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Lera25 [3.4K]
1 year ago
8

PLEASE HELP! Question is on photo! Thank you!!

Mathematics
1 answer:
ipn [44]1 year ago
7 0

We conclude that the relation presented in the picture is a function with domain: - 4 ≤ x < 1 and range: - 4 ≤ x ≤ 5. (Correct choices: B, C, H)

<h3>How to determine the domain and range of a relation and if a relation is a function</h3>

Herein we have a relation between two variables, x and y, relations involve two sets: an input set called domain and an output set called range. A relation is a function if and only if every element of the domain is related to only one element from the range.

Graphically speaking, the horizontal axis corresponds with the domain, whereas the vertical axis is for the set of the range. According to the previous concepts, we conclude that the relation presented in the picture is a function with domain: - 4 ≤ x < 1 and range: - 4 ≤ x ≤ 5. (Correct choices: B, C, H)

To learn more on functions: brainly.com/question/12431044

#SPJ1

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What is the slope of a line that passes through the points (-4,0) and (0,4)?
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F divides HJ in the ratio 3:1. If H = (-11,7) and J = (5,3), what are the<br> coordinates of F?
Hatshy [7]

Answer:

The coordinates of HF are (1, 4)

Step-by-step explanation:

The parameters of the line are;

The coordinate of the end points are H = (-11, 7), and J = (5, 3)

The ratio by which the point F divides the line = 3:1

The segments in the line are HF, and FJ

Therefore;

The fraction of the length of HJ that is represented by HF = 3/(3 + 1) × HJ = 3/4 × HJ

HF = 3/4 × HJ

Which gives the coordinates of the point F as follows;

Coordinate of F = (-11 +(5 - (-11))×3/4, 7 + (3 - 7)×3/4) =  (1, 4)

The coordinates of F are (1, 4)

We check the length of HF, from the equation for the length to of a line to get;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

l_{HF} = \sqrt{\left (4-7  \right )^{2}+\left (1-(-11)  \right )^{2}} = \sqrt{\left (-3  \right )^{2}+\left (12  \right )^{2}} = 3\cdot \sqrt{17}

Similarly, we check the length of HJ, to get;

l_{HF} = \sqrt{\left (3-7  \right )^{2}+\left (5-(-11)  \right )^{2}} = \sqrt{\left (-4  \right )^{2}+\left (16  \right )^{2}} = 4\cdot \sqrt{17}

The length of HF = 3·√(17)

The length of HJ = 4·√(17)

Therefore, from HF = 3/4× HJ, we have;

HF = 3/4 × 4·√(17) = 3·√(17)

Therefore, the coordinates of HF are (1, 4)

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Step-by-step explanation:

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