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SCORPION-xisa [38]
4 years ago
12

If the polynomial 3 + 2 + 9x – 3 is divided by the expression x – 2, what is the degree of the first term of the quotient?

Mathematics
1 answer:
Musya8 [376]4 years ago
3 0
The answer is D i think welcome:)
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42°

Step-by-step explanation:

(7x - 2) \degree = (6x + 18) \degree\\ ..(exterior \: alternate \:  \angle s) \\ 7x - 2 = 6x + 18 \\ 7x - 6x = 18 + 2 \\ x = 20 \\  \\ (7x - 2) \degree +  \angle \: 1 = 180 \degree \\..(straight \: line \:  \angle s) \\ (7 \times 20 - 2) \degree+  \angle \: 1 = 180 \degree  \\ 138 \degree+  \angle \: 1 = 180 \degree  \\  \angle \: 1 = 180 \degree - 138 \degree \\  \angle \: 1 = 42\degree

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A running circuit is in the shape of a triangle with lengths of 6km, 6.5km and 7km. What are the sizes of the angles (in minutes
Rudik [331]

A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to 180^{o}. The <em>measures</em> of the internal <u>angles</u> of the <u>triangle</u> given in the question are A = 52.6^{o}, B = 59.4^{o}, and C = 68^{o}.

A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to 180^{o}.

Considering the given question, let the <u>sides</u> of the triangle be: a = 6 km, b = 6.5 km, and c = 7 km.

Apply the <em>Cosine rule</em> to have:

c^{2} = a^{2} + b^{2} - 2ab Cos C

So that;

7^{2} = 6^{2} + (6.5)^{2} - 2(6 * 6.5) Cos C

49 = 36 + 42.25 - 78Cos C

78 Cos C = 78.25 - 49

               = 29.25

Cos C = \frac{29.25}{78}

         = 0.375

C = Cos^{-1} 0.375

   = 67.9757

C = 68^{o}

Apply the <em>Sine rule</em> to determine the <u>value</u> of B,

\frac{b}{Sin B} = \frac{c}{Sin C}

\frac{6.5}{Sin B} = \frac{7}{Sin 68}

SIn B = \frac{6.5 *Sin 68}{7}

         = 0.861

B = Sin^{-1} 0.861

   = 59.43

B = 59.4^{o}

Thus to determine the value of A, we have;

A + B + C = 180^{o}

A + 59.4^{o} + 68^{o} = 180^{o}

A = 180^{o} - 127.4

  = 52.6

A = 52.6^{o}

Therefore the <u>sizes</u> of the <em>internal angles</em> of the triangle are: A = 52.6^{o}, B = 59.4^{o}, and C = 68^{o}.

For more clarifications on applications of the Sine and Cosine rules, visit: brainly.com/question/14660814

#SPJ1

8 0
2 years ago
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