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vichka [17]
3 years ago
10

Can someone answer this question please please help me I really need it if it’s correct I will mark you brainliest .

Mathematics
2 answers:
Keith_Richards [23]3 years ago
8 0

Quarter circle: A = 1/4 x 3.14 x 6^2 = 28.26

Big rectangle: A = 4 x 12 = 48

Small rectangle: A = 2 x 6 = 12

A = 28.26 + 48 + 12

A = 88.26 square yards

Hope this helps! :)

Semenov [28]3 years ago
4 0

Answer:

88.26 square yards

Step-by-step explanation:

Big rectangle: A = 4 x 12 = 48

Quarter circle: A = 1/4 x 3.14 x 6^2 = 28.26

Small rectangle: A = 2 x 6 = 12

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Larry Mitchell invested part of his $25,000 advance at 2% annual simple interest and the rest at 3% annual simple interest. If h
alexandr402 [8]

Answer:

  • $1000 at 3%
  • $24000 at 2%

Step-by-step explanation:

Let x represent the amount invested at 3%. Then the total interest earned is ...

  0.03x +0.02(25000 -x) = 510

  0.01x = 10 . . . . . . . . . . . . subtract 500, collect terms

  x = 1000

Mitchell invested $1,000 at 3% and $24,000 at 2%.

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3 years ago
Evaluate the function f(x)=2x-2 when x equals negative 1<br> A. -4<br> B. -3<br> C. 0<br> D. 4
Alona [7]
F(x)= -4. 
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8 0
4 years ago
Put the differential equation 9ty+ety′=yt2+81 into the form y′+p(t)y=g(t) and find p(t) and g(t). p(t)= help (formulas) g(t)= he
alexgriva [62]

Answer:

p(t) = \frac{9t^{3} + 729t  - 1}{e^{t}(t^{2} + 81) }

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And

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Step-by-step explanation:

Given that,

The differential equation is -

9ty + e^{t}y' = \frac{y}{t^{2} + 81 }

e^{t}y' + (9t - \frac{1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t(t^{2} + 81 ) - 1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t^{3} + 729t  - 1}{t^{2} + 81 } )y = 0\\y' + [\frac{9t^{3} + 729t  - 1}{e^{t}(t^{2} + 81) } ]y = 0

By comparing with y′+p(t)y=g(t), we get

p(t) = \frac{9t^{3} + 729t  - 1}{e^{t}(t^{2} + 81) }

g(t) = 0

And

The differential equation 9ty + e^{t}y' = \frac{y}{t^{2} + 81 } is  linear and homogeneous.

6 0
3 years ago
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Answer:

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Step-by-step explanation:

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