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Leto [7]
3 years ago
10

PLEASE HELP MY ASSIGNMENT IS DUE IN 3 HOURS!!!!

Mathematics
1 answer:
dybincka [34]3 years ago
8 0

Answer:

8.9 Feet

Step-by-step explanation:

a^2+b^2 = c^2

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Sally's dog ate 2 out of 6 dog treatsWhat fraction of the dog treats was left? Write the fraction in simplest form.
Yuliya22 [10]

Answer:

2/3

Step-by-step explanation:

you need to subtract 2/6 from 6/6 and you get 4/6.

the top and bottom number can be divided by 2. and you get 2/3

hopes this helps

Please mark me brainliest. It would help(✿◠‿◠)

6 0
3 years ago
Read 2 more answers
Consider circle H with a 6 centimeter radius. If the length of minor arc ST is
alexandr402 [8]
Given that the circle radius of 6 cm and a minor arc ST = 11pi/2. the measure of the angle rst can be solve using the formula s = ra
where s is the lenght of the minor arc
r is the radius of the circle 
and a is the measure of the angle in radians
s = ra
a = s/r
a = (11pi/2) / 6
a = 11pi/12 
or
a = 165 degrees is the measure of angle RST
6 0
4 years ago
Mickey and Minnie had dinner at Mice are Nice restaurant. Their bill was $42.78. If they tip their waitress 18%, how much tip wi
LuckyWell [14K]

Answer:

The total bill with tip is $50.48

The tip is $7.70

Step-by-step explanation:

I promise you that is the answer!

6 0
3 years ago
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Given AG bisects CD, IJ bisects CE, and BH bisects ED. Prove KE = FD.
OverLord2011 [107]

When a line is bisected, the line is divided into equal halves.

See below for the proof of \mathbf{KE \cong FD}

The given parameters are:

  • <em>AC bisects CD</em>
  • <em>IJ bisects CE</em>
  • <em>BH bisects ED</em>

<em />

By definition of segment bisection, we have:

  • \mathbf{CK \cong KE}
  • \mathbf{EF \cong FD}
  • \mathbf{CE \cong ED}

By definition of congruent segments, the above congruence equations become:

  • \mathbf{CK = KE}
  • \mathbf{EF = FD}
  • \mathbf{CE = ED}

By segment addition postulate, we have:

  • \mathbf{CE = CK + KE}
  • \mathbf{ED = EF + FD}

Substitute \mathbf{ED = EF + FD} in \mathbf{CE = ED}

\mathbf{CK + KE = EF + FD}

Substitute \mathbf{CK = KE} and \mathbf{EF = FD}

\mathbf{KE + KE = FD + FD}

Simplify

\mathbf{2KE = 2FD}

Apply division property of equality

\mathbf{KE = FD}

By definition of congruent segments

\mathbf{KE \cong FD}

Read more about proofs of congruent segments at:

brainly.com/question/11494126

6 0
3 years ago
A plane flies x mph. How far can it go in y hours?
wel
The answer is xy hours.
8 0
3 years ago
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