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3 years ago

What is a nonlinear function

Mathematics

2 answers:

Tema [17]3 years ago

8 0

Answer: A set of dots that arent in a line-esc shape.

(Think of a scatter plot)

I'm bad at explaining, sorry.

Anettt [7]3 years ago

5 0

The best way to define this is *any highest degree term greater than 1⃣*.

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How do i write the number 640,739 in expanded form

Luda [366]

600000+ 40000+ 700+ 30+ 9

4 0

2 years ago

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a line with slope 3 passes through point (0,10). What is the y-coordinate of the point on the line with x-coordinate 2

skad [1K]

The y-coordinate is 16

<h3><u>Solution:</u></h3>

Given that a line with slope 3 passes through point (0, 10)

To find the y-coordinate of the point on the line with x-coordinate 2

Which means the point is (2, y)

Let us find the required y co-ordinate using slope formula

<em><u>The slope of line is given as:</u></em>

For a line containing points $(x_1 , y_1)$ and $(x_2 , y_2)$ is given as:

$\text {slope}=m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\text {Here } x_{1}=0 ; y_{1}=10 ; x_{2}=2 ; y_{2}=y$

Given that slope "m" = 3

Substituting the values we get,

$\begin{array}{l}{3=\frac{y-10}{2-0}} \\\\ {3=\frac{y-10}{2}} \\\\ {6=y-10} \\\\ {y=10+6=16}\end{array}$

Thus the y-coordinate is 16

8 0

3 years ago

Find the value of angle "A" and angle "B"

Harman [31]

The answer is 41

Because 90-49

8 0

2 years ago

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How can I solve for X?

Natali [406]

Since we know the angle 250 and a circle is 360 degrees we can solve for x by subtracting 250 from 360.

360-250=110 degrees.

Tada!

4 0

2 years ago

A triangle with vertices at A(0, 0), B(0, 4), and C(6, 0) is dilated to yield a triangle with vertices at A′(0, 0), B′(0, 10), a

sineoko [7]

ANSWER

The scale factor is
$2.5$

EXPLANATION

The given triangle has vertices,

$A(0,0),B(0,4),\:and\:C(6, 0)$ .

The vertices of the image triangle is,

$A'(0,0),B'(0,10),\:and\:C'(15, 0)$ .

The scale factor is given by

$k = \frac{image \: length}{object \: length}$

So we can use any of the corresponding sides to determine the scale factor,

$k = \frac{|A'B'|}{ |AB|}$

$k = \frac{ |10 - 0| }{ |4 - 0|}$

$k = \frac{ |10| }{ |4 |} = \frac{10}{4} = 2.5$

Or

$k = \frac{|A'C'|}{ |AC|}$

$k = \frac{ |15 - 0| }{ |6 - 0|}$

$k = \frac{ |15| }{ |6|} = \frac{15}{6} = 2.5$

Or

$k=\frac{|B'C'|}{|BC|}$

$k = \frac{\sqrt{(15 - 0)^2+(0-10)^2 }}{\sqrt{(6 - 0)^2+(0-4)^2}}$

$k = \frac{ 5\sqrt{13}}{2\sqrt{13}} = \frac{5}{2} = 2.5$

The correct answer is C

3 0

2 years ago