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3 years ago

a horse trots in a circle at the end of a 4m rope. About how far does the horse trotted after completing ten circles?

2 answers:

7 0

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Find 1 round
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1 round of the circle = πD
1 round of the circle = π(2x4)
1 round of the circle =8π

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Find 10 rounds
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10 round of the circle = 8π x 10
10 round of the circle = 80π
10 round of the circle = 251.33 m

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Answer: 251.33 m
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LenaWriter [7]3 years ago

5 0

251.33m. First find the distance around one circle then times by ten for all the circles.
Circumference of each circle=2x(pi)xr
=2(Pi)x10
=25.13
For ten circles=25.13x10
=251.33

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One side of a rectangle is three times the other. Find the longer side if the perimeter is<br> 29

guajiro [1.7K]

Answer:

Answer: 10.875

Step-by-step explanation:

Let the short side be x.

Then the long side is 3x.

There are two opposite sides of each length, so the perimeter is

3x + 3x + x + x = 8x

The perimeter is 29, so we get the equation

8x = 29

Solve for x by dividing both sides by 8.

x = 29/8

x = 3.625

The longer side is 3x.

3x = 3(3.625) = 10.875

Answer: 10.875

3 0

2 years ago

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GuDViN [60]

The answer is 7 because yeah

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2 years ago

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Find the unknown value 18 is 50% of

nasty-shy [4]

Question : 18 is 50% of what number?

Answer : 36.

How : well if you multiply 18 times 2 the answer is 36. Since 18 is half of 36, being 50%

-procklown

6 0

3 years ago

Luna simplifies the expression (2x-4)-(3x+2) correctly. which expression represents luna's result

pickupchik [31]

Okay, so first you have to factor out that negative throughout the parenthesis.
(2x-4)-3x-2
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2x-3x-4-2
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5 0

2 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

1 year ago