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3 years ago

12

Joni translates a right triangle 2 units down and 4 units to the right. How does the orientation of the image of the triangle co

mpare with the orientation of the preimage?
I will make it 100 pts if you help

2 answers:

babunello [35]3 years ago

8 0

it is transfered over an axes

andrew-mc [135]3 years ago

6 0

Not completely sure of what this question is asking, but the orientation is the same. The actual triangle moved diagonally.

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If trapezoid JKLM is translated according to the rule (x,y)—> (x+2, y - 6), what are the coordinates of point L

guapka [62]

Answer:

<em> ( x, y) is translation '2' units right and '6' units down </em>

<em> (x ,y))—> (x +2 ,y-6)</em>

Step-by-step explanation:

<u><em>Explanation</em></u>:-

Type of transformation change to co-ordinate point

vertical translation up 'd' units (x,y) changes to (x , y+d)

<em>vertical translation down 'd' units (x,y) changes to (x , y-d)</em>

Horizontal translation left 'c' units (x,y) changes to (x-c , y)

Horizontal translation right 'c' units (x,y) changes to (x+c , y)

Now by use above table

given data ( x, y) is translation '2' units right and '6' units down

(x ,y))—> (x +2 ,y-6)

<u><em>Final answer:-</em></u>

<em> ( x, y) is translation '2' units right and '6' units down </em>

<em> (x ,y))—> (x +2 ,y-6)</em>

3 0

3 years ago

The positive acute angle formed by the _____ side of an angle in standard position and the x-axis is called a reference angle.

Helga [31]

Answer:

option-C

terminal

Step-by-step explanation:

We know that

reference angle is between terminal side and x-axis

so, the other side will be terminal position

so, we can write as

The positive acute angle formed by the <u>terminal</u> side of an angle in standard position and the x-axis is called a reference angle.

So,

option-C

terminal

7 0

3 years ago

A 4 pound bag of sugar contains 454 one teaspoon servings and costs $3.49. A batch of muffins uses 3/4 cup of sugar. How many ba

nika2105 [10]

I don't know the answer

6 0

3 years ago

A lamina with constant density rho(x, y) = rho occupies the given region. Find the moments of inertia Ix and Iy and the radii of

jenyasd209 [6]

Answer:

Ix = Iy = $\frac{ρπR^{4} }{16}$

Radius of gyration x = y = $\frac{R}{4}$

Step-by-step explanation:

Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.

Mass of disk = ρπR2

Moment of inertia about its perpendicular axis is $\frac{MR^{2} }{2}$ . Moment of inertia of quarter disk about its perpendicular is $\frac{MR^{2} }{8}$ .

Now using perpendicular axis theorem, Ix = Iy = $\frac{MR^{2} }{16}$ = $\frac{ρπR^{4} }{16}$ .

For Radius of gyration K, equate MK2 = MR2/16, K= R/4.

3 0

3 years ago

What is the sum of 20x^2-10x-30

ivann1987 [24]

Answer:

The sum of the roots is 0.5

Step-by-step explanation:

<u><em>The correct question is</em></u>

What is the sum of the roots of 20x^2-10x-30

we know that

In a quadratic equation of the form

$ax^{2} +bx+c=0$

The sum of the roots is equal to

$-\frac{b} {a}$

in this problem we have

$20x^{2} -10x-30=0$

so

$a=20\\b=-10\\c=-30$

substitute

$-\frac{(-10)} {20}=0.5$

<u><em>Verify</em></u>

Find the roots of the quadratic equation

The formula to solve a quadratic equation is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

$a=20\\b=-10\\c=-30$

substitute

$x=\frac{-(-10)\pm\sqrt{-10^{2}-4(20)(-30)}} {2(20)}$

$x=\frac{10\pm\sqrt{2,500}} {40}$

$x=\frac{10\pm50} {40}$

$x=\frac{10+50} {40}=1.5$

$x=\frac{10-50} {40}=-1$

The roots are x=-1 and x=1.5

The sum of the roots are

$-1+1.5=0.5$ ----> is ok

5 0

3 years ago