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3 years ago

13

In a dilation the image is always similar to its pre image true or false

2 answers:

Korvikt [17]3 years ago

7 0

Answer: True

The image is similar to the preimage as it retains the same shape. The corresponding sides are proportional. The corresponding angles are congruent. The image is not congruent to the preimage if the scale factor is some other value than 1.

mestny [16]3 years ago

3 0

That is true. The image is always similar, the size of the figure can change but its shape will not.

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Plz help me i'll give points and brainliest

RoseWind [281]

Step-by-step explanation:

-1,6 is -1 left 6 up

1,-6 is 1 right -6 down

6,-1 is 6 right -1 down

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2 years ago

Find the area of the shaded region 6cm 6cm

Mashutka [201]

Hope this helps! the answer is option (A). if you still don’t understand after looking at my solution feel free to clarify ya

4 0

2 years ago

Find the value of the expression 63 - 21 + (3 x 11)

bixtya [17]

Answer: First solve the parentheses using order of operations. Then add and subtract from left to right.

Demo:

(3 * 11) = 33

Equation now looks like:

63 - 21 + 33

63 - 21 = 42

Equation now looks like:

42 + 33

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2 years ago

Read 2 more answers

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erma4kov [3.2K]

$72.

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6 0

3 years ago

Can please someone help me? The question is in the picture. Thank you

maxonik [38]

Answer:

C. $9\pi$

Step-by-step explanation:

We proceed to solve for $x$ by the Pythagorean Theorem:

$(x+1)^{2} = (x-1)^{2} + (2\cdot x - 4)^{2}$ (1)

$x^{2} + 2\cdot x + 1 = (x^{2}-2\cdot x + 1) + (4\cdot x^{2}-16\cdot x + 16)$

$x^{2} + 2\cdot x + 1 = 5\cdot x^{2}-18\cdot x + 17$

$4\cdot x^{2} -20\cdot x + 16 = 0$

$(2\cdot x)^{2} -10\cdot (2\cdot x) + 16 = 0$

$(2\cdot x -8)\cdot (2\cdot x -2) = 0$

There are two different solutions:

$x_{1} = 4$ , $x_{2} = 1$

A solution of $x$ is considered realistic when the length of every side of the triangle is a not negative number.

$x_{1} = 4$

$UV = 2\cdot (4) - 4$

$UV = 4$

$WU = 4 - 1$

$WU = 3$

$WV = 4 + 1$

$WV = 5$

$x_{2} = 1$

$UV = 2\cdot (1) - 4$

$UV = -2$

$WU = 1 - 1$

$WU = 0$

$WV = 1 + 1$

$WV = 2$

The only realistic set of solutions is for $x = 4$ , and the radius of the circle is represented by the line segment $WU$ : $r = 3$

And the area of the circle ( $A$ ) is calculated from the following formula:

$A = \pi\cdot r^{2}$

$A = \pi\cdot (9)$

$A = 9\pi$

The correct answer is C.

6 0

2 years ago