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2 years ago

How do you decide where to put a point when you are graphing?

Mathematics

1 answer:

elena55 [62]2 years ago

4 0

Step-by-step explanation:

When deciding where to put an ordered pair on a graph its important to understand how ordered pairs work.

An ordered pair is split up into to pieces

(x,y) or number example (3,5)

When graphing the x coordinate will fall on your horizontal line or the line that looks like the horizon.

When graphing the y coordinate it will fall on your vertical line or the line that goes right up into the sky like a rocket.

To graph (3,5)

Go to the right 3 and up 5

Sometimes helps to use both fingers to keep your spot. Slide both fingers until they collide and that's where you would plot your point!

Hope this helps if you have an questions just comment and ill try to explain further.

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What is the formula for finding the length of a line on the cartesian plane.

Nata [24]

Answer:

Step-by-step explanation:

distance = square root [ $(x2-x1)^{2}$ + $(y2-y1)^{2}$ ]

where you have two points on the line

Point 1 = P1 = (x1,y1)

point 2 = P1 =(x2,y2)

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3 years ago

Need help with this

siniylev [52]

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3 years ago

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Otrada [13]

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3 years ago

A circle has the equation (x−13)2+(y−1)2=4.

baherus [9]

An equation of circle has form

$(x-x_0)^2+(y-y_0)^2=r^2 \\ where\: x_0\: and\: y_0\: are\: the\: center$

So, the center is (13;1)

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3 0

3 years ago

Perform the indicated operations. Write the answer in standard form, a+bi.<br> 5-3i / -2-9i

Vsevolod [243]

$\huge \boxed{\mathfrak{Answer} \downarrow}$

$\large \bf\frac { 5 - 3 i } { - 2 - 9 i } \\$

Multiply both numerator and denominator of $\sf \frac{5-3i}{-2-9i} \\$ by the complex conjugate of the denominator, -2+9i.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2-9i\right)\left(-2+9i\right)}) \\$

Multiplication can be transformed into difference of squares using the rule: $\sf\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$ .

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2\right)^{2}-9^{2}i^{2}}) \\$

By definition, i² is -1. Calculate the denominator.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{85}) \\$

Multiply complex numbers 5-3i and -2+9i in the same way as you multiply binomials.

$\large \bf \: Re(\frac{5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9i^{2}}{85}) \\$

Do the multiplications in $\sf5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9\left(-1\right)$ .

$\large \bf \: Re(\frac{-10+45i+6i+27}{85}) \\$

Combine the real and imaginary parts in -10+45i+6i+27.

$\large \bf \: Re(\frac{-10+27+\left(45+6\right)i}{85}) \\$

Do the additions in $\sf-10+27+\left(45+6\right)i$ .

$\large \bf Re(\frac{17+51i}{85}) \\$

Divide 17+51i by 85 to get $\sf\frac{1}{5}+\frac{3}{5}i \\$ .

$\large \bf \: Re(\frac{1}{5}+\frac{3}{5}i) \\$

The real part of $\sf \frac{1}{5}+\frac{3}{5}i \\$ is $\sf \frac{1}{5} \\$ .

$\large \boxed{\bf\frac{1}{5} = 0.2} \\$

3 0

3 years ago