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MakcuM [25]
3 years ago
15

What is the measure of angle B if angle BCD is 100

Mathematics
1 answer:
JulijaS [17]3 years ago
8 0

The triangle is isosceles so the angle measurements of angle A and C is 80 degrees. So:

180 - (80 + 80)= 20

Answer: 20 degrees

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Round 1,393 to the nearest thousand
myrzilka [38]
The answer is 1,000 4 or below you round down and 5 or higher you round up
6 0
3 years ago
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PLEASE HURRY IM TIMED Andy drove 102 miles on Monday morning. This was
konstantin123 [22]

Answer:

420

Step-by-step explanation:

6 0
3 years ago
Refer to Expression 1 and Expression 2.
TiliK225 [7]

Expression 1 is 25

6*7-3^2*9+4^3

First simplify all the exponents to get 6*7 - 9 * 9 + 64 and then multiply the two numbers to get 42 - 81 + 64 and just add = 25

For expression 2, if you put the parantheses around (5+4)*2+6-2*2-1, it'll be 19

(5+4)*2+6-2*2-1

9*2+6-2*2-1

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24-5

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Remember order of operations

5 0
3 years ago
I need help with this problem, if anyone could help ASAP, that would be much appreciated. In the figure below, mROP = 125° Find
VARVARA [1.3K]

Answer:

mRP = 125°

mQS = 125°

mPQR = 235°

mRPQ = 305°

Step-by-step explanation:

Given that

  • mROP = 125°
  • ∠ROP is a central angle

Then:

  • measure of arc RP, mRP = mROP = 125°

Given that

  • ∠QOS and ∠ROP are vertical angles

Then:

  • mQOS = mROP = 125°
  • measure of arc QS, mQS = mROP = 125°

Given that

  • ∠QOR and ∠SOP are vertical angles

Then:

  • mQOR = mSOP

Given that

  • The addition of all central angles of a circle is 360°

Then:

mQOS + mROP + mQOR + mSOP = 360°

250° + 2mQOR = 360°

mQOR = (360°- 250°)/2

mQOR = mSOP = 55°

And (QOR and SOP are central angles):

  • measure of arc QR, mQR = mQOR = 55°
  • measure of arc SP, mSP = mSOP = 55°

Finally:

measure of arc PQR, mPQR = mQOR + mSOP + mQOS = 55° + 55° + 125° = 235°

measure of arc RPQ, mRPQ = mROP + mSOP + mQOS = 125° + 55° + 125° = 305°

6 0
3 years ago
6x + 4y=10 12x + y= -29
Olin [163]

Answer: x=-3    y=7.

Step-by-step explanation:

\displaystyle\\\left \{ {{6x+4y=10\ \ \ \ \ (1)} \atop {12x+y=-29\ \ \ \ (2)}} \right.\\  We\  multiply\  both\  sides\  of\  equation\  (1)\  by\  -2:\\\left \{ {{(-2)*6x+(-2)*4y=(-2)*10} \atop {12x+y=-29}} \right. \\\\\left \{ {{-12x-8y=-20} \atop {12x+y=-29}} \right. \\We\  summarize\  equations\  (1)\  and\  (2):\\-7y=-49\\-7*y=-7*7\\Divide\  both\  sides\  of\  the\  equation\  by\  -7:\\y=7.\\We\  substitute\  the\  value\  of\  y\  into \ equation\  (1):\\6x+4*7=10\\6x+28=10\\6x+28-28=10-28\\6x=-18\\6*x=6*(-3)\\Divide\  both\  sides\  of\  the\  equation\  by\  6:\\x=-3.

3 0
2 years ago
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