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telo118 [61]
3 years ago
12

taron buys fencing for his square dog pen that measures 9 feet per side. How many inches of fencing does Tim buy altogether?

Mathematics
2 answers:
liberstina [14]3 years ago
8 0
He has 432 inches altogether because  since a square has 4 sides, 9 x 4= 36 feet then you have to convert it to inches so 36 x 12= 432 inches.
kobusy [5.1K]3 years ago
8 0
9×4=36. a square has four sides and if each side is 9 inches it is 36 inches
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( m + 4 ) ( m + 1 )
love history [14]
(m+4)(m+1)

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2 years ago
Is the expression 8 x 85 equivalent to (8 X 835? Explain.<br><br>pls help I'm dying lol<br>​
Varvara68 [4.7K]

8 \times 835 = 6680

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2 years ago
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The area of the entire figure below is 111 square unit.
Ivahew [28]

The striped rectangle has a total area of 69,375 cm².

<h3>How to calculate the area of the striped rectangle?</h3>

To find the area of the striped rectangle we must carry out the following procedures.

Find the area of each segment of the rectangle, that is, what is the area of each of the 24 squares that make up the rectangle, for that we divide the 111cm² of the total area into 24.

  • 111cm² ÷ 24 = 4.625cm²

According to the above, we infer that each square has an area of 4.625cm².

On the other hand, to find the area of the striped rectangle we must take into account that it is 5 squares long by 3 squares high, that is, its area is 15 squares.

  • 3 × 5 = 15

Finally, each square has an area of 4.625cm², so to find the total area of the striped area we must multiply the area of each square by the number of squares.

15 × 4.625cm² = 69.375cm²

Learn more about areas in: brainly.com/question/27683633

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7 0
2 years ago
Using b=1+r, what does b=
alisha [4.7K]

Answer:

Adding 12 to the circle area is equal to the square area.

Or

s2 = 12 + A

Where

s = side of square

A = area of circle

So

s2 = 12 + 36

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4 0
2 years ago
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
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