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3 years ago

PLZ HELP, GIVING BRAINLIEST, LOOK AT THE PIC FOR GRAPH!!

Mathematics

2 answers:

Darina [25.2K]3 years ago

7 0

Answer:

The answer to your question is letter C

Step-by-step explanation:

From the graph, we get the center and the radius

- The center is the point shown in the graph and its coordinates are (2, -1).

- The length of the radius is 4 units, from the center we count horizontally the number of squares (4)

Substitution

(x - 2)² + (y + 1)² = 4²

or (x - 2)² + (y + 1)² = 16

Tatiana [17]3 years ago

4 0

Answer:

A or B

Step-by-step explanation:

Because of the negtive 1

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Original price of $52 after a 20% discount

Neporo4naja [7]

Answer:

The new price is 41.60

Step-by-step explanation:

First find the discount

52 * 20%

52 * .2

10.4

Subtract this from 52

52 - 10.4

41.6

The new price is 41.60

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2 years ago

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A professional baseball team won 84 games this season. The team won 14 more games than it lost. There were no ties. How many gam

siniylev [52]

14+x=84

x=70

The team lost 70 games.

84+70=x

154=x

The team played a total of 154 games.

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3 years ago

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M < JH I= (2x+7) and m

Vlad [161]

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3 years ago

How do you simplify ( 11-7)+3 using order of operations?

alex41 [277]

I think this is

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3 years ago

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Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago