Mathematics, 17.10.2020 19:01 suwwu

Prove that segment DE is parallel to AB

Answers: 3

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Answer from: Quest

step-by-step explanation:

when the line is solid:

it's either < or >

when the line is dashed:

it's either < or >

this > inequality sign means:

greater thanmore thanaboveexceeds

this < inequality sign means:

less thanfewer thanbelow

this > inequality sign means:

greater than/equal tono less thanat leastminimum of

this < inequality sign means:

less than/equal tono more thanat mostmaximum of

about the problem:

since the line is solid, that means it's either < or >

srry i'm not done ! i'll

Which inequality is represented by the graph? which test point holds true for the inequality?

Answer from: Quest

the correct option is b

step-by-step explanation:

Can somebody me figure out the probability?

Answer from: Quest

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation:

Basic circle problem given: measure of arc ed = 50°, m∠edk = 80° find: ∠dsk

Answer from: Quest

Answer: f(x) = -2x^3 +3x^2 +11x -6see attachedan infinite number. since the magnitude of the leading coefficient is not specified, it may be any negative number. (we have chosen the smallest magnitude integer that makes all coefficients be integers.)step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.

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Prove that segment DE is parallel to AB
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