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2 years ago

A regular pentagon has a perimeter of 5x+15 what is the length of one side

Mathematics

1 answer:

dsp732 years ago

6 0

Answer:

plz give full question

its just half

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Eric has a spinner that is divided into different colored section that are the same size and shape on any spin the theoretical p

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It occurred 4 more times than expected

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3 years ago

The total surface of the cube measures 54ft. How long is its edge?

Fantom [35]

Answer:

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Step-by-step explanation:

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3 years ago

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Student scientific calculator 3×10^5 is displayed as 3 EE 5 in that case how was the panic of 80,000 and 5000 be displayed on Th

faltersainse [42]

Hello,
We know that
3×10^5= 3 EE 5
Change 80,000 to the scientific notation
8×10^4
Doing the same like 3×10^5, I can get 8 EE 4
Change 5000 to the scientific notation
5×10^3
Then it will appeared as 5 EE 3. Hope it help!

4 0

2 years ago

Estimate the product of 2.28 × 5.59 using compatible numbers.

Ierofanga [76]

The quotient of 2.28 × 5.59 using compatible numbers is approximately 12.75

<h3>What are products?</h3>

Quotients are result derived from the multiplication of two rational or integers. Given the expression

2.28 × 5.59

Convert to fraction

2.28 × 5.59 = 228/100 ÷ 559/100
2.28 × 5.59= 127452/10000

2.28 × 5.59= 12.75

Hence the quotient of 2.28 × 5.59 using compatible numbers is approximately 12.75

Learn more on product here: brainly.com/question/673545

#SPJ1

3 0

2 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