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3 years ago

Endpoint 1: (-1,2) Midpoint: (9,-6) Endpoint 2=

Mathematics

1 answer:

Vlad [161]3 years ago

7 0

Answer:

(19 , -14)

Step-by-step explanation:

Find the distance in between each x & y for a coordinate.

Let: (x₁ , y₁) = (-1 , 2)

Let: (x₂ , y₂) = (9 , -6)

From x₁ ⇒ x₂: 9 - (-1) = 10

From y₁ ⇒ y₂: -6 - 2 = -8 = 8*

*Remember that distance cannot be negative, but for the sake of this question, we will leave it as -8.

The distance between the x points are in intervals of 10. The distance between the y points are in intervals of 8. Add 10 & subtract 8 to their respective numbers to get endpoint 2:

(9 (+ 10) , -6 (- 8)) = (19 , -14)

Endpoint 2 = (19 , -14)

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A square napkin is folded in half on the diagonal and placed on the diameter of a round plate (see diagram below). If the folded

Crank

Answer:

$A=81(\pi-1)\ in^2$

Step-by-step explanation:

step 1

Find the area of the plate

The area of a circle is given by the formula

$A=\pi r^{2}$

we have

$r=18/2=9\ in$ ---> the radius is half the diameter

substitute

$A=\pi (9)^{2}\\A=81\pi\ in^2$

step 2

Find the area of the square napkin folded (is a half of the area of the square napkin)

we know that

The diagonal of the square is the same that the diameter of the plate

Applying Pythagorean theorem

$D^2=2b^2$

where

b is the length side of the square

we have

$D=18\ in$

substitute

$18^2=2b^2$

solve for b^2

$b^2=162\ in^2$ -----> is the area of the square

Divide by 2

$162/2=81\ in^2$

step 3

Find the area of the space on the plate that is NOT covered by the napkin

we know that

The area of the space on the plate that is NOT covered by the napkin, is equal to subtract the area of the square napkin folded (is a half of the area of the square napkin) from the area of the plate

$A=(81\pi-81)\ in^2$

simplify

$A=81(\pi-1)\ in^2$

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3 years ago

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o-na [289]

Answer:

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2 years ago

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andrezito [222]

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3 years ago

PLEASE HELP!! URGENT!! i will mark brainliest if its right!! In the figure below, ∠DEC ≅ ∠DCE, ∠B ≅ ∠F, and DF ≅ BD. Point C is

zvonat [6]

Answer:

See below.

Step-by-step explanation:

This is how you prove it.

<B and <F are given as congruent.

This is 1 pair of congruent angles for triangles ABC and GFE.

<DEC and <DCE are given as congruent.

Using vertical angles and substitution of transitivity of congruence of angles, show that angles ACB and GEF are congruent.

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3 years ago