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Alinara [238K]
3 years ago
8

Find the domain and range of each relation. Then determine if the relation is a function. D={-2, -1, 3, 5}, R={-7 ,0 ,4} Functio

n?
Mathematics
1 answer:
poizon [28]3 years ago
5 0

Answer:

D = \{-2,-1,3,5\} -- Domain

R = \{-7,0,4\} -- Range

It is a function

Step-by-step explanation:

Given

D = \{-2,-1,3,5\}

R = \{-7,0,4\}

Required

State the domain and range

Determine if the relation is a function

From the question, we have the domain and range as:

D = \{-2,-1,3,5\} -- Domain

R = \{-7,0,4\} -- Range

Next, is to determine if the relation is a function or not.

Yes, it is a function.

The number of elements in the domain is 4

The number of elements in the range is 3

<em>When the domain has more elements than the range, this is called a many-to-one function, and it is a valid type of function.</em>

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*WILL GIVE BRAINLIEST*
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Answer:

$50 and

$30

Step-by-step explanation:

Question 1. Principal = $500

Rate = 5%

Time = 2yesrs

Interest = PRT /100

= 500 x 5 x 2 /100

= 5000/100

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Rate = 3%

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3 years ago
A circle passes through points A(7,4), B(10,6), C(12,3). Show that AC must be the diameter of the circle.
Artist 52 [7]

so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.

in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.

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now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}

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Step-by-step explanation:

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