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3 years ago

9m 7.2m 3m find the volume

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

5 0

Answer:

The volume of the prism is 97.2 m³

Step-by-step explanation:

Let's recall that the volume of a triangular prism is base * height and the base is a triangle, then the formula is as follows:

Volume = 1/2 (base of the triangle * height of the triangle) * height of the prism

Replacing with the real values, we have:

Volume = 1/2 (7.2 * 9) * 3

Volume = 32.4 * 3 = 97.2 m³

The volume of the prism is 97.2 m³

You might be interested in

For all three water parks the cost of a function of numbers of right compare the functions for all three water parks in terms of

Aleks [24]

Answer:

See Explanation

Step-by-step explanation:

Given

See attachment for complete question

First, we determine the cost function for all the three rides.

Ride A

From the graph, we have the following points

$(x_1,y_1) = (0,8)$

$(x_2,y_2) = (2,12)$

Calculate slope

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{12-8}{2-0}$

$m = \frac{4}{2}$

$m =2$ --- This represents the rate per ride

The equation is the calculated as:

$y = m(x - x_1) + y_1$

So, we have:

$y = 2(x - 0) + 8$

$y = 2(x) + 8$

$y = 2x + 8$

So, the cost function is:

$C(x) =2x + 8$

Calculate the cost of admission i.e. x=0

$C(0) = 2*0+8 = 8$

So, we have:

$C(0) = 8$ --- Admission Charge

$C(x) =2x + 8$ --- Cost function

$m =2$ --- Rate per ride

Ride B

From the table, we have the following points

$(x_1,y_1) = (0,12)$

$(x_2,y_2) = (4,15)$

Calculate slope

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{15-12}{4-0}$

$m = \frac{3}{4}$

$m = 0.75$ --- This represents the rate per ride

The equation is the calculated as:

$y = m(x - x_1) + y_1$

So, we have:

$y = 0.75(x - 0) + 12$

$y = 0.75(x) + 12$

$y = 0.75x + 12$

So, the cost function is:

$C(x) = 0.75x + 12$

Calculate the cost of admission i.e. x=0

$C(0) = 0.75*0 + 12=12$

So, we have:

$C(0) =12$ --- Admission Charge

$C(x) = 0.75x + 12$ --- Cost function

$m = 0.75$ --- Rate per ride

Ride C

No additional fee;

So, the cost function is;

$C(x) = 30$

In summary, we have:

Ride A

$C(x) =2x + 8$ --- Cost function

$m =2$ --- Rate per ride

Ride B

$C(x) = 0.75x + 12$ --- Cost function

$m = 0.75$ --- Rate per ride

Ride C

$C(x) = 30$ --- Cost function

By comparison

Ride A has the highest rate per ride of (#2), followed by ride B with a rate of #0.75 per ride.

Ride C has no charges per ride

The impact on the total cost is that:

Ride A: People that opt for ride A will pay the least to get admitted (i.e #8) but they pay the most (i.e. #2) per each ride they take

Ride B: People that opt for ride B will pay #12 to get admitted, but they pay 0.75 per each ride they take

For A and B, the overall cost depends on the number of rides taken.

Ride C: Irrespective of the number of rides taken, people that opt for ride C will pay the same flat fee of #30

4 0

3 years ago

A $300 bike is marked down to $175 what is the percent markdown to the nearest whole percent?

LUCKY_DIMON [66]

Answer:

58% due to dividing 300 by 100 then getting that answer (3) then divide 175 by 3 which equals one percent and you get 58.333 so round that to the nearest whole and you get 58%

8 0

4 years ago

Read 2 more answers

15.30 find the inverse laplace transform of: 1. (a) f1(s) = 6s 2 8s 3 s(s 2 2s 5) 2. (b) f2(s) = s 2 5s 6 (s 1) 2 (s 4) 3. (c) f

EleoNora [17]

The solution of the inverse Laplace transforms is mathematically given as

$f_{1}(t)=e^{-t}\sin (2 t)$

$f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}$

$f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)$

<h3>What is the inverse Laplace transform?</h3>

Generally, the equation for the function is mathematically given as

$F_{1}(s)=\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}$

By Applying the Partial fractions method

$\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}=\frac{A}{s}+\frac{B s+C}{s^{2}+2 s+5}$

$6 s^{2}+8 s+3=A\left(s^{2}+2 s+5\right)+(B s+C) s$

$\begin{aligned}&3=5 A \\&A=\frac{3}{5}\end{aligned}$

Considers s^2 coefficient

$\begin{aligned}&6=A+B \\&B=6 \cdot A \\&B=\frac{27}{5}\end{aligned}$

Consider s coeffici ent

$\begin{aligned}&8=2 A+C \\&C=8-2 A \\&C=\frac{34}{5}\end{aligned}$

Putting these values into the previous equation

$&F_{1}(s)=\frac{3}{5 s}+\frac{27 s+34}{5\left(s^{2}+2 s+5\right)} \\\\&F_{1}(s)=\frac{3}{5 s}+\frac{27(s+1)}{5\left((s+1)^{2}+4\right)}+\frac{7 \times 2}{10\left((s+1)^{2}+4\right)}$

By taking Inverse Laplace Transforms

$f_{1}(t)=\frac{3}{5}+\frac{27}{5} e^{-t} \cos (2t) + \frac{7}{10}\\\\$

$f_{1}(t)=e^{-t}\sin (2 t)$

For B

$F_{2}(s)=\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}$

By Applying Partial fractions method

$\begin{aligned}&\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\\\&s^{2}+5 s+6=A(s+1)(s+4)+B(s+4)+C(s+1)^{2}\end{aligned}$

at s=-1

$1-5+6=3 B \\\\B=\frac{2}{3}$

at s=-4

$&16-20+6=9 C \\\\&9 C=2 \\\\&C=\frac{2}{9}$

at s^2 coefficient

1=A+C

A=1-C

A=7/9

inputting Variables into the Previous Equation

$\begin{aligned}&F_{2}(s)=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\&F_{2}(s)=\frac{7}{9(s+1)}+\frac{2}{3(s+1)^{2}}+\frac{2}{9(s+4)}\end{aligned}$

By taking Inverse Laplace Transforms

$f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}$

For C

$F_{3}(s)=\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}$

Using the strategy of Partial Fractions

$\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}=\frac{A}{s+1}+\frac{B s+C}{s^{2}+4 s+8}$

$10=A\left(s^{2}+4 s+8\right)+(B s+C)(s+1)$

S=-1

10=(1-4+8) A

A=10/5

A=2

Consider constants

10=8 A+C

C=10-8 A

C=10-16

C=-6

Considers s^2 coefficient

0=A+B

B=-A

B=-2

inputting Variables into the Previous Equation

$&F_{3}(s)=\frac{2}{s+1}+\frac{-2 s-6}{\left((s+2)^{2}+4\right)} \\\\&F_{3}(s)=\frac{2}{s+1}-\frac{2(s+2)}{\left((s+2)^{2}+4\right)}-\frac{2}{\left((s+2)^{2}+4\right)}$

Inverse Laplace Transforms

$f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)$