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2 years ago

14

a garden is 18 feet 3 inches long and 10 feet 8 inches wide. the amount of fencing needed to enclose the garden is Need help and

will mark brainlist

1 answer:

Anna [14]2 years ago

5 0

Answer:

$\boxed{57ft 10in}$

Step-by-step explanation:

<em>Hey there!</em>

Well if the length is 18 and 3 inches we need to find the sum of both lengths,

18ft 3in + 18ft 3in = 36ft 6in

Width- 10ft 8in + 10ft 8in = 20ft 16 in -> 21ft 4in

l + w = 57ft 10in

<em>Hope this helps :)</em>

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40 steps from 2nd to 3rd base

Step-by-step explanation:

18 - 7 + 3 - 5 + 11 = 20, which is halfway, so double that to get to 3rd base.

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2 years ago

If r is the midpoint of qs rs=2x-4, st= 4x-1 and rt = 8x-43 find qs

sammy [17]

Answer:

$QS=68\ units$

Step-by-step explanation:

step 1

Find the value of x

we know that

r is the midpoint of qs

so

QR=RS

QS=QR+RS------> QS=2RS -----> equation A

RT=RS+ST ----> equation B

see the attached figure to better understand the problem

Substitute the given values in the equation B and solve for x

$8x-43=(2x-4)+(4x-1)$

$8x-43=6x-5$

$8x-6x=43-5$

$2x=38$

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step 2

Find the value of RS

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substitute the value of x

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step 3

Find the value of QS

Remember equation A

QS=2RS

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2 years ago

The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

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Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

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If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

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Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

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Answer:

c

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Answer:

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using a graphing tool

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2 years ago

Read 2 more answers