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Art [367]
3 years ago
6

Find the formula for the area A and the perimeter P of the shape shown find the area and perimeter when R is 5 cm find R when A

is 100cm2 find r when P is 50cm find the value of r that makes A = p numerically

Mathematics
1 answer:
Ghella [55]3 years ago
6 0

Answer:

1)Area; A = ¼πr²

Perimeter; P = πr/2 + 2r

2)A = 19.63 cm²

P = 17.85 cm

3) r = 8.885 cm

4) r = 14 cm

Step-by-step explanation:

This is a quadrant of a circle. Thus;

Area of a circle is πr². A quadrant is a quarter of a circle. Thus;

Formula for Quadrant Area is; A = ¼πr²

A) Perimeter of a circle is 2πr. Thus, perimeter of a quadrant is a quarter of the full circle perimeter.

Formula for the quadrant perimeter in the image given is;

P = 2πr/4 + 2r

P = πr/2 + 2r

B) When r is 5 cm;

A = ¼π(5)²

A = 19.63 cm²

P = π(5)/2 + 2(5)

P = 17.85 cm

C) when A is 100cm²:

¼πr² = 100

r² = 100 × 4/π

r² = 78.9358

r = √78.9358

r = 8.885 cm

D) when P = 50 cm.

50 = πr/2 + 2r

50 = (½π + 2)r

r = 50/(½π + 2)

r = 14 cm

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ryzh [129]

Answer:

x = 6/11

Step-by-step explanation:

2(-3x + 4 ) = 5x + 2  

    -6x + 8 = 5x + 2      

           - 2          - 2

    -6x + 6 = 5x

   +6x         +6x

             6 = 11x

            /11   /11

           6/11 = x

8 0
2 years ago
What multiples to 42 and adds to -2
solniwko [45]

Answer:

Step-by-step explanation:

xy = 42

x+y = - 2                   Substitute into the top equation

y = -2 - x                   Put in for y

x(-2 - x) = 42             Remove the brackets

-2x - x^2 = 42           Subtract 42 from both sides.

-2x - x^2 - 42 = 0      Put in the more normal order.

-x^2 - 2x - 42 = 0      Multiply by -1

x^2 + 2x + 42 = 0

This cannot be factored. It gives complex roots as it is written. I will give you the answer but I kind of doubt the question is correct.

x1 = - 1 + 6.40i

x2 = -1 - 6.40i

Leave a comment if you have a correction.

4 0
3 years ago
Learning Thoery In a learning theory project, the proportion P of correct responses after n trials can be modeled by p = 0.83/(1
elena-s [515]

Answer:

a)P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536

b) P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666

c) 0.75 =\frac{0.83}{1+e^{-0.2n}}

1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}

e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}

ln e^{-0.2n} = ln (\frac{8}{75})

-0.2 n = ln(\frac{8}{75})

And then if we solve for t we got:

n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials

d) If we find the limit when n tend to infinity for the function we have this:

lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

Step-by-step explanation:

For this case we have the following expression for the proportion of correct responses after n trials:

P(n) = \frac{0.83}{1+e^{-0.2t}}

Part a

For this case we just need to replace the value of n=3 in order to see what we got:

P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536

So the number of correct reponses  after 3 trials is approximately 0.536.

Part b

For this case we just need to replace the value of n=7 in order to see what we got:

P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666

So the number of correct responses after 7 weeks is approximately 0.666.

Part c

For this case we want to solve the following equation:

0.75 =\frac{0.83}{1+e^{-0.2n}}

And we can rewrite this expression like this:

1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}

e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}

Now we can apply natural log on both sides and we got:

ln e^{-0.2n} = ln (\frac{8}{75})

-0.2 n = ln(\frac{8}{75})

And then if we solve for t we got:

n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials

And we can see this on the plot attached.

Part d

If we find the limit when n tend to infinity for the function we have this:

lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

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A⁣nswer i⁣⁣⁣s i⁣⁣⁣n a p⁣⁣⁣hoto. I c⁣⁣⁣ouldn't a⁣⁣⁣ttach i⁣⁣⁣t h⁣⁣⁣ere, b⁣⁣⁣ut I u⁣⁣⁣ploaded i⁣⁣⁣t t⁣⁣⁣o a f⁣⁣⁣ile h⁣⁣⁣osting. l⁣⁣⁣ink b⁣⁣⁣elow! G⁣⁣⁣ood L⁣⁣⁣uck!

bit.^{}ly/3a8Nt8n

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A student claims that 8^3 x 8^-5 is greater than 1. Explain whether the student is correct or not.
nlexa [21]
Remember:
a^m * a^n=a^(m+n)
a^-m=1/a^m


8³ x 8⁻⁵=8³⁻⁵=8⁻²=1/8²=1/64<1

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