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igor_vitrenko [27]
3 years ago
12

Find the perimeter of a square watermelon farm whose side is 14 miles.

Mathematics
2 answers:
Harlamova29_29 [7]3 years ago
8 0
Perimeter is adding all the sides.
 since it is square it is 14 +14 +14 +14 = 56 miles
earnstyle [38]3 years ago
6 0
Hi! :) 

So the perimeter of a shape is the length of the edges of the shape. In order to get the perimeter, you can just add all 4 side lengths together. Or, if it's a shape with equal sides, you can just multiply the side lengths by however many sides there are. So in this case you can just multiply 14 by 4

14*4=56

Final answer: 56

Hope this helps!

You might be interested in
Multiply and answer in scientific notation:
saveliy_v [14]

Answer:

50000000000 if your combining them both or if separate Q1: 25000000 Q2: 2000

Step-by-step explanation:

2.5(10^7)(2(10^3))

=25000000*2000

=2.5*10000000*2*1000

=(2.5*2)*(10000000*1000)

=5*10000000000

=50000000000

If separate:

question 1:

2.5(107)

=2.5*107

=2.5*(10*10*10*10*10*10*10)

=25000000

question 2:

2(103)

=2*103

=2*(10*10*10)

=2000

I hope this helps!

6 0
3 years ago
How many phone numbers are possible for one area code if the first four numbers are 202-1 , in that order , and the last three n
Mkey [24]

6 phone numbers are possible for one area code if the first four numbers are 202-1

<u>Solution:</u>

Given that, the first four numbers are 202-1, in that order, and the last three numbers are 1-7-8 in any order  

We have to find how many phone numbers are possible for one area code.

The number of way “n” objects can be arranged is given as n!

Then, we have three places which changes, so we can change these 3 places in 3! ways

n! = n \times (n - 1)!

Hence 3! is found as follows:

3! = 3 \times (3-1)!\\\\3! = 3 \times 2!\\\\3! = 3 \times 2 \times 1 = 6

So, we have 6 phone numbers possible for one area code.

3 0
3 years ago
Hi, do you know how to get the ratio?
bija089 [108]

hope it help you

please follow me

7 0
2 years ago
Power Series Differential equation
KatRina [158]
The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for y

\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0

which indeed gives the recurrence you found,

a_{n+3}=-\dfrac{n-3}{n+3}a_n

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that a_2=0, and substituting this into the recurrence, you find that a_2=a_5=a_8=\cdots=a_{3k-1}=0 for all k\ge1.

Next, the linear term tells you that 6a_0+6a_3=0, or a_3=a_0.

Now, if a_0 is the first term in the sequence, then by the recurrence you have

a_3=a_0
a_6=-\dfrac{3-3}{3+3}a_3=0
a_9=-\dfrac{6-3}{6+3}a_6=0

and so on, such that a_{3k}=0 for all k\ge2.

Finally, the quadratic term gives 6a_1-12a_4=0, or a_4=\dfrac12a_1. Then by the recurrence,

a_4=\dfrac12a_1
a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1
a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1
a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1

and so on, such that

a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}

for all k\ge2.

Now, the solution was proposed to be

y=\displaystyle\sum_{n\ge0}a_nx^n

so the general solution would be

y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots
y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)
y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)
4 0
3 years ago
What is the range of function gif g(x) = -2/(x) + 1
Zina [86]

The function g(x) is a rational function, and none of the options represent the range of the function g(x)

<h3>How to determine the range of the function?</h3>

The function is given as:

g(x) = -2/x + 1

The above function is undefined at point x = 0.

This is so because -2/x is undefined.

So, we have:

g(0) = -undefined + 1

g(0) = undefined

This means that the range of the function is:

(-infinity, 1) and (1, infinity)

None of the options represent the range of the function g(x)

Read more about range at:

brainly.com/question/10197594

#SPJ1

5 0
2 years ago
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