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MrRa [10]
3 years ago
13

Solve the equation. 3/2 + b = 7/4 (stylised in fractions) b= ?

Mathematics
1 answer:
gtnhenbr [62]3 years ago
6 0

3/2 + b = 7/4

First rewrite 3/2 to have the same denominator as 7/4:

3/2 x 2 = 6/4

Now you have 6/4 + b = 7/4

To find b subtract 6/4 from 7/4:

b = 7/4 - 6/4

b = 1/4

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You would subtract.

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So Sandra bought more candy. Hope this helps!



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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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gogolik [260]
The answer will be -32
8 0
3 years ago
What is D and R? Please answer I need help.
Setler79 [48]

Answer:

  31. D: {8, 4, 0, -4}; R: {2, -1}; yes

  32. D: {-1, 2, 7}; R: {-4, -3, -2, 0}; no

  33. D: {-4, -3, -1, 2, 3, 5}; R: {-3, -2, 0, 3, 5}; yes

  34. D: (-∞, ∞); R: (-∞, 4]; yes

  35. D: [0, 5]; R: [-2, 3]; no

Step-by-step explanation:

In this context, D means "domain" and R means "range." The domain of a function is the list of input values for which the function is defined. For ordered pairs, it is the first number of the pair. For an x-y table, it is the list of x-values. For a graph, it is the possible values of x.

A relation is a <em>function</em> only if there are no repeated values in the domain (2 or more outputs for the same input.)

The range of a function is the list of output values produced by the function. For ordered pairs, it is the second number of the pair. For an x-y table, it is the list of y-values. For a graph, it is the possible values of y.

__

31. D: {8, 4, 0, -4}

    R: {2, -1}

Function: yes

Domain and range values don't need to be repeated. Often, they're listed in order from lowest to highest. Here, we have listed them in order of occurrence in the function definition.

__

32. D: {-1, 2, 7}

      R: {-4, -3, -2, 0}

Function: no

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33. D: {-4, -3, -1, 2, 3, 5}

     R: {-3, -2, 0, 3, 5}

Function: yes

__

34. D: (-∞, ∞)

     R: (-∞, 4]

Function: yes

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35. D: [0, 5]

     R: [-2, 3]

Function: no . . . . . . there are 2 y-values for most x-values

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