Which equation has no solution?

4(x + 3) + 2x = 6(x + 2)
5 + 2(3 + 2x) = x + 3(x + 1)
5(x + 3) + x = 4(x + 3) + 3
4 + 6(2 + x) = 2(3x + 8)

Answer:

Step-by-step explanation:

x² + 2x = x(x + 2)

[tex]\sf \dfrac{x +5}{x + 2}-\dfrac{x+1}{x^2+2x}=\dfrac{x + 5}{x +2}-\dfrac{x+1}{x(x+2)}[/tex]

LCM = x(x+2)

                        [tex]\sf =\dfrac{(x+5)*x}{(x+2)*x}-\dfrac{x+1}{x(x+2)}\\\\=\dfrac{x*x + 5*x}{x^2+2x}-\dfrac{x+1}{x^2+2x}\\\\=\dfrac{x^2+5x - (x+1)}{x^2+2}\\\\=\dfrac{x^2+5x -x - 1}{x^2+2x)}\\\\=\dfrac{x^2+4x-1}{x^2+2x}[/tex]

Answer: [tex]\pm1,\pm3,\pm5, \pm15[/tex]

Step-by-step explanation:

We can list the possible rational roots of this polynomial using the Rational Root Theorem. This theorem states that all the possible rational roots of an equation follow the structure [tex]\frac{p}{q}[/tex], where p is any of the factors of the constant term and q is any of the factors of the leading coefficient.

In this example, -15 is the constant term and 1 is the leading coefficient ([tex]x^4[/tex] has a coefficient of 1).

The factors of -15 are [tex]\pm1,\pm3,\pm5, \pm15[/tex], while the factors of 1 are [tex]\pm1[/tex]. p is can be any one of the factors of -15, while q can be any of the factors of 1.

[tex]\frac{\pm1,\pm3,\pm5, \pm15}{\pm1}[/tex]

The possible roots can be any of the numbers on the top divided by any of the numbers on the bottom. Since dividing by 1 or -1 won't change any of the numbers on the top, the rational roots of this function are [tex]\pm1,\pm3,\pm5, \pm15[/tex].

Using the law of sines, it is found that the distance between the kite and the dog is of 284.77 meters.

Suppose we have a triangle in which:

The lengths and the sine of the angles are related as follows:

[tex]\frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}[/tex]

For the situation described, we have that:

Hence:

[tex]\frac{\sin{42^\circ}}{h} = \frac{\sin{87^\circ}}{425}[/tex]

Applying cross multiplication:

[tex]h = 425\frac{\sin{42^\circ}}{\sin{87^\circ}}[/tex]

h = 284.77 meters.

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The decimal notation for the given value in scientific notation is 215400.

Very large or very small numbers are denoted in scientific notation to simplify complex calculations.

The scientific notation is represented by: a × 10ⁿ

Wher a - any number or decimal number

n - the power of 10; it is an integer number

To convert this into decimal notation,

The given number is in the scientific notation,

2.154 × 10⁵

Since the power of 10 is positive, the decimal point in the number 2.154 moves towards the right. I.e,

2.154 × 10⁵ = 215400

Here power is 5. So, the decimal point moved five places to the right.

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The annual interest rate is 21%

The given parameters are

Loan Amount, P = $300

Interest, I = $63

Number of years, T = 1

The annual interest rate is calculated as

I = PRT

Substitute the known values in the above equation

63 = 300 * R * 1

Evaluate the product

300R = 63

Divide through by 300

R = 21%

Hence, the annual interest rate is 21%

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The slopes of the two diagonals of the rhombus are: A. b - f/a - e and -a + e/b - f.

To find the slope, the formula used is: change in y/change in x.

If two lines are perpendicular to each other, their slope values will be negative reciprocals.

In a rhombus, the two diagonals in the rhombus bisects each other at angle 90 degrees. This means that the two diagonals of a rhombus are perpendicular to each other. Therefore, the slope of the diagonals of any rhombus would be negative reciprocals.

The slope of the diagonals of the rhombus given would therefore be negative reciprocals to each other.

Given two endpoints of one of the diagonals as:

(a, b) = (x1, y1)

(e, f) = (x2, y2)

Slope (m) = change in y / change in x = b - f/a - e

The negative reciprocal of b - f/a - e is -a + e/b - f.

Therefore, the slopes of the two diagonals are: A. b - f/a - e and -a + e/b - f.

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The denominator cannot be zero, so

[tex]e^x - 9 = 0 \implies e^x = 9 \implies x = \ln(9)[/tex]

is not in the domain of [tex]f(x)[/tex].

[tex]\ln(x)[/tex] is defined only for [tex]x>0[/tex], and we have

[tex]\ln(x+1) > 0 \implies e^{\ln(x+1)} > e^0 \implies x+1 > 1 \implies x>0[/tex]

so there is no issue here.

By the same token, we need to have

[tex]x+1 > 0 \implies x > -1[/tex]

Taking all the exclusions together, we find the domain of [tex]f(x)[/tex] is the set

[tex]\left\{ x \in \Bbb R \mid x > 0 \text{ and } x \neq \ln(9)\right\}[/tex]

or equivalently, the interval [tex](0,\ln(9))\cup(\ln(9),\infty)[/tex].

Answer: -3 ≤ x ≤ -1

Step-by-step explanation:

1 ≤ 3 - 4x ≤ 9

1 + 3 ≤ - 4x ≤ 9 + 3; Add 3 on all sides

4 ≤ -4x ≤ 12

1 ≤ -x ≤ 3; Divide 4 on all sides

-1 ≥ x ≥ -3; Multiply -1 on all sides(FYI: When multiplying or dividing negative numbers in inequalities, make sure to reverse the signs as well)

Answer:

2

Step-by-step explanation:

1≤3-4x≤9

subtract 3

1-3≤3-4x-3≤9-3

-2≤-4x≤6

divide by 2

-1≤-2x≤3

multiply by -1

1≥2x≥-3

or

-3≤2x≤1

divide by 2

[tex]-\frac{3}{2} \leq \frac{2x}{2} \leq \frac{1}{2} \\-\frac{3}{2} \leq x\leq \frac{1}{2} \\[/tex]

length of interval

[tex]=\frac{1}{2} -(\frac{-3}{2} )\\=\frac{1}{2} +\frac{3}{2} \\=\frac{1+3}{2} \\=\frac{4}{2} \\=2[/tex]

The radius of the circle of the circle equation (x + 3/8)^2 + y^2 = 1 is 1 unit

The circle equation of the graph is given as:

(x + 3/8)^2 + y^2 = 1

The general equation of a circle is represented using the following formula

(x - a)^2 + (y - b)^2 = r^2

Where the center of the circle is represented by the vertex (a, b) and the radius of the circle is represented by r

By comparing the equations (x - a)^2 + (y - b)^2 = r^2 and (x + 3/8)^2 + y^2 = 1, we have the following comparison

(x - a)^2 = (x + 3/8)^2

(y - b)^2 = y^2

1 = r^2

Rewrite the last equation as follows:

r^2= 1

Take the square root of both sides of the equation

√r^2 = √1

Evaluate the square root of 1

√r^2 = 1

Evaluate the square root of r^2

r = 1

Hence, the radius of the circle of the circle equation (x + 3/8)^2 + y^2 = 1 is 1 unit

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[tex]\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}[/tex]

◉ [tex]\large\bm{ -4}[/tex]

[tex]\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}[/tex]

Before performing any calculation it's good to recall a few properties of integrals:

[tex]\small\longrightarrow \sf{\int_{a}^b(nf(x) + m)dx = n \int^b _{a}f(x)dx + \int_{a}^bmdx}[/tex]

[tex]\small\sf{\longrightarrow If \: a \angle c \angle b \Longrightarrow \int^{b} _a f(x)dx= \int^c _a f(x)dx+ \int^{b} _c f(x)dx }[/tex]

So we apply the first property in the first expression given by the question:

[tex]\small \sf{\longrightarrow\int ^3_{-2} [2f(x) +2]dx= 2 \int ^3 _{-2} f(x) dx+ \int f^3 _{2} 2dx=18}[/tex]

And we solve the second integral:

[tex]\small\sf{\longrightarrow2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot(3 - ( - 2)) }[/tex]

[tex]\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} 2dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot5 = 2 \int^3_{-2} f(x)dx10 = }[/tex]

Then we take the last equation and we subtract 10 from both sides:

[tex]\sf{{\longrightarrow 2 \int ^3_{-2} f(x)dx} + 10 - 10 = 18 - 10}[/tex]

[tex]\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 8}[/tex]

And we divide both sides by 2:

[tex]\small\longrightarrow \sf{\dfrac{2 { \int}^{3} _{2} }{2} = \dfrac{8}{2} }[/tex]

[tex]\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx=4}[/tex]

Then we apply the second property to this integral:

[tex]\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 4}[/tex]

Then we use the other equality in the question and we get:

[tex]\small\sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx = 8 + 2 \int ^3_{-2} f(x)dx = 4}[/tex]

[tex]\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =4}[/tex]

We substract 8 from both sides:

[tex]\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx -8=4}[/tex]

• [tex]\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =-4}[/tex]

The equations of the three altitudes of triangle ABC include the following:

A triangle can be defined as a two-dimensional geometric shape that comprises three (3) sides, three (3) vertices and three (3) angles only.

A slope is also referred to as gradient and it's typically used to describe both the ratio, direction and steepness of the function of a straight line.

Mathematically, the slope of a straight line can be calculated by using this formula;

[tex]Slope, m = \frac{Change\;in\;y\;axis}{Change\;in\;x\;axis}\\\\Slope, m = \frac{y_2\;-\;y_1}{x_2\;-\;x_1}[/tex]

Also, the point-slope form of a straight line is given by this equation:

y - y₁ = m(x - x₁)

Assuming the following parameters for triangle ABC:

For the equation of altitude AM, we have:

Slope of BC = (2 - 8)/(4 - 0)

Slope of BC = -6/4

Slope of BC = -3/2

Slope of AM = -1/slope of BC

Slope of AM = -1/(-3/2)

Slope of AM = 2/3.

The equation of altitude AM is given by:

y - y₁ = m(x - x₁)

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

3y = 2(x + 2)

3y = 2x + 4

3y - 2y - 4 = 0.

For the equation of altitude BN, we have:

Slope of CA = (2 - 0)/(4 - (-2))

Slope of CA = 2/6

Slope of CA = 1/3

Slope of BN = -1/slope of CA

Slope of BN = -1/(1/3)

Slope of BN = -3.

The equation of altitude BN is given by:

y - y₁ = m(x - x₁)

y - 8 = -3(x - 0)

y - 8 = -3x

y + 3x - 8 = 0.

For the equation of altitude CL, we have:

Slope of AB = (8 - 0)/(0 - (-2))

Slope of AB = 8/2

Slope of AB = 4

Slope of CL = -1/slope of AB

Slope of CL = -1/4

The equation of altitude CL is given by:

y - y₁ = m(x - x₁)

y - 2 = -1/4(x - 4)

4y - 2= -(x - 4)

4y - 2= -x + 4

4y + x - 2 - 4 = 0.

4y + x - 6 = 0.

In conclusion, we can infer and logically deduce that the equations of the three altitudes of triangle ABC include the following:

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Answer:

675

Step-by-step explanation:

100%-32%=68%

459/0.68= 675

Hope this helps! :)

Answer:

3

Step-by-step explanation:

2+ (-2/3)^2 ÷ 1/3

make -2/3 postive because its being raised by an even exponent

you never divide by fractors so make 1/3 into 3 and make it multiplication

2+ (2/3)^2 × 3

raise the fraction to the power of 2

2+ (4/9)× 3

cancel out the GCF

2+ 4/3

add

10/3

simplify

3.3 ≅ 3

Answer: B

Step-by-step explanation:

The horizontal change is 6.

The vertical change is 5.

So, the distance is [tex]\sqrt{6^2 + 5^2}=\sqrt{61}[/tex]

If you do in fact mean [tex]f(1)=f(6)=0[/tex] (as opposed to one of these being the derivative of [tex]f[/tex] at some point), then integrating twice gives

[tex]f''(x) = -\dfrac1{x^2}[/tex]

[tex]f'(x) = \displaystyle -\int \frac{dx}{x^2} = \frac1x + C_1[/tex]

[tex]f(x) = \displaystyle \int \left(\frac1x + C_1\right) \, dx = \ln|x| + C_1x + C_2[/tex]

From the initial conditions, we find

[tex]f(1) = \ln|1| + C_1 + C_2 = 0 \implies C_1 + C_2 = 0[/tex]

[tex]f(6) = \ln|6| + 6C_1 + C_2 = 0 \implies 6C_1 + C_2 = -\ln(6)[/tex]

Eliminating [tex]C_2[/tex], we get

[tex](C_1 + C_2) - (6C_1 + C_2) = 0 - (-\ln(6))[/tex]

[tex]-5C_1 = \ln(6)[/tex]

[tex]C_1 = -\dfrac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)[/tex]

Then

[tex]\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}[/tex]

From the constraint, we have

[tex]x+y+z=9 \implies z = 9-x-y[/tex]

so that [tex]s[/tex] depends only on [tex]x,y[/tex].

[tex]s = g(x,y) = xy + y(9-x-y) + x(9-x-y) = 9y - y^2 + 9x - x^2 - xy[/tex]

Find the critical points of [tex]g[/tex].

[tex]\dfrac{\partial g}{\partial x} = 9 - 2x - y = 0 \implies 2x + y = 9[/tex]

[tex]\dfrac{\partial g}{\partial y} = 9 - 2y - x = 0[/tex]

Using the given constraint again, we have the condition

[tex]x+y+z = 2x+y \implies x=z[/tex]

so that

[tex]x = 9 - x - y \implies y = 9 - 2x[/tex]

and [tex]s[/tex] depends only on [tex]x[/tex].

[tex]s = h(x) = 9(9-2x) - (9-2x)^2 + 9x - x^2 - x(9-2x) = 18x - 3x^2[/tex]

Find the critical points of [tex]h[/tex].

[tex]\dfrac{dh}{dx} = 18 - 6x = 0 \implies x=3[/tex]

It follows that [tex]y = 9-2\cdot3 = 3[/tex] and [tex]z=3[/tex], so the only critical point of [tex]s[/tex] is at (3, 3, 3).

Differentiate [tex]h[/tex] again and check the sign of the second derivative at the critical point.

[tex]\dfrac{d^2h}{dx^2} = -6 < 0[/tex]

for all [tex]x[/tex], which indicates a maximum.

We find that

[tex]\max\left\{xy+yz+xz \mid x+y+z=9\right\} = \boxed{27} \text{ at } (x,y,z) = (3,3,3)[/tex]

The second derivative at the critical point exists

[tex]$\frac{d^{2} h}{d x^{2}}=-6 < 0[/tex] for all x, which suggests a maximum.

Given, the constraint, we have

x + y + z = 9

⇒ z = 9 - x - y

Let s depend only on x, y.

s = g(x, y)

= xy + y(9 - x - y) + x(9 - x - y)

= 9y - y² + 9x - x² - xy

To estimate the critical points of g.

[tex]$&\frac{\partial g}{\partial x}[/tex] = 9 - 2x - y = 0

[tex]$&\frac{\partial g}{\partial y}[/tex] = 9 - 2y - x = 0

Utilizing the given constraint again,

x + y + z = 2x + y

⇒ x = z

x = 9 - x - y  

⇒ y = 9 - 2x, and s depends only on x.

s = h(x) = 9(9 - 2x) - (9 - 2x)² + 9x - x² - x(9 - 2x) = 18x - 3x²

To estimate the critical points of h.

[tex]$\frac{d h}{d x}=18-6 x=0[/tex]

⇒ x = 3

It pursues that y = 9 - 2 [tex]*[/tex] 3 = 3 and z = 3, so the only critical point of s exists at (3, 3, 3).

Differentiate h again and review the sign of the second derivative at the critical point.

[tex]$\frac{d^{2} h}{d x^{2}}=-6 < 0[/tex]

for all x, which suggests a maximum.

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Option A. The total share expense that the company would share would be given as 512,000

These are the necessary expenses that are needed for the smooth functioning of a particular business that are not within the confinement of the O and M agreement. It has to do with shared facilities.

The total number of the employees that are knwon to satistfy condition are given as

800 * 0.8

= 640

The options that are estmated that would be exercised

This is given as the employees * share option

= 640×50

=32000.

The total shae for the company would be gotten as

= 32000× $16

This gives us $512,000.

Hence it can be concluded that the total share of the company if they have 80 percent meeting the condition is $512000.

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Answer:

12. standard deviation =6.22972...

. variance = 38.80952...

By solving a linear equation, we conclude that you bought 4 buns and 5 cakes.

We will define the two variables:

We know that you spent exactly £4.35, then we only need to solve the linear equation:

x*0.40 + y*0.55 = 4.35

We need to solve that equation for x and y, such that the values of x and y can only be positive whole numbers.

We can rewrite the equation as:

y*0.55 = 4.35 - x*0.40

y = (4.35 - x*0.40)/0.55

Now we only need to evaluate it in different values of x, and see for which value of x, the outcome y is also a whole number.

We will see that for x = 4, we have:

y = (4.35 - 4*0.40)/0.55 = 5

So you bought 4 buns and 5 cakes.

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40p + 55p = 95p

435p ÷ 95p = 4.57...

So, the sum of the two pairs can go into £4.35 four times.

That meant if I increase the buns & cakes four times, it will help me get closer to the sum of £4.35 & work out which was extra

(40 x 4 = 160) & (55 x 4 = 220)

160 + 220= 380p or £3.80

Adding them up I get 380 pence & that's 55p short to getting to the sum of 435p or £4.35

(435-380=55p)

Thus, the extra is the cake, meaning you bought 4 buns and 5 cakes.

Hope this helps!

Answer:

Step-by-step explanation:

A new baby grew:

Total inches during two months:

Answer:

See below

Step-by-step explanation:

Distance = rate * time

              = 65 m/hr * 2 1/4 hr = 146.25 miles

rate = distance / time

for Declan :   rate =  146.25 miles / (2 hrs + 35/60 min) = 56.61 mph

Answer: 25 in³

Step-by-step explanation:

We can calculate the volume of the pyramid by first calculating the volume of a prism with the same dimensions, and then dividing by 3 (all pyramids a volume that's [tex]\frac{1}{3}[/tex] of the volume of a prism with the same dimensions).

The volume of a prism is its base area times its height. The base would be a square, so its area is 5², which is 25. The height is 3 inches, making the prism's volume 75 in³.

The volume of the pyramid would be one-third of this value, which is 75[tex]75\div3[/tex] which is 25 in³.

Answer: Look in step-by-step explanation

Step-by-step explanation:

Area has a unit of cm^2 or m^2, whilst Volume has a unit of cm^3 or m^3

Using this information, we can see that we have to square both sides of the similarity ratio for the area ratio and cube both sides of the similarity ratio for the volume ratio
For example, question 2 states that the similarity ratio is 3:6, so the area ratio is 3^2:6^2 or 9:36 and the volume ratio is 3^3:6^3 = 27:216

You can do the rest from here

The solution for the simultaneous equations 2x + y - 2z = 12, x + 2y +z =18 and 2x - y + 2z =16 are x = 7, y = 4 and z = 3

Linear equations are equations that have constant average rates of change, slope or gradient

A system of linear equations is a collection of at least two linear equations.

In this case, the system of equations is given as

2x + y - 2z = 12,

x + 2y +z =18

2x - y + 2z =16

Eliminate y and z in the equations by adding the first and the third equation together

This gives

2x + y - 2z = 12,

+

2x - y + 2z =16

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

4x = 28

Divide both sides by 4

x = 7

Substitute x = 7 in x + 2y +z =18 and 2x - y + 2z =16

7 + 2y +z =18

2(7) - y + 2z =16

This gives

2y + z = 11

-y + 2z= 2

Multiply -y + 2z= 2 by 2

-2y + 4z= 4

Add -2y + 4z= 4 and 2y + z = 11

5z = 15

Divide by 5

z = 3

Substitute z = 3 in -y + 2z= 2

-y + 2*3= 2

Evaluate

y = 4

Hence, the solution for the simultaneous equations 2x + y - 2z = 12, x + 2y +z =18 and 2x - y + 2z =16 are x = 7, y = 4 and z = 3

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The options Patel has to solve the quadratic equation 8x² + 16x + 3 = 0 is x = –1 Plus or minus StartRoot StartFraction 5 Over 8 EndFraction EndRoot.

8x² + 16x + 3 = 0

8x² + 16x = -3

8(x² + 2x) = -3

8(x² + 2x + 1) = -3 + 8

8(x² + 1) = 5

(x² + 1) = 5/8

(x + 1) = ± √5/8

x = -1 ± √5/8

Therefore,

x = –1 Plus or minus StartRoot StartFraction 5 Over 8 EndFraction EndRoot

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Explanation:

There are 25 choices for president. Then we have 25-1 = 24 choices for the secretary, and 24-1 = 23 choices for treasurer. This countdown (25,24,23) is because any given person cannot serve in more than one position.

Multiply out those values: 25*24*23 = 13800

Side note: you could use the nPr permutation formula as an alternative path. Plug in n = 25 and r = 3.

Answer:

10, -7, -35, 5

Step-by-step explanation:

f(9) = -5(-9 + 7) = 10

0 = -5(x + 7)

0 = x + 7

x = -7

f(0) = -5(0 + 7) = -35

-60 = -5(x + 7)

12 = x + 7

x = 5

Answer: 40

Step-by-step explanation:

Sub x=4 into 3x²-2x+1

∴ 3(4)²-2(4)+1

=3(16)-8+1

=48-8+1

=40

Answer:

41

Step-by-step explanation:

substitute the x to 4

so 3(4)² - 2(4) + 1 = 41

The total number of common tangents that can be drawn to the circles is 1

The tangent lines of a circle are the lines drawn, that touch the circle at only one point

The complete question is added as an attachment

From the attached figure, we have the following highlights:

The one point of intersection is the only point where both circles can have common tangents

Since there is only one point of intersection, then the number of common tangents on the circles is 1

Hence, the total number of common tangents that can be drawn to the circles is 1

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Answer:

The volume of the solid is  [tex]1,021m^3[/tex]

Step-by-step explanation:

1. Cylinder

For find the volume of a cylinder we use the next formula:

[tex]V_{cylinder} = \pi h r^2 = \pi \cdot 9m\cdot (5m)^2 \approx 706.86m^3[/tex]

Then the volume of the cylinder is equal to [tex]706.86m^3[/tex]

2. Cones

For find the volume of a cone we use the next formula:

[tex]V_{cone} = \pi \frac{h}{3} r^2 = \pi \cdot \frac{6m}{3} \cdot (5m)^2 \approx 157.08m^3[/tex]

However there are two cones so we have to multiply the volume of one cone for two and we get the total volume for the cones which is [tex]314.16m^3[/tex]

3. Sum of the volumes

Finally we sum the volumes of the cylinder and the cones for get the final result

[tex]V_{cylinder} + V_{cones} = 706.86m^3 + 314.16m^3 \approx 1021m^3[/tex]

So approximating the result is [tex]1021m^3[/tex]