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3 years ago

1/2r = 7/2. solve for r

Mathematics

2 answers:

Yuri [45]3 years ago

6 0

Answer:

r = 7

Step-by-step explanation:

hope this was helpful let me know you need the steps! :)

kvv77 [185]3 years ago

3 0

Answer: 1.75 or 1 3/4

Step-by-step explanation:

Because first you need to get R by it’s self by dividing it by 1/2 then if you divided it by one side you have to divide it by the other side so you get r=1.75 or 1 3/4.

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Given f(x)=(x+5)(x-2), what are the x-intercepts?

Yuri [45]

Answer:

The intercepts are -5 and 2

Step-by-step explanation:

Using the zero product property

f(x)=(x+5)(x-2)

0 =(x+5)(x-2)

0 = (x+5) 0=(x-2)

x = -5 x = 2

The intercepts are -5 and 2

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If Shane and Susan both push on a box with a force of 25N, but in opposite directions, what will happen to the box?

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Since the forces exerted on the box are equal and opposite in direction, the net force will be 0, so the box will not move.
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3 years ago

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Can you give me the Answers for Estimating 92 × 68

Elodia [21]

Answer:

The answer is 6256 of this question

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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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ella [17]

Answer:

The Answer is A.) -4

Step-by-step explanation:

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