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Brilliant_brown [7]
3 years ago
14

What are the domain and range of this function y=(x+3)^2 -5

Mathematics
2 answers:
n200080 [17]3 years ago
8 0

Answer: Choice A

Note: the range should be [-5, \infty). See explanation below.

===================================================

We can plug in any real number for x to get some output for y. The domain is the set of all real numbers in which we say (-\infty, \infty) which is interval notation. It represents the interval from negative infinity to positive infinity.

The range is the set of possible outputs. The smallest output possible is y = -5 which occurs at the vertex (3,-5). We can get this y value or larger. So we can describe the range as the set of y values such that y \ge -5 and that translates to the interval notation [-5, \infty).

The square bracket says "include this endpoint" while the curved parenthesis says to exclude the endpoint. Your teacher mistakenly wrote (-5, \infty) for choice A, when they should have written [-5, \infty)

I think either your teacher made a typo or somehow the formatting messed up. Either way, choice A is the closest to the answer.

Tatiana [17]3 years ago
4 0

Option A is the correct choice .

Domain will be → all real numbers .

Range will be → y belongs to R : y greater than or equal to -5 .

→ (x+3)^2 It means 0 to infinity

So (x+3)^2-5= (-5, infinity)

→ Range = (-5, infinity )

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A running circuit is in the shape of a triangle with lengths of 6km, 6.5km and 7km. What are the sizes of the angles (in minutes
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A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to 180^{o}. The <em>measures</em> of the internal <u>angles</u> of the <u>triangle</u> given in the question are A = 52.6^{o}, B = 59.4^{o}, and C = 68^{o}.

A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to 180^{o}.

Considering the given question, let the <u>sides</u> of the triangle be: a = 6 km, b = 6.5 km, and c = 7 km.

Apply the <em>Cosine rule</em> to have:

c^{2} = a^{2} + b^{2} - 2ab Cos C

So that;

7^{2} = 6^{2} + (6.5)^{2} - 2(6 * 6.5) Cos C

49 = 36 + 42.25 - 78Cos C

78 Cos C = 78.25 - 49

               = 29.25

Cos C = \frac{29.25}{78}

         = 0.375

C = Cos^{-1} 0.375

   = 67.9757

C = 68^{o}

Apply the <em>Sine rule</em> to determine the <u>value</u> of B,

\frac{b}{Sin B} = \frac{c}{Sin C}

\frac{6.5}{Sin B} = \frac{7}{Sin 68}

SIn B = \frac{6.5 *Sin 68}{7}

         = 0.861

B = Sin^{-1} 0.861

   = 59.43

B = 59.4^{o}

Thus to determine the value of A, we have;

A + B + C = 180^{o}

A + 59.4^{o} + 68^{o} = 180^{o}

A = 180^{o} - 127.4

  = 52.6

A = 52.6^{o}

Therefore the <u>sizes</u> of the <em>internal angles</em> of the triangle are: A = 52.6^{o}, B = 59.4^{o}, and C = 68^{o}.

For more clarifications on applications of the Sine and Cosine rules, visit: brainly.com/question/14660814

#SPJ1

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