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3 years ago

Quadrilateral ABCD was dilated with the origin as the center of dilation to create quadrilateral EFGH on the graph below.

Mathematics

2 answers:

ivann1987 [24]3 years ago

5 0

Answer:

the scale factor is 2.5

Step-by-step explanation:

A dilation of 2.5 because side AB is 4 units and EF is 10 units and if you divide 10/4 will get you with your scale factor.

Nezavi [6.7K]3 years ago

4 0

I think the answers is 2.5

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Y - 3 = -2 ( x + 5 ) in standard form

agasfer [191]

Answer:

2x + y = -7

Step-by-step explanation:

Standard form is Ax + By = C

A represents the coefficient with x

B represents the coefficient with y

C represents the constants

1 - Rewrite to make it easier to work it out

y - 3 = -2 ( x + 5 )

2 - Distribute

y - 3 = -2x - 10

3 - Get the variables on one side and the constants on the other

y - 3 = -2x - 10

+ 2x + 2x

2x + y - 3 = -10

+ 3 + 3

2x + y = -7

Hope this helps,

7 0

3 years ago

Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable

bekas [8.4K]

Answer:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

$u=\frac{y}{x}$

Multiply both sides by x:

$ux=y$

Second step: Differentiate the substitution:

$u'x+u=y'$

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

$u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}$

$u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}$

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of $x^2$ so I'm going to divide top and bottom by $x^2$ :

$u'x+u=\frac{u+u^2}{3+u^2}$

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

$u'x=\frac{u+u^2}{3+u^2}-u$

Find a common denominator: Multiply second term on right hand side by $\frac{3+u^2}{3+u^2}$ :

$u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}$

Combine fractions while also distributing u to terms in ( ):

$u'x=\frac{u+u^2-3u-u^3}{3+u^2}$

$u'x=\frac{-u^3+u^2-2u}{3+u^2}$

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

$u' \frac{3+u^2}{-u^3+u^2-2u}x=1$

Divide both sides by x:

$\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}$

Reorder the top a little of left hand side using the commutative property for addition:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0

3 years ago

A box is filled with 4 red cards, 4 green cards, and 6 brown cards. A card is chosen at random from the box. What is the probabi

atroni [7]

The probability that the card chosen will not be green is 5/7

5 0

4 years ago

Write the equation of the line that passes<br> through the points (3, 1) and (-2, 6).

tia_tia [17]

Answer:

Equation is Y=mx+c (that's how I know it, it may differ)

Step-by-step explanation:

First get the slope which is M and you just gotta do the second y - the first y from the brackets and divide them by the second x - the first x

6-1 /-2-3 = 5/-5

= -1

So now u gotta find c, C= Y-mx

And u have got two brackets, which have an x and y, and u now have m.

Pick one of the brackets like (-2,6) and put them in the equation so it will be : C =6+1*-2

So C =4

The equation now is : Y = - 1x + 4

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2 years ago

What Greek mathematician is known as the father of Geometry?

Luden [163]

C. Euclid is known as the father of Geometry

6 0

4 years ago

Read 2 more answers