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Naya [18.7K]
3 years ago
10

In △MNO, m = 20, n = 14, and m∠M = 51°. How many distinct triangles can be formed given these measurements?

Mathematics
2 answers:
alexandr402 [8]3 years ago
3 0
<span> The answer will be 2 There is only one distinct triangle possible, with m∠N ≈ 33°.</span>
olga2289 [7]3 years ago
3 0

Answer: B is your answer  (:


Step-by-step explanation:


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A soda can had a diameter of 6 cm . What is the circumference of the can?
Alchen [17]
Hello, if the diameter is 6cm, the circumference would be 18.84cm - hope this helps!
4 0
2 years ago
Consider the cases SSS, AAS, ASA, SAS, and SSA. For which of these cases are you unable to solve the triangle using only the Law
qwelly [4]

Answer:

  • SSS and SAS

Step-by-step explanation:

To use the law of sines we need to know two angles and one side or two sides and one angle which should be opposite to one of known sides.

  • SSS - <u>unable</u> to solve as no angle is known
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  • ASA - two angles and the included side, able to find the missing angle solve the rest
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7 0
3 years ago
Solve the system. x – 4y – z = 21 6x – 3y – z = –4 –x +2y –5z = 19 Answer: x =      , y =      , z =      
FrozenT [24]

Consider the given equations:

-x+2y-5z = 19  (Equation 1)

6x-3y-z= -4    (Equation 2)

x-4y-z=21        (Equation 3)

Adding equations 1 and 3, we get

-x+2y-5z+x-4y-z = 19+21

-2y-6z = 40

So, we get -y -3z = 20 (Equation 4)

Multiplying equation 3 by '6', we get

6x - 24y-6z = 126       (Equation 5)

Subtracting equation 5 from equation 2, we get

(6x-3y-z)-(6x - 24y-6z) = -4-126

6x-3y-z-6x + 24y+6z) = -4-126

21y+5z = -130     (Equation 6)

Multiplying equation 4 by '21' and adding it to equation 6, we get

-21y-63z+21y+5z = 420-130

-58z = 290

So, z = -5

Since, -y-3z = 20 -y+15=20 y=-5

So, y=-5

Now, x-4y-z=21 x-4(-5)-(-5)=21 x+20+5=21 x+25=21 x= -4

So, x = -4

Therefore, x = -4, y= -5 and z= -5 are the solutions to the given equations.

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lys-0071 [83]

Answer:

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Step-by-step explanation:

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