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schepotkina [342]
3 years ago
9

Quadrilateral EFGH was dilated by a scale factor of 2 from the center (1, 0) to create E'F'G'H'. Which characteristic of dilatio

ns compares segment F'H' to segment FH?
A). A segment that passes through the center of dilation in the pre-image continues to pass through the center of dilation in the image.

B). A segment that does not pass through the center of dilation in the pre-image is parallel to its corresponding segment in the image.

C). A segment in the image has the same length as its corresponding segment in the pre-image.

D). A segment that does not pass through the center of dilation in the pre-image is perpendicular to its corresponding segment in the image.

Mathematics
1 answer:
jok3333 [9.3K]3 years ago
3 0

Answer:

  B).  A segment that does not pass through the center of dilation in the pre-image is parallel to its corresponding segment in the image

Step-by-step explanation:

Both A and B are true of dilations. However, since FH does not pass through the center of dilation, statement B is applicable to FH and F'H'.

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amm1812

S_n=\dfrac{a_1+a_n}{2}\cdot n\\n=16\\a_1=1\\a_{16}=?\\\\a_n=a_1+(n-1)\cdot d\\d=6\\a_{16}=1+(16-1)\cdot6=1+90=91\\\\S_{16}=\dfrac{1+91}{2}\cdot6=46\cdot6=276

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3 years ago

jekas [21]

Given: In the given figure, there are two equilateral triangles having side 50 yards each and two sectors of radius (r) = 50 yards each with the sector angle θ = 120°

To Find: The length of the park's boundary to the nearest yard.

Calculation:

The length of the park's boundary (P) = 2× side of equilateral triangle + 2 × length of the arc

or,                    (P) = 2× 50 yards + 2× (2πr) ( θ ÷360°)

or,                    (P) = 2× 50 yards + 2× (2×3.14× 50 yards) ( 120° ÷360°)

or,                    (P) = 100 yards + 2× (2×3.14× 50 yards) ( 120° ÷360°)

or,                    (P) = 100 yards + 209.33 yards

or,                    (P) = 309.33 yards ≈309 yards

Hence, the option D:309 yards is the correct option.

7 0
3 years ago
Read 2 more answers
What is the equation, in standard form, of a parabola that models the values in the table?
love history [14]

Answer:

<u>Y=4x^2+3x-6</u>

Step-by-step explanation:

For the standard form equation to model the values in the table, each value of x in the table should give the matching the y value when substituted into the equation. We will test each equation:

<u>Y=3x^2+4x-6 for (-2,4)</u>

Y=3(-2)^2+4(-2)-6=3(4)+-8-6=12+-8-6=-2\\

This does not give 4 as the answer and is not a solution.

<u>Y=4x^2+3x-6 for (-2,4)</u>

Y=4(-2)^2+3(-2)-6=4(4)+-6-6=16+-6-6=-4\\

This does give 4 as the answer and is a possible solution.

<u>Y=4x^2-3x-6 for (-2,4)</u>

Y=4(-2)^2-3(-2)-6=4(4)+6-6=16+6-6=16\\

This does not give 4 as the answer and is not a solution.

<u>Y=-4x^2-3x-6 for (-2,4)</u>

Y=-4(-2)^2-3(-2)-6=-4(4)+6-6=-16+6-6=-16\\

This does not give 4 as the answer and is not a solution.

The only possible solution is <u>Y=4x^2+3x-6</u>

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Where they at bruhhh
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Determine whether the quadrilateral is a parallelogram justify your answer.
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SOLUTION:
From the figure, all 4 angles are congruent. Since each pair of opposite angles are congruent, the quadrilateral is a parallelogram by Theorem 6.10.
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