A farmer with 4000 meters of fencing wants to enclose a rectangular plotthat borders a straight river. If the farmer does not fence the side along theriver, what is the largest rectangular area that can be enclosed

Answer:

3x - y +5=0

Step-by-step explanation:

Slope:-

=> m = 2-8/-1-1 = -6/-2

=> m = 3

Using point slope form :-

=> y - 2 = 3 ( x - (-1))

=> y - 2 = 3x + 3

=> 3x - y +5 = 0

Answer:

y=3x+5

Step-by-step explanation:

first find the gradient

m=y2-y1/x2-x1

from the coordinates you are given x one is -1,x two is1,y one is 2,y two is 8

8-2/1-(-1)

=3

therefore the gradient is 3

after finding the gradient use the equation

y-y1=m(x-x1)

substitute x1,y1 and the gradient..y and x are constant

y-2=3(x-(-1))

y-2=3x+3

y=3x+3+2

y=3x+5

Answer:

SA=BA+LA and SA=BA+ph

Step-by-step explanation:

I just looked it up

The correct formulas to find the surface area of a right prism are:

SA = BA + LA and SA = BA + ph.

Options (A), and (D) are the correct answer.

A prism is a three-dimensional object.

There are triangular prism and rectangular prism.

We have,

The correct formulas to find the surface area of a right prism are:

1)

SA = BA + LA, where SA is the total surface area, BA is the area of the two identical bases, and LA is the lateral area (the sum of the areas of all the rectangular sides).

2)

SA = BA + ph, where SA is the total surface area, B is the area of one base, p is the perimeter of the base, h is the height of the prism, and ph is the area of all the rectangular sides (the lateral area).

Therefore,

The correct formulas to find the surface area of a right prism are:

SA = BA + LA and SA = BA + ph.

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Answer:

[tex]perimetro = \dfrac{(3 + \sqrt{3})m}{2}[/tex]

Step-by-step explanation:

triángulo 30°-60°-90°

[tex] 1~:~\sqrt{3}~:~2 [/tex]

[tex] 1m~:~m\sqrt{3}~:~2m [/tex]

[tex] \dfrac{m}{2}~:~\dfrac{m\sqrt{3}}{2}~:~m [/tex]

[tex]perimetro = \dfrac{m}{2} + \dfrac{m\sqrt{3}}{2} + m[/tex]

[tex]perimetro = \dfrac{m}{2} + \dfrac{m\sqrt{3}}{2} + \dfrac{2m}{2}[/tex]

[tex]perimetro = \dfrac{m + m\sqrt{3} + 2m}{2}[/tex]

[tex]perimetro = \dfrac{3m + m\sqrt{3}}{2}[/tex]

[tex]perimetro = \dfrac{(3 + \sqrt{3})m}{2}[/tex]

Step 1: Distribute

5x + 10 - 3x > 7 - 4x + 12

Step 2: Combine like terms

2x + 10 > -4x + 19

Step 3: Move all variable terms to one side, and all constants to the other

6x + 10 > 19

6x > 9

Step 4: Divide

x > 9 / 6

x > 1.5

Hope this helps!

Answer:

103 is my answer.

Step-by-step explanation:

hope it helps

Answer:

[tex]y=|x-(-2)|+2[/tex]

Step-by-step explanation:

Hope this is helpful.

The required absolute function is given as y = |x - 2| + 2.

Given that,
From the graph, y = |x - h| + k, values of h and k is to be determined,

Functions are the relationship between sets of values. e g y=f(x), for every value of x there is its exists in a set of y. x is the independent variable while Y is the dependent variable.

Here,
In the graph and function, h represents the shift on the x-axis, which 2 units left so the value of h is 2, while k is given the shift of the function over the y-axis which 2 units up means that k = 2

Thus, The required absolute function is given as y = |x - 2| + 2.

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(a⁴-3a²b²+b⁴)/(a²-ab-b²)

Let me know if there is something wrong to my answer ^_^

Answer:

hope it will helpfulll to youuu

Answer:

3,4

Step-by-step explanation:

Answer: The question is not complete. the figure is attached below.

The polygon is a rectangle.

Step-by-step explanation:

Line 1 goes through (-2, 3)

hence the slope of line 1 = -3/2

Line 2 goes through (0, 3) (2, 0)

hence the slope of line 2 = -3/2

Line 3 goes through (0, 1) (-3, 1)

hence the slope of line 3 = 2/3

Line 4 goes through (3, 0) (0, -2)

Hence the slope of line 4 = 2/3

therefore -3/2 * 2/3 = -1

Line1 + Line3      Line1 + Line4       Line2 + Line3        Line2+Line4

because, the distance between Line1 and Line2 is not equal to that Line3 and Line4.

The polygon is a rectangle polygon.

Answer:

3

Step-by-step explanation:

A = [tex]\frac{base1 + base2}{2}[/tex] x h     Formula

[tex]\frac{base1 + 5}{2}[/tex] x 20 = 80      Substitution

[tex]\frac{base1 + 5}{2}[/tex] = 4                 Work  (divide by 20, multiply by 2, subtract 5)

base1 + 5 = 8

base1 = 3                  Solution

Answer:

4

Step-by-step explanation:

For this, you need to find the scale factor of two sides that are already given to you.

So, we will have to use the hypotenuse and one of the legs to make sure there is an accurate scale factor.

Hypotenuse:

15 / 5 = 3

Leg:

6 / 2 = 3

____________

So the scale factor is 3.

Using the leg (with the x) we need to divide 12 by the scale factor (3) to give us what x is equal to.

12 / 3 = 4

So, the answer is 4.

Answer:

Perpendicular, CD has a slope of -5/2

Step-by-step explanation:

Parallel lines have the same slope, perpendicular have slopes that are negative reciprocals, or in other words if one slope is m then the other's slope is -1/m and if it is neither fo those then it is neither parallel or perpendicular.  Also,  if two lines have the same slope you have to make sure they are not the same line.  

Now, to find the slope  when you have two points you use the formula (y2 - y1)/(x2 - x1), where the points are (x1, y1) and (x2, y2).  You can make either point point 1 or point 2.  Just make sure x1 and y1 are from one point and x2 and y2 are from the other point.  

So the two lines are AB and CD.  Since the question asks I will start with CD.  The points are (1,8) and (3,3)  So I will call C point 1 and D point 2.  

(y2 - y1)/(x2 - x1)

(3 - 8)/(3 - 1)

-5/2

Thankfully this is lowest terms.

Now for AB.  A will be point 1 and B will be point 2

(y2 - y1)/(x2 - x1)

(4-2)/(5-0)

2/5

So they are not the same, so what if you take negative 1 over one of the slopes.  

well. -1/(2/5) = -5/2

So these are perpendicular.  

If you would like an explanation for how to find the full equation of a line I can explain that.  

Answer:

2×5×7+2×5×2+2×7×2

70+20+28

108cm^2

[tex] \huge\boxed{\mathfrak{Answer}}[/tex]

[tex]l = 5 \: cm \\ w = 2 \: cm \\ h = 7 \: cm[/tex]

The formula to find SA => 2lh + 2lw + 2hw

[tex]SA => 2lh + 2lw + 2hw \\ = 2 \times 5 \times 7 + 2 \times 5 \times 2 + 2 \times 7 \times 2 \\ = 70 + 20 + 28 \\ = 118 \: \: cm {}^{2} [/tex]

=> The surface area of the rectangular prism is 118 cm².

Answer:

:

Example Question #1:

Step-by-step explanation:

Which of the following is the equation of a line that is parallel to the line 4x – y = 22 and passes through the origin?

Possible Answers:

4x – y = 0

(1/4)x + y = 0

4x + 8y = 0

4x = 8y

y – 4x = 22

Correct answer:

4x – y = 0

Explanation:

We start by rearranging the equation into the form y = mx + b (where m is the slope and b is the y intercept); y = 4x – 22

Now we know the slope is 4 and so the equation we are looking for must have the m = 4 because the lines are parallel. We are also told that the equation must pass through the origin; this means that b = 0.

In 4x – y = 0 we can rearrange to get y = 4x. This fulfills both requirements.

', .

Answer:

Because it is a right triangle one angle is obviously 90°. The other two are approximately 36.87° and 53.13°.

hope it helps you

have a good day

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Step-by-step explanation:

hope it helps thank you❤

Respuesta:

12 horas

Explicación paso a paso:

Dado que :

A Mar del Plata = cada 2 horas

A Necochea = cada 4 horas

A San Clemente = cada 6 horas

Número de horas que tardarán los 3 autobuses en salir al mismo tiempo, si ambos salieron ahora;

Tome el mínimo común múltiplo de los 3 autobuses:

Múltiplos de:

2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22

4: 4, 8, 12, 16, 20, 24, 28, 32

6: 6, 12, 18, 24, 30, 36, 42

El mínimo común múltiplo de 2 horas, 4 horas y 6 horas es 12;

Por lo tanto, los 3 autobuses saldrán a la misma hora en las próximas 12 horas:

Answer:

1) -4(3-5x) >= –6x+9

–12+20x >= -6x+9

The choose (D)

2 ) The choose (A)

3) The choose (A) step 1

4) 1/2(n-4)-3=3-(2n+3)

n=2

The choose ( B)

5) 2.5 (5x-4)=10+4(1.5+0.5x)

x=2

The choose (C)

1. -12 + 20 > -6x + 9

2. simplify by combining like terms

3. step 1

4. n=2

5. 2

If “the notation arcsin x represents the inverse function to sine” is true when the sine function is to the interval [ -pi/2, pi/2 ].

The sine function is one of the three primary functions in trigonometry, the others being cosine, and tan functions.

The confusion is compounded by the fact that we use the notation sin⁻¹(x) interchangeably with arcsin(x), and we call it the inverse sine.

Here is a counterexample disproving your given statement:

Let x = π.

The value π is in the domain of sin(x).

arcsin[sin(π)] = arcsin(0) = 0

arcsin[sin(π)] ≠ π

Therefore arcsin(x) is not the inverse of sin(x).

If you restrict the sine function to the interval [ -pi/2, pi/2 ]. Otherwise, arcsin will not be a function.

Hence, If “the notation arcsin x represents the inverse function to sine” is true when the sine function is to the interval [ -pi/2, pi/2 ].

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Answer:

It's d or the 4th graph

Step-by-step explanation:

A graph which represents the overall equation (inequality) represented by this scenario.

A graph can be defined as a type of chart that is commonly used for the graphical representation of data points on both the horizontal and vertical lines of a cartesian coordinate, which are the x-coordinate (x-axis) and y-coordinate (y-axis).

Let the variable x represent the total number of guests.

Let the variable y represent the amount of meat

Next, we would write an inequality that represents an amount of meat that must be 2 pounds fewer than one-third (1/3) the total number of guests. Since the customer decides they want at least that amount of meat, the inequality symbol that must be used is greater than (>).

y > 1/3(x) + 2

y > x/3 + 2

In conclusion, the y-value for this inequality graph (see attachment) intersects the x-axis at 0 and 6 with the boundary line shaded upward and dashed to indicate that it is not part of the solution, which must be represented with a greater than (>) inequality symbol.

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Answer:

17

Step-by-step explanation:

Complete question is;

Given a circle with (4. -3) and (2, 1) as the endpoints of the diameter.​ Write the equation of the circle.

Answer:

x² + y² - 6x + 2y + 5 = 0

Step-by-step explanation:

The end points of the diameter are;

(4. -3) and (2, 1).

Thus, the centre coordinates will be the midpoint of the diameter endpoints.

Thus;

Centre coordinates = ((4 + 2)/2), ((-3 + 1)/2) = (3, -1)

Diameter;

d = √(1 - (-3))² + (2 - 4)²)

d = √20

d = 2√5

Radius = ½ × diameter

Thus;

r = ½ × 2√5

r = √5

Equation of a circle is;

(x - a)² + (y - b)² = r²

Where;

(a, b) are coordinates of the centre of the circle

r is radius.

Thus;

(x - 3)² + (y - (-1))² = (√5)²

x² - 6x + 9 + y² + 2y + 1 = 5

x² + y² - 6x + 2y + 10 = 5

x² + y² - 6x + 2y + 10 - 5 = 0

x² + y² - 6x + 2y + 5 = 0

Answer:

ummm u did not add the statements to the question

Step-by-step explanation:

but a polygon ABCD might not be a rectangle if:

Answer:

40

Step-by-step explanation:

A+B+C = 31

Add 3 years to each age

A+3  +B+3  + C+3  = 31 +3+3+3

They will be

A+3  +B+3  + C+3  = 40

Answer:

it will be 40

Step-by-step explanation:

If they are altogether 31 years old now in 3 years we just add 9 thus it is 40

Answer:

The tank is 10cm high

Step-by-step explanation:

Given

[tex]L=60cm[/tex] -- length

[tex]W=60cm[/tex] -- width

[tex]x = \frac{1}{5}[/tex] --- water lever

[tex]Addition = 24L[/tex]

Required

The height of the tank

Let y represents the remaining fraction before water is added.

So:

[tex]y + x = 1[/tex]

Make y the subject

[tex]y = 1 - x[/tex]

[tex]y = 1 - \frac{1}{5}[/tex]

Solve

[tex]y = \frac{5 - 1}{5}[/tex]

[tex]y = \frac{4}{5}[/tex]

Represent the volume of the tank with v

So:

[tex]y * v = 24L[/tex]

Make v the subject

[tex]v = \frac{24L}{y}[/tex]

Substitute: [tex]y = \frac{4}{5}[/tex]

[tex]v = \frac{24L}{4/5}[/tex]

[tex]v = 30L[/tex]

Represent the height of the tank with h;

So, the volume of the tank is:

[tex]v = lwh[/tex]

Make h the subject

[tex]h = \frac{v}{lw}[/tex]

Substitute values for v, l and w

[tex]h = \frac{30L}{60cm * 50cm}[/tex]

Convert 30L to cm^3

[tex]h = \frac{30*1000cm^3}{60cm * 50cm}[/tex]

[tex]h = \frac{30000cm^3}{3000cm^2}[/tex]

[tex]h = 10cm[/tex]

Answer:y=-5+1

Step-by-step explanation:

Answer:

143

Step-by-step explanation:

to find the area we multiply half of the sum of the bases with height

(18+8)/2*11 = 143

Answer:

The length and width that maximize the area are:

W = 2*√8

L = 2*√8

Step-by-step explanation:

We want to find the largest area of a rectangle inscribed in a semicircle of radius 4.

Remember that the area of a rectangle of length L  and width W, is:

A = L*W

You can see the image below to see how i will define the length and the width:

L = 2*x'

W = 2*y'

Where we have the relation:

4 = √(x'^2 + y'^2)

16 = x'^2 + y'^2

Now we can isolate one of the variables, for example, x'

16 - y'^2 = x^'2

√(16 - y'^2) = x'

Then we can write:

W = 2*y'

L = 2*√(16 - y'^2)

Then the area equation is:

A = 2*y'*2*√(16 - y'^2)

A = 4*y'*√(16 - y'^2)

If A > 1, like in our case, maximizing A is the same as maximizing A^2

Then if que square both sides:

A^2 = (4*y'*√(16 - y'^2))^2

      = 16*(y'^2)*(16 - y'^2)

      = 16*(y'^2)*16 - 16*y'^4

      = 256*(y'^2) - 16*y'^4

Now we can define:

u = y'^2

then the equation that we want to maximize is:

f(u) = 256*u - 16*u^2

to find the maximum, we need to evaluate in the zero of the derivative:

f'(u) = 256 - 2*16*u = 0

      u = -256/(-2*16) = 8

Then we have:

u = y'^2 = 8

solving for y'

y' = √8

And we know that:

x' = √(16 - y'^2) = √(16 - (√8)^2) = √8

And the dimensions was:

W = 2*y' = 2*√8

L = 2*y' = 2*√8

These are the dimensions that maximize the area.

Answer:

Solution given:

Let there be a point P(x, y) equidistant from

A(-3, 2) and B(0,4),

so PA = PB,

[tex]\sqrt{(x+3)²+(y-2)²}=\sqrt{(x-0)²+(y-4)²}[/tex]

squaring both side

[tex](\sqrt{(x+3)²+(y-2)²})^{2}=(\sqrt{(x-0)²+(y-4)²})²[/tex]

x²+6x+9+y²-4y+4=x²+y²-8y+16

x²+6x+y²-4y-x²-y²+8y=16-4-9

6x-4y+8y=3

6x-4y=3 is a required locus

Actually:

A locus is a curve or other figure formed by all the points satisfying a particular equation of the relation between coordinates, or by a point, line, or surface moving according to mathematically defined conditions.