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photoshop1234 [79]
3 years ago
7

Need help with this please!

Mathematics
1 answer:
Rashid [163]3 years ago
7 0

Answer:

y=x^2

Step-by-step explanation:

The given relation is

D={(x,y):(3,9),(5,25),(-2,4)}

We can see that;

3\to 3^2=9

5\to 5^2=25

-2\to (-2)^2=4

Hence in general,

x\to x^2

Therefore the equation for the relation is;

y=x^2

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The system of equations y=-2x+3 and y= x is graphed.What is the solution to the system of equations?
Evgen [1.6K]

Answer:

(1,1) x=1 y=1

Step-by-step explanation:

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Which expression has the sum of twenty six twentieths? (4 points) Selected:a. two fifths plus four tenthsThis answer is incorrec
patriot [66]

Answer: b. four fifths plus two fourths

Step-by-step explanation: Change the fractions to their equivalents in 20ths-- like finding the common denominator.

Divide the original denominator into 20, then multiply the result by the original numerator to get the new numerator.

4/5  <em>20/5 = 4    4×4 = 16   so 4/5 is equivalent to 16/20</em>

2/4   <em>20/4 = 5    5×2 = 10   so  2/4 is equivalent to 10/20   </em>

Add:    16/20 + 10/20 = 26/20

8 0
3 years ago
A slope field produces...
balandron [24]

It's A. B doesn't really make sense ("derivative of the differential equation" is somewhat nonsensical, "derivative of an equation" is not meaningful).

More to the point: Slope fields are used to visualize solutions to differential equations of the form

<em>y'</em> = <em>f(x</em>, <em>y)</em>

You take some point (<em>x</em>, <em>y</em>) and evaluate <em>y'</em> at the point. This gives the slope of the line tangent to the particular solution to the DE that passes through the point (<em>x</em>, <em>y </em>).

Sample several points and evaluate <em>y'</em> at those points and you get several different slopes.

Simple example:

<em>y'</em> = <em>x</em> ² - <em>y</em> ² = (<em>x</em> + <em>y </em>) (<em>x</em> - <em>y</em> )

Let's take the points (1, 1), (-1, 0), and (2, -2), at which we get slopes

<em>y'</em> = <em>f</em> (1, 1) = 0

<em>y'</em> = <em>f</em> (-1, 0) = 1

<em>y'</em> = <em>f</em> (2, -2) = 0

From here, you can get particular solutions that pass through a certain point by interpolating the slopes of the tangents to the solution. I've attached a slope field for the example here at the points listed above. (In the order red, green, black). Each light gray arrow in the background shows the slope of the tangent line.

If you're still unsure how the slope field is generated, I suggest looking up videos on the subject. The process is a bit difficult to describe without a dynamic visual aid.

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3 years ago
Which graph shows a line that is perpendicular to line AB?
kherson [118]
The 3rd slot is the answer you are looking for
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3 years ago
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