Given r= 10 and e=0.57, find the Cartesian coordinates. Check your answer.

Nautical flags are used to represent letters of alphabets. The flag for the letter O consists of a yellow right triangle and a r

liraira [26]

Answer:

The length of a side of the flag is 7.2inches

Step-by-step explanation:

Let length = x

Let hypotenuse = x+3

Using Pythagoras' Theorem:

x^2 + x^2 = (x+3)^2

x^2 + x^2 = x^2 + 6x + 9

2x^2 = x^2 + 6x + 9

×^2 - 6x - 9 = 0

$x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a}$

x=[-(-6)+sqrt.(-6)^2-4(1)(-9)]/2(1)

×=[6+sqrt.72]/2

×=7.2inches

×=[-(-6)+sqrt.(-6)^2-4(1)(-9)]/2(1)

x=[6-sqrt.72]/2

×=-1.2(rej)

(PS. sqrt is square root)

(Correct me if i am wrong)

6 0

3 years ago

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Is 11/12 greater or less than 2/3?

Sauron [17]

False beacause 2/3 is bigger than 11/12. Think about it like pizza 2/3 pieces is better cuz there bigger & 11/12 are smaller pieces.

8 0

3 years ago

Simpify the expression.<br>5+4z-2z<br><br>Simpilfied expression

Dafna11 [192]

Answer:

2z + 5

Step-by-step explanation:

you have to combine like terms

subtract 4z - 2z and you get 2z

the number with the variable next to it goes first not the constant

2z + 5

the variable is z and the constant is 5 because it doesn't have variable next to it

hope this makes sense and helps :)

tell me if you have any questions

4 0

2 years ago

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What are you required to do when you want to use a discount membership to save money?

solong [7]

When you want to use a discount membership to save money you are required to purchase a membership. A popular place that people often purchase memberships from are Costco or Sam's Club.

8 0

3 years ago

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For what value of constant c is the function k(x) continuous at x = 0 if k =

nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x) is continuous, we take the limit of the parameter as follows:

$\mathbf{ \lim_{x \to 0^-} k(x) = \lim_{x \to 0^+} k(x) = 0 }$

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= c }$

Provided that:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }$

Using l'Hospital's rule:

$\mathbf{\implies \lim_{x \to 0} \ \ \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}= \lim_{x \to 0} sec \ x \ tan \ x = 0}$

Therefore:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}=0 }$

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

#SPJ1

5 0

2 years ago