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3 years ago

The function f(x) = -square root x is shown on the graph.

Mathematics

2 answers:

konstantin123 [22]3 years ago

6 0

Answer:

O The range of the function is all real numbers greater

than or equal to 0.

Step-by-step explanation:

The function f(x) = -square root x is shown on the graph.

Which statement is correct?

ТУ

The domain of the function is all real numbers less

than or equal to - 1

O The range of the function is all real numbers greater

than or equal to 0.

The range of the function is all real numbers less

than or equal to 0.

The domain of the function is all real numbers less

than or equal to 0.

4 5 6 7 X

-2

-5

balu736 [363]3 years ago

6 0

Answer:

C:domain: all real numbers

range: all real numbers greater than 1

Step-by-step explanation:

Edge 2021

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3 years ago

Which equation shows 2x – y = 6 converted to slope-intercept form?

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Question 2 of 5

jeka57 [31]

The number of solutions in each equation are as follows:

1 solution: 4^x = 2^{-x}
2 solution: 3/2x + 2 = 2^{x} + 1 and 3x + 1 = 2^{-x}.
No solution: 4^x + 2 = 3^x - 1 and 2x - 5 = 3^{x} + 2.

<h3>How to determine the number of solutions?</h3>

In order to determine the number of solutions, we would split the single equation to two different equations and then plot a graph, so as to reveal their solutions.

This ultimately implies that, the number of solutions is equal to the point of intersection between the lines of the equations plotted on a graph.

In conclusion, the number of solutions in each equation are as follows:

1 solution: 4^x = 2^{-x}
2 solution: 3/2x + 2 = 2^{x} + 1 and 3x + 1 = 2^{-x}.
No solution: 4^x + 2 = 3^x - 1 and 2x - 5 = 3^{x} + 2.

Read more on number of solutions here: brainly.com/question/12558210

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2 years ago

State the type of angles and solve for the given variable.

Zanzabum

Answer:

option c is correct,

as they are correspondence angles their values are same so,

x + 25° = 55°

→ x = 55° - 25°

→ x = 30° ans

hope this answer helps you dear...take care and may u have a great day ahead!

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