distance d between time t.

[tex]v=\frac{d}{t}[/tex]  and [tex]t*v=d[/tex]

The distance that Hal travels when traveling with the wind is:

[tex](2\ hours)(v + c) = 600[/tex] miles

Where v is the speed of Hal and c is the wind speed.

The distance when traveling against the wind is:

[tex](3\ hours)(v-c) = 600[/tex] miles

Now we solve the first equation for v

[tex](2)(v + c) = 600[/tex]

[tex]2v + 2c = 600[/tex]

[tex]2v= 600-2c[/tex]

[tex]v= 300-c[/tex]

Now we substitute the value of v in the second equation and solve for c

[tex]3((300-c)-c) = 600[/tex]

[tex]3(300-2c) = 600[/tex]

[tex]900-6c = 600[/tex]

[tex]-6c = 600-900[/tex]

[tex]-6c = -300[/tex]

[tex]6c = 300[/tex]

[tex]c = 50\ miles/h[/tex]

Then:

[tex]v= 300-(50)[/tex]

[tex]v= 250\ miles/h[/tex]

The maximum speed of Hal's airplane in still air is:

[tex]v= 250\ miles/h[/tex]

The speed of the wind

[tex]c = 50\ miles/h[/tex]

Answer:

(n+9) (2p+q)

Step-by-step explanation:

2p(n + 9) + q(n + 9) =

Factor out the term (n+9)

(n+9) (2p+q)

This is completely factor

Answer: -1 is the slope of the red line,

Step-by-step explanation: The slope of parallel lines are always the same. Hope this helps!

Answer:

the slope would be -1

Step-by-step explanation:

Answer:

1. [tex]\frac{x^2+4x-7}{x-1}[/tex]

2. [tex]\frac{x^2-2x+7}{x-1}[/tex]

3. [tex]\frac{2x^2-x-7}{x-1}[/tex]

4. [tex]\frac{2x^2-3x+7}{x-1}[/tex]

Step-by-step explanation:

1. [tex](x+5) + \frac{-2}{x-1}[/tex]

Taking LCM

[tex]=\frac{(x-1)(x+5)+(-2)}{x-1}\\ Solving:\\=frac{x(x+5)-1(x+5)-2}{x-1} \\=frac{x^2+5x-1x-5-2}{x-1} \\Adding\,\,like\,\,terms:\\=\frac{x^2+4x-7}{x-1}[/tex]

2. [tex]x-1 +\frac{6}{x-1}[/tex]

Taking LCM and solving

[tex]=\frac{(x-1)(x-1)+6}{x-1}\\=\frac{(x(x-1)-1(x-1)+6}{x-1}\\=\frac{x^2-1x-1x+1+6}{x-1}\\Adding\,\,like\,\,terms:\\=\frac{x^2-2x+7}{x-1}[/tex]

3. [tex](2x+1)+\frac{-6}{x-1}[/tex]

Taking LCM and solving:

[tex]=\frac{(2x+1)(x-1)-6}{x-1} \\=\frac{2x(x-1)+1(x-1)-6}{x-1} \\=\frac{2x^2-2x+1x-1-6}{x-1}\\Adding\,\,like\,\,terms:\\=\frac{2x^2-x-7}{x-1}[/tex]

4. [tex](2x-1)+\frac{6}{x-1}[/tex]

Taking LCM and solving:

[tex]=\frac{(2x-1)(x-1)+6}{x-1} \\=\frac{2x(x-1)-1(x-1)-6}{x-1} \\=\frac{2x^2-2x-1x+1+6}{x-1}\\Adding\,\,like\,\,terms:\\=\frac{2x^2-3x+7}{x-1}[/tex]

Answer:

I'm pretty sure that is correct.

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

Use exponents property:

[tex]\dfrac{x^a}{x^b}=x^{a-b}[/tex]

1. Note that

[tex]\dfrac{x^7}{x^{11}}=x^{7-11}=x^{-4}=\dfrac{1}{x^4}[/tex]

and

[tex]\dfrac{y^6}{y^8}=y^{6-8}=y^{-2}=\dfrac{1}{y^2}[/tex]

2. Now

[tex]\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\sqrt{\dfrac{5}{x^4y^2}}=\dfrac{\sqrt{5}}{\sqrt{x^4}\sqrt{y^2}}=\dfrac{\sqrt{5}}{x^2y}[/tex]

because [tex]x>0,\ y>0[/tex]

Answer:

2

Step-by-step explanation:

The function is given as  [tex]f(x)=\frac{3}{(x-11)(x+4)}[/tex]

Vertical asymptotes occur when the denominator is set to 0.

Thus,

(x-11)(x+4) = 0

x = 11 or x = -4

Hence, there are 2 vertical asymptotes

Answer: 2

Step-by-step explanation:

A P E X

Answer:

Option C

Step-by-step explanation:

we know that

The altitude of a three-dimensional object is equal to the height of the object, is the perpendicular distance of the base to the other base or the perpendicular distance of the base to the apex of the object

therefore

A segment that is perpendicular to the planes containing the two bases

The statement which best describes an altitude of a three-dimensional object is: C. a segment that is perpendicular to the planes containing the two bases.

Altitude is also referred to as an elevation and it can be defined as the vertical distance (height) above the surface of a plane.

In Geometry, the altitude of a three-dimensional object is characterized by the following:

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Answer:

Step-by-step explanation:

15/25. Divide both sides by a common number which is 5 3/5 is the final answer.

The result of the given mathematical expression is [tex]\frac{3}{5}[/tex] or 0.6.

"A mathematical expression is a finite combination of symbols that is well-formed according to rules that depend on the context."

Given mathematical expression is

= (15 ÷ 25)

[tex]= \frac{15}{25}\\= \frac{3}{5}\\= 0.6[/tex]

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Answer:

[tex](0, \infty)[/tex]

Step-by-step explanation:

By definition all the exponential functions of the form

[tex]f (x) = a (b) ^ x[/tex]

Where a is the main coefficient and b is the base they have range [tex](0, \infty)[/tex]

Whenever [tex]a> 0[/tex] and b> 0.

In this case the function is:

[tex]f (x) = 11(\frac{1}{3}) ^ x[/tex]

Note that for this function [tex]a = 11> 0[/tex] and [tex]b =\frac{1}{3}>0[/tex]

Therefor the range is: [tex](0, \infty)[/tex]

A relation is a function if you associate exactly one output for every input. This means that, when you choose a value for x, there must be only one correspondent value for y. This only happens in the top-right parabola.