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liubo4ka [24]
2 years ago
8

How much fencing is required to enclose a circular garden whose radius is 14 m? Use 3.14 pi

Mathematics
1 answer:
ipn [44]2 years ago
7 0

Answer:

87.92 m

Step-by-step explanation:

Formula for the circumference of a circle (which is what you're finding with this problem) is pi d

Pi being 3.14 and d being the diameter

Since the radius is 14 the diameter is 2 times that , 28m

Now just multiply 28 and 3.14 which gives you 87.92

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If the line passing through the points (a, 1) and (−2, 6) is parallel to the line passing through the points (−12, 9) and (a + 2
iren2701 [21]

\frac{6 - 1}{ - 2 - a}  =  \frac{9 - 1}{ - 12 - a - 2}  \\  \frac{5}{ - 2 - a}  =  \frac{8}{ - 14 - a}  \\ 16 + 8a = 70 + 5a \\ 3a = 54 >  > a = 18
8 0
3 years ago
What is the prduct of 33 and 6925?
liubo4ka [24]

Answer:

228525-Look below for steps:)

Step-by-step explanation:

Step 1:

6925

*   33

______

Step 2:

5*3=15

carry the one

3*2+1=7

9*3=27

carry the 2

6*3+2=20

SO your first number is 20775 but we are NOT done yet!

add a zero below 5

5*3=15

carry the one

3*2+1=7

9*3=27 carry the 2

then 6*3+2=20

then add 207750+20775

which equals....

228525

so 228525 is your answer!!!

<em>I really do hope this made sense!</em>

<em>Have a great day!</em>

<em>- Hailey: D</em>

<em>(NOTHING IS COPIED AND PASTED!!!!!)</em>

3 0
3 years ago
2.943÷2.7 any help???
Zinaida [17]

The answer will be 1.09. Hope this will help you!

6 0
3 years ago
Read 2 more answers
In the diagram below of a parallelogram STUV, SV=x+3, VU 2x-1, and TU=4x-3. What is the length of SV?
Katena32 [7]

Answer:

SV=5\ units

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

In a parallelogram opposites sides are parallel and congruent

so

In this problem

SV=TU ---> by opposite sides

substitute the given values

x+3=4x-3

solve for x

4x-x=3+3\\3x=6\\x=2

Find the length of SV

SV=x+3

substitute the value of x

SV=2+3=5\ units

5 0
3 years ago
If 8 identical blackboards are to be divided among 4 schools,how many divisions are possible? How many, if each school mustrecei
MAXImum [283]

Answer:

There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.

Step-by-step explanation:

Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.

 The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is {11 \choose 3} = 165 . As a result, we have 165 ways to distribute the blackboards.

If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is {7 \choose 3} = 35. Thus, there are only 35 ways to distribute the blackboards in this case.

4 0
3 years ago
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