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4 years ago

Find the area of the shaded figure

Mathematics

1 answer:

Rasek [7]4 years ago

3 0

Answer:

Step-by-step explanation: first start with the size on the outside and add it together then subtract 6 in and thats should help

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PLEASE HELP ASAP!! DUE TODAY

SpyIntel [72]

Answer:

Step-by-step explanation:

when 1/2 inch equals 4 feet the room is 2.25 inches, then

2 1/4 = 8/4 + 1/4 = 9/4

9/4 inches is the scale distance of the room

each 1/2 inch is 4 feet so divide 9/4 by 1/2 then multiply that by 4 to find the distance of the room

9/4 ÷ 1/2 = 9/4

9/4 * 2/1 = 18/4

18/4 = 9/2

9/2 * 4 feet = 9/2 * 4/1

9/2 * 4/1 = 36/2

36/2=18

the room is 18 feet across

then if the scale is 2/3 of an inch is 4 feet , then

2/3 * 9/4 = 18/12

18/12 = 3 inches is the scale size of Quinto's room

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3 years ago

Which number has a 3 with a value that is 100 times greater than the value of the 3 in 20.342?

rjkz [21]

36.25 ..................................................................

3 0

3 years ago

Read 2 more answers

Solve write or draw to explain Saul and Luisa each scored 167 points on computer games How many points did they score together

nirvana33 [79]

Answer:167 + 167 = 334

Step-by-step explanation:

7 0

3 years ago

The Laplace Transform of a function f(t), which is defined for all t > 0, is denoted by L{f(t)} and is defined by the imprope

lesya692 [45]

(1) D

$L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt$

Integrate by parts, taking

$u = t \implies \mathrm du=\mathrm dt$

$\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}$

Then

$L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}$

(2) A

$L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}$

7 0

3 years ago

Please could you help me!!!!!

Degger [83]

Answer:

Step-by-step explanation:

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3 years ago