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natita [175]
3 years ago
9

Please answer now correct answer fast

Mathematics
1 answer:
Vikki [24]3 years ago
8 0

Answer:

Area = 112.1 m^2

Step-by-step Explanation:

Given:

∆WXY

m < X = 130°

WY = x = 31 mm

m < Y = 26°

Required:

Area of ∆WXY

Solution:

Find the length of XY using Law of Sines

\frac{w}{sin(W)} = \frac{x}{sin(X)}

X = 130°

x = WY = 31 mm

W = 180 - (130 + 26) = 24°

w = XY = ?

\frac{w}{sin(24)} = \frac{31}{sin(130)}

Multiply both sides by sin(24) to solve for x

\frac{w}{sin(24)}*sin(24) = \frac{31}{sin(130)}*sin(24)

w = \frac{31*sin(24)}{sin(130)}

w = 16.5 mm (approximated)

XY = w = 16.5 mm

Find the area of ∆WXY

area = \frac{1}{2}*w*x*sin(Y)

= \frac{1}{2}*16.5*31*sin(26)

= \frac{16.5*31*sin(26)}{2}

Area = 112.1 m^2 (to nearest tenth).

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Find the first three terms in the expansion , in ascending power of x , of (2+x)^6 and obtain the coefficient of x^2 in the expa
Nataly_w [17]

Answer:

The first 3 terms in the expansion of (2 + x)^{6} , in ascending power of x are,

64 , 192 \times x^{1} {\textrm{  and  }}240 \times x^{2}

coefficient of x^{2} in the expansion of (2+x - x^{2})^{6} = (240 - 192) = 48

Step-by-step explanation:

(2+x)^{6}

= \sum_{k=0}^{6}(6_{C_{k}} \times x^{k} \times 2^{6 - k})

= 6_{C_{0}} \times x^{0} \times 2^{6}  + 6_{C_{1}} \times x^{1} \times 2^{5} + 6_{C_{2}} \times x^{2} \times 2^{4} + terms involving higher powers of x

= 64 + 192 \times x^{1} + 240 \times x^{2} + terms involving higher powers of x

so, the first 3 terms in the expansion of (2 + x)^{6} , in ascending power of x are,

64 , 192 \times x^{1} {\textrm{  and  }}240 \times x^{2}

Again,

(2+x - x^{2})^{6}

= \sum_{k=0}^{6}(6_{C_{k}} \times (2 + x)^{k} \times (-x^{2})^{6 - k})

Now, by inspection,

the term x^{2} comes from k =5 and k = 6

for k = 5, the coefficient of  x^{2}  is , (-32) \times 6 = -192

for k = 6 , the coefficient of x^{2} is, 6_{C_{2}} \times 2^{4} = 240

so,   coefficient of x^{2} in the final expression = (240 - 192) = 48

3 0
3 years ago
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