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3 years ago

What is the translation from quadrilateral IJK to quadrilateral I'J’K’

Mathematics

1 answer:

liberstina [14]3 years ago

6 0

Answer:

The translation from triangle IJK to triangle I'J'K' is $T_{(2, 6)}$ which is 2 units to the right and 6 units up

Step-by-step explanation:

The coordinates of triangle JKI are;

J has coordinates (1, - 1)

K has coordinates (1, - 4)

I has coordinates (-3, - 2)

While, the coordinates of translation triangle J'K'I' are;

J' has coordinates (3, 5)

K' has coordinates (3, 2)

I' has coordinates (-1, 4)

Which give the translation as follows

Translation in the y-coordinate (y values);

For J = 5 - (-1) = 6

For K = 2 - (-4) = 6

For I = 4 - (-2) = 6

Translation in the x-coordinate (x values);

For J = 3 - 1 = 2

Therefore, the translation from triangle IJK to triangle I'J'K' is T(2, 6) which is 2 units to the right and 6 units up

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Work out radius of 1.43 x 10^5

Gennadij [26K]

Answer:

Step-by-step explanation:

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3 years ago

Point B is on line AC so that AB : BC = 2 : 1. Point D is on line AB so that AD : DB = 3 : 2. Find AD : DC

Leya [2.2K]

Firstly, we will draw figure

Let's assume

length of AC=x

we have

AB : BC = 2 : 1

so,

$AB=\frac{2}{3} x$

$BC=\frac{1}{3} x$

Point D is on line AB

and

$AB=\frac{2}{3} x$

AD : DB = 3 : 2

so, we get

$AD=\frac{3}{5} *\frac{2}{3} x$

$AD=\frac{2}{5} x$

$DB=\frac{2}{5} *\frac{2}{3} x$

$DB=\frac{4}{15} x$

now, we can locate these values

Firstly, we will find DC

DC=DB+BC

now, we can plug values

$DC=\frac{4}{15} x+\frac{1}{3} x$

$DC=\frac{3x}{5}$

we have got

$AD=\frac{2}{5} x$

now, we can find ratio

$\frac{AD}{DC} =\frac{\frac{2}{5} x}{\frac{3x}{5}}$

now, we can simplify it

$\frac{AD}{DC} =\frac{2}{3}$

so,

AD:DC=2:3...........Answer

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3 years ago

1) Which circles are congruent?

alexdok [17]

Hello!

1) A and D

The radius is the distance from the center of a circle to any part of the edge. Both A and D have a radius of three, regardless of what direction they go.

2) 5"

In circle C we see that there is a line segment that is 5" that goes from the center to the edge, or the radius.

3) C

The different variables are representing different points on the circles. The circle that contains a line connecting C and L is circle C.

4) BN

Circle B contains a radius that is the distance from B to N, or BN.

5) EF

The diameter is the distance of a straight line that was pass through the middle of a circle, or its width. The diameter is double the radius. In circle A we see that line EF passes through the center put goes all the way across the circle.

6)IJ

Line IJ isn't even linear, so it could not be the radius. Both AG and BH are straight lines that represent the radius.

7) 6"

The diameter is always twice the radius. In circle A the radius is 3, so the diameter is 6.

8) A and D

Just as we said in the first answer, A and D both have a radius that measures 3 inches.

I hope this helps!

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3 years ago

What is this problem

sergeinik [125]

Answer:

$m=A-\frac{2n}{Z}$

Step-by-step explanation:

Multiply by 2:

2n = Z(A - m)

Divide by Z:

$\frac{2n}{Z} =A-m$

Subtract A:

$-m=\frac{2n}{Z}-A$

Multiply by -1:

$m=A-\frac{2n}{Z}$

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3 years ago

CAN SOMEONE PLEASE EXPLAIN THIS FOR ME PLEASE HELP

Nata [24]

The equation that describes the given polynomial is:

P(x) = x²*(x - 2)²*(x + 2)

<h3>How to find the equation for the polynomial?</h3>

Remember that a polynomial with a leading coefficient A and roots:

x₁, x₂, x₃,...,xₙ

The equation that describes the polynomial is:

P(x) = a*(x - x₁)*(x - x₂)*...*(x - xₙ)

In this case, we know that the leading coefficient is 1, and the roots are:

2, 2, 0, 0, -2.

(the first two appear two times because have a multiplicity of two).

Then we can write the equation of the polynomial as:

P(x) = 1*(x - 2)*(x - 2)*(x - 0)*(x - 0)*(x + 2)

P(x) = x²*(x - 2)²*(x + 2)

Learn more about polynomials:

brainly.com/question/4142886

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1 year ago