A water trough is in the shape of a trapezoidal prism with the dimensions shown below.

A). find total volume of water that the trough can hold

Answer:

It will be 2 feet and 9.6 inches wide

Step-by-step explanation:

First you divide 24 inches(2 feet) by 5 inches to find out the rate unit rate of change. Your answer would be 4.8. Then you multiply 4.8 by 7 to figure out how wide it would be, which gives you 33.6 inches(or 2 feet and 9.6 inches)

Answer:

Option b (1/2,1/3)

Step-by-step explanation:

we have

2x-3y=0 ------> equation A

4x+6y=4 -----> equation B

Solve the system by elimination

Multiply equation A by 2 both sides

2(2x-3y)=2(0)

4x-6y=0 ------> equation C

Adds eqution B and equation C

4x+6y=4

4x-6y=0

-------------

4x+4x=4+0

8x=4

x=1/2

Find the value of y

2x-3y=0

2(1/2)-3y=0

1-3y=0

3y=1

y=1/3

the solution is the point (1/2,1/3)

Answer:

The correct answer is option A.  5 units

Step-by-step explanation:

Points to remember

Distance formula

Length of a line segment with end points (x1, y1) and (x2, y2) is given by,

Distance = √[(x2 - x1)² + (y2 - y1)²]

It is given that, two points are

(5, –2) and (5, 3)

To find the distance

(x1, y1) = (5, -2)  and (x2, y2) = (5, 3)

Distance = √[(x2 - x1)² + (y2 - y1)²]

 = √[(5 - 5)² + (3 - -2)²]

 = √[(5)²  = 5

Therefore the correct option is Option A  5 units

Answer:

8

Step-by-step explanation:

The number is a, another number is b.

a = b + 18 So, b=a - 18

(a+b) + ab

= a + a - 18 + a (a - 18)

= 2a - 18 + a^2 - 18a

{ ax^2 + bx + c }

= a^2 -1 6a - 18 {a = 1b = -16 }

When a = b/-2a = -16/-2*1 = 8

the a^2 - 1ba - 18 is minimum,

So the number is 8

[tex]\boxed{c^2+d^2+2cd}[/tex]

Hope this helps.

r3t40

It would be 49,009 because LA of a cone is height (50) times the radius (120) and that equals about 49,009.

Answer:

48984 m^2

Step-by-step explanation:

The height(h) of cone is given by: 50 m.

Diameter of cone is: 240 m.

Also radius(r) of cone is:240/2=120 m.

I'll do both.

[tex]

5x-97>-34\Rightarrow x>\frac{-34+97}{5}=\frac{63}{5}=\boxed{12.6} \\

\boxed{x\in(12.6, \infty)} \\ \\

2x+31<29\Longrightarrow x<29-31=\boxed{-2} \\

\boxed{x\in(-\infty, -2)}

[/tex]

[tex]2\log4 + \log3 -\log2 + \log5= \\ \\ = \log 2^4+ \log3 -\log2 + \log5 = \\ \\ = \log 16+ \log3 -\log2 + \log5 = \\ \\ = \log\Big(16\cdot 3:2\cdot 5\Big) = \log\Big(\dfrac{16\cdot 3\cdot 5}{2}\Big) = \log(8\cdot 3\cdot 5) = \\ \\ =\log120[/tex]

Opposite of adding three is subtracting three or adding -3

Answer:

₹[tex]9,600[/tex]

Step-by-step explanation:

we know that

The simple interest formula is equal to

[tex]I=P(rt)[/tex]

where

I is the amount interest Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

[tex]t=4.5\ years\\ P=?\\ I=1,512\\r=0.035[/tex]

substitute in the formula above

[tex]1,512=P(0.035*4.5)[/tex]

[tex]P=1,512/(0.035*4.5)=9,600[/tex]

Answer:

D. [tex](x-2y)(x-y)(x+y)[/tex]

Step-by-step explanation:

In the polynomial [tex]x^3-2x^2y-xy^2+2y^3[/tex] group first two terms and second two terms:

[tex](x^3-2x^2y)+(-xy^2+2y^3)[/tex]

First two terms have common factor [tex]x^2[/tex] and last two terms have common factor [tex]y^2,[/tex] hence

[tex](x^3-2x^2y)+(-xy^2+2y^3)=x^2(x-2y)+y^2(-x+2y)[/tex]

In brackets you can see similar expressions that differ by sign, so

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

Now use formula

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

You get

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

Answer:

96 meatballs are required

Step-by-step explanation:

We know that the formula for he catering business is

[tex]m=4a+2c[/tex]

Where

c= the number of children, m=the number of meat balls, a= the number of adults

So if [tex]c = 8[/tex]  and [tex]a = 20[/tex] then

[tex]m = 4(20) + 2(8)[/tex]

[tex]m = 80 + 16[/tex]

[tex]m = 96[/tex]

Finally  96 meatballs are required for the party

Answer:

439cor.to 3 Sig. fig.

Step-by-step explanation:

(5+12)×6×14-(5÷2)²π×14

Answer:

This answer would be 2,463.01 "D"

Step-by-step explanation:

The approximate size of the cover will be 2463.01 square feet

'A cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.'

According to the given problem,

The size of the cover of the tank = area of  the circle of the tank                                (approximately)

r = 28 feet

Area of the circle = [tex]\pi r^{2}[/tex]

                             = [tex]\pi *28^{2}[/tex]

                             = [tex]2463.01[/tex]

Hence, we have concluded that in order to cover the tank, which is a circular area, the area of the cover has to be approximately 2463.01 square feet.

Learn more about cylinder here:

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a term will be the expression between the + or - signs

[tex]\bf \stackrel{\stackrel{one}{\downarrow }}{x^2}+\stackrel{\stackrel{two}{\downarrow }}{xy}-\stackrel{\stackrel{three}{\downarrow }}{y^2}+\stackrel{\stackrel{four}{\downarrow }}{5}[/tex]

Answer:

40cm

Step-by-step explanation:

1. 10000cm^3/25cm= 400cm^2

2. 400cm^2/10 cm= 40cm

Final answer:

To find the height of the box, divide the volume of the box by the area of the base. In this case, the height is 40 cm.

Explanation:

To find the height of the box, we need to divide the volume of the box by the area of the base. The volume is given as 10000 cm³ and the base has dimensions 25 cm by 10 cm. So, the area of the base is 25 cm * 10 cm = 250 cm². Now, we can find the height by dividing the volume by the area: Height = Volume / Area = 10000 cm³ / 250 cm² = 40 cm.

Answer: circumference: 22π  area: 121π

Step-by-step explanation:

area of a circle is [tex]\pi r^2[/tex]

circumference is [tex]2\pi r[/tex]

41° because it’s a right angle and if 90=2y+8, y=41

Answer:

41

Step-by-step explanation:

Answer:

[tex]5,890.5\ ft^{2}[/tex]

Step-by-step explanation:

step 1

Find the area that cover each individual head of a sprinkler system

The area of the circle is equal to

[tex]A=\pi r^{2}[/tex]

we have

[tex]r=25\ ft[/tex]

assume

[tex]\pi=3.1416[/tex]

substitute

[tex]A=(3.1416)(25)^{2}[/tex]

[tex]A=1,963.5\ ft^{2}[/tex]

step 2

Find the area that covers 3 sprinkler heads

[tex](3)*1,963.5=5,890.5\ ft^{2}[/tex]

Answer:

5890.5 ft²

Step-by-step explanation:

did it on edg

[tex]\bf \stackrel{\textit{using the exponential model}}{N=2^D}~\hspace{7em}\begin{array}{ccll} \stackrel{days}{D}&\stackrel{\$}{N}\\ \cline{1-2} 1&2^1\implies 2\\ 2&2^2\implies 4 \end{array}[/tex]

so, using that exponential model, the 1st output value works, but the second value of 2² does not give us 8 as output.

let's check the linear model using slopes to get the equation.

[tex]\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{8})\qquad \impliedby \begin{array}{|cc|ll} \cline{1-2} D&N\\ \cline{1-2} 1&2\\ 2&8\\ \cline{1-2} \end{array}[/tex]

[tex]\bf slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{8-2}{2-1}\implies \cfrac{6}{1}\implies 6 \\\\\\ \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=6(x-1) \\\\\\ y-2=6x-6\implies y=6x-4[/tex]

now, using that model, x = 6, then y = 6(6) - 4, or y = 32.

After giving 93 centimeters (which is 0.93 meters) of his 214 meters of rope to his brother, Ravi has 213.07 meters of rope left.

This question involves the concept of subtraction in the measurement unit of meters and centimeters. Initially, Ravi has 214 meters of rope. If he gives 93 centimeters of rope to his brother, keep in mind that 1 meter equals 100 centimeters. So, 93 centimeters is equal to 0.93 meters.

So, to find out how much rope Ravi is left with, you subtract the amount he gave to his brother from the original amount he had: 214 meters - 0.93 meters = 213.07 meters. This is the amount of rope Ravi has left after giving some to his brother.

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Answer:

156.25% increase (56.25% added)

Step-by-step explanation:

Area formula of circle: A = πr²

1. Area when rope = 20 ft:

Plug in: A = π(20)²

Multiply: A = 400π ft²

2. Area when rope = 25 ft:

Plug in: A = π(25)²

Multiply: A = 625 ft²

3. Percent increase:

Increase compared to original: 625/400 = 1.5625 = 156.25%

Answer:

56.25%

Step-by-step explanation:

We are given that a horse is tethered on a 20 ft rope. If the rope is lengthened to 25 ft, we are to percentage by which its grazing area increases.

Grazing area with 20 ft rope = [tex]\pi \times 20^2[/tex] = [tex]400\pi[/tex]

Grazing area with 25 ft rope = [tex]\pi \times 25^2[/tex] = [tex]625\pi[/tex]

Percentage by which area increases = [tex] \frac { 6 2 5 \pi - 400 \pi } { 400 \pi } \times 100 [/tex] = 56.25%

Answer:

[tex]175\ in^{2}[/tex]

Step-by-step explanation:

step 1

Find the scale factor

we know that

If two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube

Let

z----> the scale factor

x----> volume of the larger solid

y----> volume of the smaller solid

[tex]z^{3}=\frac{x}{y}[/tex]

we have

[tex]x=125\ in^{3}[/tex]

[tex]y=27\ in^{3}[/tex]

substitute

[tex]z^{3}=\frac{125}{27}[/tex]

[tex]z=\frac{5}{3}[/tex]

step 2

Find the surface area of the larger solid

we know that

If two figures are similar, then the ratio of its surface areas is equal to the scale factor squared

Let

z----> the scale factor

x----> surface area of the larger solid

y----> surface area of the smaller solid

[tex]z^{2}=\frac{x}{y}[/tex]

we have

[tex]z=\frac{5}{3}[/tex]

[tex]y=63\ in^{2}[/tex]

substitute

[tex](\frac{5}{3})^{2}=\frac{x}{63}[/tex]

[tex]x=\frac{25}{9}*63=175\ in^{2}[/tex]

Answer:

Part A)  Is a directly proportional

Part B) The constant of proportionality is [tex]k=0.25[/tex]

Part C) The equation is [tex]y=0.25x[/tex]

Part D) [tex]62.5[/tex]  acres of soybeans

Step-by-step explanation:

Part A) Are the number of acres of corn and the number of acres of soybeans directly proportional or inversely proportional?

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form [tex]y/x=k[/tex]

Let

x-----> the number of acres of corn

y---->  the number of acres of soybeans

so

[tex]y/x=k[/tex]  

if the number of acres of corn increases then the number of acres of soybeans increases

if the number of acres of corn decreases then the number of acres of soybeans decreases

therefore

The relationship is a directly proportional

Part B) What is the constant of proportionality?

we have that

For x=200, y=50

substitute

[tex]y/x=k[/tex]

[tex]k=50/200[/tex]

[tex]k=0.25[/tex]

Part C) Let x equal the number of acres of corn and y equal the number of acres of soybeans. Write an equation to show this relationship.

The linear equation that represent the direct variation is equal to

[tex]y/x=k[/tex]

we have

[tex]k=0.25[/tex]

substitute

[tex]y/x=0.25[/tex]

[tex]y=0.25x[/tex]

Part D) How many acres of soybeans can the farmer plant this year if he plants 250 acres of corn?

For x=250

Find the value of y

substitute in the linear equation the value of x and solve for y

[tex]y=0.25(250)=62.5[/tex] -----> acres of soybeans

Answer:

∴The distance CA =  = 2√2

Step-by-step explanation:

Find the distance CA:

The distance between two points (x₁,y₁),(x₂,y₂) = d

The coordinates of point C = (-2,2)

The coordinates of point A = (0,0)

The distance CA distance between C and A

∴The distance CA =  = 2√2

The distance[tex]\( C'B' \) is \( \sqrt{10} \)[/tex] units.

after applying the transformation[tex]\( T: (x, y) \rightarrow (x + 2, y + 1) \)[/tex], the distance between points [tex]\( C' \)[/tex] and [tex]\( B' \)[/tex] is [tex]\( \sqrt{10} \)[/tex] units.

The distance [tex]\( C'B' \)[/tex] is [tex]\( \sqrt{10} \)[/tex] units.

To find the distance [tex]\( C'B' \)[/tex], we first need to find the coordinates of points [tex]\( C' \)[/tex] and [tex]\( B' \)[/tex] after applying the transformation [tex]\( T: (x, y) \rightarrow (x + 2, y + 1) \) to points \( C \)[/tex] and [tex]\( B \).[/tex]

Given the coordinates of [tex]\( C \)[/tex] and [tex]\( B \)[/tex] as[tex]\( C(1, 2) \)[/tex] and [tex]\( B(4, 3) \)[/tex] respectively, we apply the transformation to each point:

For point [tex]\( C \):[/tex]

[tex]\[ C'(x', y') = (x + 2, y + 1) = (1 + 2, 2 + 1) = (3, 3) \][/tex]

For point [tex]\( B \):[/tex]

[tex]\[ B'(x', y') = (x + 2, y + 1) = (4 + 2, 3 + 1) = (6, 4) \][/tex]

Now, we use the distance formula to find the distance between [tex]\( C' \)[/tex]and [tex]\( B' \):[/tex]

[tex]\[ C'B' = \sqrt{(x'_2 - x'_1)^2 + (y'_2 - y'_1)^2} \][/tex]

[tex]\[ C'B' = \sqrt{(6 - 3)^2 + (4 - 3)^2} \][/tex]

[tex]\[ C'B' = \sqrt{(3)^2 + (1)^2} \][/tex]

[tex]\[ C'B' = \sqrt{9 + 1} \][/tex]

[tex]\[ C'B' = \sqrt{10} \][/tex]

Thus, the distance [tex]\( C'B' \)[/tex] is[tex]\( \sqrt{10} \)[/tex] units.

In conclusion, after applying the transformation [tex]\( T: (x, y) \rightarrow (x + 2, y + 1) \)[/tex], the distance between points [tex]\( C' \)[/tex] and [tex]\( B' \)[/tex] is [tex]\( \sqrt{10} \)[/tex] units.

Complete question

Using the transformation T: (x, y) (x + 2, y + 1), find the distance named. Find the distance C'B’

Check the picture below.

Answer:

It's a Right Triangle

Answer:

(b, c)

Step-by-step explanation:

The mid-point of two vertices or point is calculated by adding the respective coordinates of those points and then dividing by two. The x-coordinates will be added and then divided by 2 and then y-coordinates will be added and divided by 2.

So for the given question,

P(0,0)

Q(2a,0)

And

R(2b, 2c)

Mid-point of PR =( (0+2b)/2, (0+2c)/2)

=(2b/2, 2c/2)

=(b,c)

So the mid-point of PR is (b, c)

Last option is the correct answer..

Answer:

1/4

Step-by-step explanation:

Step 1: Find the GCF. List out the factors of the numerator and the denominator. 1, 3, 9 are the factors of 9, while 1, 2, 3, 4, 6, 9, 12, 18, and 36 are the factors of 36. 9 is a common factor of both of them, so the GCF is 9.

Step 2: Divide the numerator and denominator by 9 (the GCF). 9/9 is 1. 36/9 is 4. This means that our fraction is 1/4. The fraction is in simplest form.

To solve the equation[tex]\(x^2 = 50\),[/tex] we'll take the square root of both sides. However, when we do this, we need to consider both the positive and negative square roots:

[tex]\[ x = \pm \sqrt{50} \]\[ x = \pm \sqrt{25 \cdot 2} \]\[ x = \pm 5\sqrt{2} \]So, the solutions to the equation \(x^2 = 50\) are \(x = 5\sqrt{2}\) and \(x = -5\sqrt{2}\).[/tex]

To solve the equation [tex]\(x^2 = 50\),[/tex]we'll take the square root of both sides. Remembering that the square root of a number has both positive and negative solutions, we have:

[tex]\[ x = \pm \sqrt{50} \]\[ x = \pm \sqrt{25 \cdot 2} \]\[ x = \pm 5\sqrt{2} \][/tex]

Therefore, the solutions to the equation[tex]\(x^2 = 50\) are \(x = 5\sqrt{2}\) and \(x = -5\sqrt{2}\). This means that when \(x\) is equal to either \(5\sqrt{2}\) or \(-5\sqrt{2}\), \(x^2\) will be equal to 50. These solutions represent the values of \(x\)[/tex]that satisfy the original equation and make it true.