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2 years ago

The perimeter please help

Mathematics

2 answers:

ycow [4]2 years ago

7 0

Answer:I believe it is d

Step-by-step explanation:

ollegr [7]2 years ago

5 0

Answer:

the answer is 2/3 (E)

Step-by-step explanation:

36 divided by 18 = 2

54 divided by 18 = 2

You might be interested in

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t$

= $e^{-(t-3)} u (t-3)$

Recall that:

$y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

Then

$y(t) = 7 -7e^{-t} +e^{-(t-3)} u (t-3)$

$y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)$

$\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

3 0

3 years ago

A construction crew has just finished building a road. The crew worked for 4 2/7 days. If the road is 9 kilometers long, how man

Lana71 [14]

4 2/7 equals 30/7
30/7 divided by 9 equals 10/21
So the crew built 10/21 of a km everyday

4 0

3 years ago

Read 2 more answers

I need help with this question

Nastasia [14]

Answer:

120 cm³

Step-by-step explanation:

V = (1/3)×base area×height

= (1÷3)×(4×9)×10

=120

8 0

3 years ago

Please, help thanks! I would like an explanation how to do it too, IF you dont mind.

Aneli [31]

Answer:

Phoebe has £180, Andy has £60, and Polly has £30.

Step-by-step explanation:

Ok, to start of let's establish some variables! We'll give Phoebe the variable "p", Andy the variable "a", and Polly the variable "o". By giving each person a variable we can create an equation and solve for it!

So, what do we know?

Well we know the following information...

p + a + o = 270

(Phoebe + Andy + Polly = £270)

p = 3a

(Phoebe = 3 times more than Andy)

a = 2o

(Andy = 2 times more than Polly)

To be able to solve this equation we have to solve for just one variable first.The only one we can use is "o" for Polly!

3(2o) + 2o + o = 270

6o + 2o + o = 270 (Distribute 3 to get like terms)

9o = 270 (Combine like terms [6+2+1])

o = 30 (Divide by 9 on both sides to get "o" by itself)

Now that we have how much Polly has, we have to solve for Andy and Phoebe. (To do this we will input our value, 30, where our variable is "o")

Andy

a = 2(30) (Input term)

a = 60 (Multiply by 2 to get final answer)

(Since we now know what Andy has we can input 60 into our "a" variable to solve for Phoebe)

Phoebe

p = 3(60) (Input term)

p = 180 (Multiply by 3 to get final answer)

Now! Just to check we are going to input our variables into our original equation to be sure we have the right variables.

p + a + o = 270

180 + 60 + 30 = 270!!

Hope this Helps! :)

Have any questions? Ask below in the comments and I will try my best to answer.

-SGO

4 0

2 years ago

A. 2 cm B. 2√2 cm C. 4 cm D. 4√2 cm

algol [13]

Answer:

.D. 4√2 cm

Step-by-step explanation:

With reference angle 45°

base (b) = 4 cm

hypotenuse (h) = AC

Now

Cos 45° = b /h

1/√2 = 4 / AC

AC = 4√2 cm

3 0

2 years ago