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rewona [7]
2 years ago
12

The hanger image below represents a balanced equation.

Mathematics
1 answer:
vlada-n [284]2 years ago
5 0

Answer:

18

Step-by-step explanation:

36 divided by two is 18 therefore 18+18  is 36

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Solve the triangle. B = 36°, a = 38, c = 18
Gelneren [198K]

Answer:

A ≈ 119.7°, b ≈ 25.7, C ≈ 24.3°

Step-by-step explanation:

A suitable app or calculator does this easily. (Since you're asking here, you're obviously not unwilling to use technology to help.)

_____

Given two sides and the included angle, the Law of Cosines can help you find the third side.

... b² = a² + c² - 2ac·cos(B)

... b² = 38² + 18² -2·38·18·cos(36°) ≈ 661.26475

... b ≈ 25.715

Then the Law of Sines can help you find the other angles. It can work well to find the smaller angle first (the one opposite the shortest side). That way, you can tell if the larger angle is obtuse or acute.

... sin(C)/c = sin(B)/b

... C = arcsin(c/b·sin(B)) ≈ 24.29515°

This angle and angle B add to less than 90°, so the remaining angle is obtuse. (∠A can also be found as 180° - ∠B - ∠C.)

... A = arcsin(a/b·sin(B)) ≈ 119.70485°

4 0
3 years ago
Which expression is equivalent to?????
qwelly [4]
Your first choice is the correct answer and I am 100% sure.
7 0
3 years ago
Read 2 more answers
The sum of the squares of 3 consecutive positive integers is 116. What are the numbers?
dalvyx [7]
3 divided by 116 is 37
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3 years ago
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I need help<br> what is 2+4+1<br> im desperate
uranmaximum [27]
*Quack* 7 hope this helps fellow duck!
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3 years ago
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Let V be a vector space of dimension 4. Determine if each statement is true or false. (a) Any set of 5 vectors in V must be line
son4ous [18]

Answer:

a) True

b) False

c) False

d) False

e) True

Step-by-step explanation:

a) Each basis of V has four vectors. Then any set of 5 vectors must be linear dependent (LD).

b) Suppose that \{v_1,v_2,v_3,v_4\} is a basis of V. Considere the set A=\{v_1,\lambda_1v_1,\lambda_2v_1,v_2,v_3\} where \lambda_1, \lambda_2 are scalars. The set has 5 vectors but V\neq span(A) because v_4 is not belong to A and v_4  is linear independent of v_1

c)  Suppose that \{v_1,v_2,v_3,v_4\} is a basis of V. Considere the set A=\{v_1,\lambda_1v_1,\lambda_2v_1,\lambda_3v_1\} where \lambda_1, \lambda_2,\lambda_3 are scalars. A has four nonzero vectors but isn't a basis because is a LD set.

d)  Suppose that \{v_1,v_2,v_3,v_4\} is a basis of V. Considere the set A=\{v_1,\lambda_1v_1,\lambda_2v_1\} where \lambda_1, \lambda_2 are scalars. A has 3 nonzero vectors but isn't a basis because is a LD set.

e)  Since any basis of V must have 4 elements, then a set of three vectors cannot generate V.

4 0
3 years ago
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