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amid [387]
3 years ago
6

Marlon and Michelle have to graph a function with the equation y=x2and are having to debate about which is the most important el

ement to get started, the domain or the range . which of them is correct?

Mathematics
1 answer:
Usimov [2.4K]3 years ago
5 0

Answer:

Correct answer is the domain.

Step-by-step explanation:

A domain is the input set of a function.

A range is the output set of the function.

It is more  important to know the domain of the function to be able to determine the corresponding output set so the function can be graphed. To start with the domain is very useful because every domain element has a corresponding unique output element.

Suppose you started with the range element of 9. The input set for an output of 9 is {-3,3}. This makes it hard to match up the elements. This example highlights why it is important to start with the domain rather than the range.

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Answer:

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7 0
2 years ago
Read 2 more answers
Please help
Rudiy27
It would be C since a scale factor of 2 makes the original A coordinate of (3,3) times 2 therefore the new coordinates being (6,6)
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3 years ago
What are a lot of common sixth grade math problems
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4 0
3 years ago
Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl
GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane n=(-2,-2,1), then r will have the next parametric equations:

x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3

Substitute the value of \lambda in the parametric equations:

x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u

7 0
3 years ago
ANSWER THIS QUICKLY PLEASE!!!!!!!
Effectus [21]
Hi
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2 years ago
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