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2 years ago

13) The graph

Mathematics

1 answer:

Sever21 [200]2 years ago

3 0

Hi there! g(0) means the value of g(x) at x = 0. Look at x = 0 and see which y-point makes x = 0.

See that y = 4 makes x = 0 for the graph. Thus, g(0) = 4.

Next is to find the domain and range. Domain is the set of all x-values while Range is the set of all y-values.

For domain, look at the horizontal x-axis plane and the graph. Notice how the graph infinitely goes up for both negative and positive. Since the graph is in negative and positive as well. That means the domain is all real numbers

For range, look at the vertical y-axis and the graph. Notice how when y < -2, there are no points that are part of the graph. The graph goes infinitely for range as well but there're no y-values less than -2 for the graph.

Therefore, range is y >= -2.

Answer

g(0) = 4
domain is set of all real numbers.
range is y >= -2

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Pls help me

Vladimir [108]

<h3>Learning Task 1</h3>

Correct responses;

$\displaystyle 1. \hspace{0.3 cm} 7^{-1} = \frac{1}{7}$

2. (14·a·b·c)⁰ = 1

$\displaystyle 3. \hspace{0.3 cm} 10^{-9} = \frac{1}{10^9}$

4. 5·(x·y)⁰ = 5

5. 0¹⁵ = 0

$\displaystyle 6. \hspace{0.3 cm} \frac{24 \cdot x^8 \cdot y^4}{6 \cdot x^5 \cdot y^4} = 4 \cdot x^3$

$\displaystyle 7. \hspace{0.3 cm} \left(\frac{5 \cdot x^0}{y} \right)^{-1} = \frac{y}{5}$

$8. \hspace{0.3 cm}\displaystyle \frac{120 \cdot a^5 \cdot b^6 \cdot c^{-5} }{12 \cdot a^{-2} \cdot b^0 \cdot c^{2}} = \frac{10 \cdot a^7 \cdot b^6}{c^7}$

$\displaystyle 9. \hspace{0.3 cm} \frac{(4 \cdot x)^0 \cdot y^{-5} \cdot z^{-2} }{(234 \cdot x \cdot y \cdot z)^0} = \frac{1}{y^{5} \cdot z^2}$

$\displaystyle 10. \hspace{0.3 cm} \left(\frac{(5 \cdot x \cdot y)^0}{10} \right)^{-2} = 100$

$11. \displaystyle \hspace{0.3 cm} \left(\frac{(a \cdot b)}{9} \right)^{-1} = \frac{9}{a \cdot b}$

$\displaystyle 12. \hspace{0.3 cm} \frac{9}{x^{-2}} = 9 \cdot x^2$

$13. \displaystyle \hspace{0.3 cm} \frac{12 \cdot b^{-3}}{a^{-4}} = \frac{12 \cdot a^4}{b^3}$

$14. \displaystyle \hspace{0.3 cm} \frac{1}{a^{-5 \cdot n}} =a^{5 \cdot n}$

$15. \displaystyle \hspace{0.3 cm} \left(\frac{1}{2} \right)^{-5} = 32$

<h3>Methods by which the above expressions are simplified</h3>

The given expressions expressed into non zero and non negative exponents are;

$\displaystyle 1. \hspace{0.3 cm} \mathbf{7^{-1}} = \underline{\frac{1}{7}}$

2. (14·a·b·c)⁰ = 14⁰ × a⁰ × b⁰ × c⁰ = 1 × 1 × 1 × 1 =<u> 1</u>

$\displaystyle 3. \hspace{0.3 cm} \mathbf{ 10^{-9}} = \underline{ \frac{1}{10^9}}$

4. 5·(x·y)⁰ = 5 × x⁰ × y⁰ = 5 × 1 × 1 = <u>5</u>

5. 0¹⁵ = <u>0</u>

$\displaystyle 6. \hspace{0.3 cm} \mathbf{\frac{24 \cdot x^8 \cdot y^4}{6 \cdot x^5 \cdot y^4}} = \frac{ 4 \times 6 \times x^5 \times x^3 \times y^4}{6 \times x^5 \times y^4} = \underline{4 \cdot x^3}$

$\displaystyle 7. \hspace{0.3 cm} \mathbf{\left(\frac{5 \cdot x^0}{y} \right)^{-1} }= \frac{1}{\frac{5 \times 1}{y} } = \underline{\frac{y}{5}}$

$8. \hspace{0.3 cm}\displaystyle \mathbf{\frac{120 \cdot a^5 \cdot b^6 \cdot c^{-5} }{12 \cdot a^{-2} \cdot b^0 \cdot c^{2}}} = 10 \cdot a^{5 - (-2)} \cdot b^{6 - 0} \cdot c^{-5 -2} = \underline{\frac{10 \cdot a^7 \cdot b^6}{c^7}}$

$\displaystyle 9. \hspace{0.3 cm} \mathbf{\frac{(4 \cdot x)^0 \cdot y^{-5} \cdot z^{-2} }{(234 \cdot x \cdot y \cdot z)^0}} = y^{-5} \cdot z^{-2} = \underline{\frac{1}{y^{5} \cdot z^2}}$

$\displaystyle 10. \hspace{0.3 cm} \mathbf{\left(\frac{(5 \cdot x \cdot y)^0}{10} \right)^{-2}} = \left(\frac{1}{10} \right)^{-2} = \frac{1}{\left(\frac{1}{10} \right)^2 } = \frac{1}{\frac{1}{100} } = \underline{100}$

$11. \displaystyle \hspace{0.3 cm} \mathbf{\left(\frac{(a \cdot b)}{9} \right)^{-1}} = \frac{1}{\left(\frac{(a \cdot b)}{9} \right) } = \underline{\frac{9}{a \cdot b}}$

$\displaystyle 12. \hspace{0.3 cm} \mathbf{\frac{9}{x^{-2}}} = \frac{9}{\frac{1}{x^2} } = \underline{9 \cdot x^2}$

$13. \displaystyle \hspace{0.3 cm} \mathbf{\frac{12 \cdot b^{-3}}{a^{-4}}} = 12 \cdot \frac{b^{-3}}{a^{-4}} =12 \cdot \frac{1}{\frac{b^3}{a^4} } = 12 \cdot \frac{a^4}{b^3} = \underline{\frac{12 \cdot a^4}{b^3}}$

$14. \displaystyle \hspace{0.3 cm} \mathbf{\frac{1}{a^{-5 \cdot n}}} = \frac{1}{\frac{1}{a^{5 \cdot n}} } = \underline{a^{5 \cdot n}}$

$15. \displaystyle \hspace{0.3 cm} \mathbf{\left(\frac{1}{2} \right)^{-5}} = \frac{1}{\left(\frac{1}{2} \right)^5} = 2^5 = \underline{32}$

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2 years ago

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Answer:

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Step-by-step explanation:

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Answer:

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