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kipiarov [429]
4 years ago
5

Line segments can't be extended. True False

Mathematics
1 answer:
serg [7]4 years ago
7 0
False I hope this helps 
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NEED ANSWER ASAP!!! PLEASE HELP
lys-0071 [83]

Answer:

Answer "C"

Step-by-step explanation:

I believe this because it would not make sense to compare the already known equation to the unknown equation and does make sense to compare the unknown equation to the known equation

(pls dont delete this moderators)

3 0
3 years ago
Please help me find the slope of the line which one would it be ? :) and why
valina [46]
Slope is rise/run
5 up and 2 to the left 5/2
The slope is 5/2 :)
7 0
3 years ago
17. Evaluate a. 6! b. 8P5 c. 12C4
Andreyy89
The factorial ! just means we multiply by every natural number less that the value so
6! =6×5×4×3×2×1= 720

for permutations we use the formula n!/(n-r)!
so we have 8!/(8-5)!=8!/3!=8×7×6×5×4

for combinations s we have n!/(n-r)!r!
so we have 12!/(12-4)!4!=12!/8!4!=12×11×10×9/4×3×2=11×10×9/2=99×5
7 0
3 years ago
If x-12/x+36=0, what is the value of x
tatyana61 [14]

Answer:

12

Step-by-step explanation:

x - 12 / x + 36 = 0

Multiply both sides by x + 36

x - 12 = 0 ✖ (x + 36)

x - 12 = 0

Add 12 to both sides

x - 12 + 12 = 0 + 12

x = 12

I hope this was helpful, Please mark as brainliest  

4 0
3 years ago
Read 2 more answers
Plz hurry!!!! thank you!!!!
Nikitich [7]

Answer:

Area of trapezium = 4.4132 R²

Step-by-step explanation:

Given, MNPK is a trapezoid

MN = PK and ∠NMK = 65°

OT = R.

⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).

Now, sum of interior angles in a quadrilateral of 4 sides = 360°.

⇒ x + x + 65° + 65° = 360°

⇒ x = 115°.

Here, NS is a tangent to the circle and ∠NSO = 90°

consider triangle NOS;

line joining O and N bisects the angle ∠MNP

⇒ ∠ONS = \frac{115}{2} = 57.5°

Now, tan(57.5°) = \frac{OS}{SN}

⇒ 1.5697 = \frac{R}{SN}

⇒ SN = 0.637 R

⇒ NP = 2×SN = 2× 0.637 R = 1.274 R

Now, draw a line parallel to ST from N to line MK

let the intersection point be Q.

⇒ NQ = 2R

Consider triangle NQM,

tan(∠NMQ) = \frac{NQ}{QM}

⇒ tan65° = \frac{NQ}{QM}

⇒ QM = \frac{2R}{2.1445}

QM = 0.9326 R .

⇒ MT = MQ + QT

          = 0.9326 R + 0.637 R  (as QT = SN)

⇒ MT = 1.5696 R

⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R

Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).

⇒ A = (\frac{NP + MK}{2}) × (ST)

       = (\frac{1.274 R + 3.1392 R}{2}) × 2 R

       = 4.4132 R²

⇒ Area of trapezium = 4.4132 R²

5 0
3 years ago
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