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Bingel [31]
3 years ago
14

A reflecting pool is shaped like a square with a side length of 4 meters. What is the perimeter of the​ pool? Explain how you fo

und your answer.
Mathematics
1 answer:
True [87]3 years ago
6 0

Answer:

16 meters.

Step-by-step explanation:

There are two things to keep in mind at this point, the first is that a square has all its sides equal and the second is that the perimeter in this type of figure is the sum of all the sides.

Knowing the above, we can deduce that therefore each side of the square measures 4 meters and the perimeter could be calculated since we know the value of all the sides, we know that a square has four sides:

p = 4 +4 +4 +4

p = 16

that is to say that the perimeter is equal to 16 meters.

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According to general equation for conditional probability, if P(A^B) = 2/3 and P(B) = 3/4, what is P(A|B)?
Alexxx [7]

Answer:

a

Step-by-step explanation:

ok

5 0
3 years ago
I need the answer and the explanation/ angle name pls help this is a quiz grade
Sever21 [200]

Step-by-step explanation:

The answer is :-

x and w : Dr. Alpha was meeting up with this patient in the library.

Know why:-

Because x and w are Vertically Opposite Angles! And they are always same!

\tt{Hope \ it \ Helps \ :D}

8 0
3 years ago
If 3220 = 80+7, what is the value of c?
Setler79 [48]

Answer:

Correct option is

C

36.25

Modal class =30−40

So we have, l=30,f0=12,f1=32,f2=20 and h=10

⇒  Mode=l+2f1−f0f2f1−f0×h

                  =30+2×32−12−2032−12×10

                  =30+6.25

                  =36.25

∴  Mode =36.25

8 0
3 years ago
4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.
nikitadnepr [17]

Answer:

\sum_{n=0}^9cos(\frac{\pi n}{2})=1

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}

Step-by-step explanation:

\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))

=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})

=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1

2nd

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}

=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0

3th

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))

=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}

What we use?

We use that

e^{i\pi n}=cos(\pi n)+i sin(\pi n)

and

\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}

6 0
3 years ago
Find a range of th fucunction f(x)=4x-1 for the domain{-1,0,1,2,3}
vlada-n [284]
F(x) = 4x - 1

All you have to do is substitute the values in the domain for x in the function.

f(-1) = 4(-1) - 1
f(-1) = -4 - 1
f(-1) = -5

f(0) = 4(0) - 1
f(0) = -1

f(1) = 4(1) - 1
f(1) = 4 - 1
f(1) = 3

And so on.
7 0
3 years ago
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