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2 years ago

A regular pentagon has sides of 10 cm. Find the radius of the largest circle which can

Mathematics

1 answer:

aliina [53]2 years ago

7 0

Answer is 6.88 cm

Divide the pentagon radially into five congruent, isosceles triangles. Each triangle has a vertex angle of 360°÷5=72° and altitude of 5÷tan(36°).

radius of circle = 5÷tan(36°) = 6.88 cm

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WILL GIVE BRAINLIEST! Someone please graph these, I can't graph them.

nadezda [96]

Answer:

see attached image

Step-by-step explanation:

F(x) = 2x + 5 is a linear graph because the exponent on x is 1. I tell my students that think of graphs having one less turn/corner that the value of the highest exponent. so since 2x has a exponent of 1, 1-1 =0 so it has no turns or its a straight line.

this has a slope of 2 or 2/1 or up 2 and right 1 from the y intercept which is 5

so mark 5 on the y axis and a from there go up 2 and right 1 and make another point. Join these points and you have your graph

and

g(x) = (x-5)/2 is its inverse and is found:

F(x) = 2x + 5 write it this way y = 2x + 5

now swap the x and y x = 2y - 5

solve for y

x = 2y + 5

- 5 -5

x - 5 = 2y

/2 /2

(x-5)/2 = y

it can be written as y = x/2 - 5/2

and graphed the same way as above with a 1/2 slope and -5/2 y intercept

4 0

3 years ago

Read 2 more answers

There 6 pounds of apples in one bag. Make a table to show how many pound of apples are in 6 bags. How do you your answer is easo

kotegsom [21]

Answer:

If there are 6 pounds of an apple in a bag and there are 6 bags you are going to do 6x6 which would be 36

Step-by-step explanation:

6 pounds

bags: oooooo

so 6 bags x another 6 pounds would be 36.

7 0

3 years ago

Read 2 more answers

Brenda noticed that the reptile building at the zoo was located at the point (-3, 4) on the zoo map. Name the quadrant of the ma

rodikova [14]

It is located in quadrant 2

4 0

2 years ago

Prove the following by induction. In each case, n is apositive integer. 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

To prove: $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

Find the derivative of

kirill [66]

Answer:

$\displaystyle y'(1, \frac{3}{2}) = -3$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \frac{3}{2x^2}$

$\displaystyle \text{Point} \ (1, \frac{3}{2})$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y = \frac{3}{2}x^{-2}$
Basic Power Rule: $\displaystyle y' = -2 \cdot \frac{3}{2}x^{-2 - 1}$
Simplify: $\displaystyle y' = -3x^{-3}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{-3}{x^3}$

Step 3: Solve

Substitute in coordinate [Derivative]: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1^3}$
Evaluate exponents: $\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1}$
Divide: $\displaystyle y'(1, \frac{3}{2}) = -3$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e

6 0

2 years ago