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Rus_ich [418]
3 years ago
6

Which equation is the inverse of 5 y + 4 = (x + 3) squared + one-half?

Mathematics
2 answers:
telo118 [61]3 years ago
8 0

Answer:

The correct option is the last option d.) y = negative 3 plus-or-minus StartRoot 5 x + seven-halves EndRoot

Step-by-step explanation:

the given equation is 5y + 4 =(x+3)^{2}  + \frac{1}{2}

Therefore we can write 5y\hspace{0.1cm} = (x+3)^{2} + \frac{1}{2} - 4  \hspace{0.3cm}   \Rightarrow  \hspace{0.2cm}  5y = (x+3)^{2} - \frac{7}{2}   \hspace{0.3cm} \Rightarrow \hspace{0.3cm}  y = \frac{(x+3)^{2}}{5} - \frac{7}{10}

To find the inverse of the above function let us replace x with y and y with x.

Therefore we get

x = \frac{(y+3)^{2}}{5} - \frac{7}{10}

Now we write the above equation with only y on the Left hand side and we will obtain the inverse of the given function

y = \pm \sqrt{5 (x + \frac{7}{10} )} \hspace{0.1cm} - \hspace{0.1cm} 3 \hspace{0.1cm} \Rightarrow\hspace{0.1cm}  y =  \pm \sqrt{5x + \frac{7}{2} }  - \hspace{0.1cm} 3 \hspace{0.1cm}

Therefore the correct option is the last option d.) y = negative 3 plus-or-minus StartRoot 5 x + seven-halves EndRoot , y =  \pm \sqrt{5x + \frac{7}{2} }  - \hspace{0.1cm} 3 \hspace{0.1cm}

ohaa [14]3 years ago
4 0

Answer:

y = negative 3 plus-or-minus StartRoot 5 x + seven-halves EndRoot

Step-by-step explanation:

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A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression a=a1+
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Answer:

a=\frac{3\times 7}{7}+\frac{12}{7}  (First step for obtaining a common denominator for the two fraction)

a=\frac{33}{7}\text{ m/s}^2

Step-by-step explanation:

\text{Given Expression: }a=a_1+\frac{F}{m}

where,

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a_1=3.00\text{ m/s}^2

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Now we will simplify above expression for a

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Len [333]

Answer:

The relation is not a function

The domain is {1, 2, 3}

The range is {3, 4, 5}

Step-by-step explanation:

A relation of a set of ordered pairs x and y is a function if

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  • x appears once in ordered pairs

<u><em>Examples:</em></u>

  • The relation {(1, 2), (-2, 3), (4, 5)} is a function because every x has only one value of y (x = 1 has y = 2, x = -2 has y = 3, x = 4 has y = 5)
  • The relation {(1, 2), (-2, 3), (1, 5)} is not a function because one x has two values of y (x = 1 has values of y = 2 and 5)
  • The domain is the set of values of x
  • The range is the set of values of y

Let us solve the question

∵ The relation = {(1, 3), (2, 3), (3, 4), (2, 5)}

∵ x = 1 has y = 3

∵ x = 2 has y = 3

∵ x = 3 has y = 4

∵ x = 2 has y = 5

→ One x appears twice in the ordered pairs

∵ x = 2 has y = 3 and 5

∴ The relation is not a function because one x has two values of y

∵ The domain is the set of values of x

∴ The domain = {1, 2, 3}

∵ The range is the set of values of y

∴ The range = {3, 4, 5}

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y = -8 - 11/6x

Answer:  y = -11/6x - 8

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