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3 years ago

the width of a rectangle is 4 less than half the length. if l represents the length, wich equation could be used to find the wid

th, w?

Mathematics

1 answer:

torisob [31]3 years ago

7 0

$w=\dfrac{1}{2}l-4$

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I’m confused please help

Maru [420]

<h2>Hello!</h2>

The answer is:

$Sin(A)=0.93$

To solve the problem, remember to set your calculator in "degree mode"

So, we are given that the angle A is equal to 68°

So, we have that:

$Sin(A)=Sin(68\°)=0.93$

Hence, we have that:

$Sin(A)=0.93$

Have a nice day!

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3 years ago

The distance in feet that Karina swims in a race is represented by 6d - 6, where d is the distance for

vaieri [72.5K]

The answer would be a hope it helps

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3 years ago

Some integers are irrational numbers true or false please explain

Agata [3.3K]

False, all integers are rational because they can be expressed as a fraction. Like -4 is a rational number because it can be -4/1. All integers are rational.

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2 years ago

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.

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2 years ago

PLEASE HELP ME !!!

vfiekz [6]

Answer:

-0.4 or 2.3

Step-by-step explanation:

simple

each box is 0.1 unit

7 0

2 years ago

Read 2 more answers