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riadik2000 [5.3K]
3 years ago
8

H. 539.28 - 6.42 Help pls

Mathematics
1 answer:
My name is Ann [436]3 years ago
6 0

Answer:

532.86

Step-by-step explanation:

just subtract starting with the numbers after the decimal place

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Will give brainliest
klemol [59]

Your question is calling the figure ABCD, but the image is showing PQRS


Using the letters that you wrote:

Because the figure was only rotated and not scaled , the sides with the same letters would still be the same

So AB would be the same as A'B' ( Length of AB = Length of A′B)

and

BC = B'C'

CD = C'D'

AD = A'D'


Or by the picture:

PQ = P'Q'

QR = Q'R'

RS = R'S'

PS = P'S'





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3 years ago
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point b (3,6) is dilated from the origin by a scale factor of r = 2/3 . what are the coordinates of b’
frez [133]

joke 1. 2 hunters were out in the woods right while they were following fresh tracks from a big elk one of the hunters drop to the ground and his eyes roll backwards so the other hunter whips out his phone and calls the hospital "I THINK MY FRIEND IS DEAD!" he told the doctor the doctor told him " Calm down let's make sure he's dead. so there is a moment of silence the you here a big pop the hunter says "NOW WHAT!"     (he shot his friend)

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3 0
2 years ago
What is PA <br> Enter your answer in the box
SSSSS [86.1K]

Answer:

3

Step-by-step explanation:

P is the in-center

⇒PA=PE=PD because they are in-radius of the in-circle

We know that, tangent segments drawn from a point outside the circle are always equal in length

⇒DK=EK=7.2

In right triangle PKE,

using Pythagoras' Theorem : PK^{2}=PE^{2}+KE^{2}

⇒PE^{2}=PK^{2}-KE^{2}

⇒PE=\sqrt{PK^{2}-KE^{2} }

⇒PE^{2}=\sqrt{7.8^{2}-7.2^{2}} }

⇒PE=3

Therefore, PA=3

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3 years ago
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Delicious77 [7]

Hey there!

“2 (5x + 3)”

• FIRST and ONLY STEP: DISTRIBUTE 2 to ALL OF YOUR NUMBERS in the parentheses

2(5x) + 2(3)

2(5x) = 10x ⬅️ FRONT number

2(3) = 6 ⬅️ BACK number

BOTH OF YOUR NUMBERS ARE POSITIVE SO WE DON’T HAVE TO CHANGE ANYTHING

Answer: 10x + 6 ☑️

Good luck on your assignment and enjoy your day!

~LoveYourselfFirst:)

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