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Reika [66]
3 years ago
15

If you don’t know how to solve this equation, please skip my question and answer someone else’s question(s). Using either slope-

intercept form, y = mx + b or point-slope form y - y1 = m (x - x1), solve the equation 2y = -3x + 6.
I know that the answer is y = -3/2x + 3, but I got y = 3/2x + 3 and I don’t know what I did wrong. Assignment is due in 2 hours!!

Mathematics
1 answer:
Vsevolod [243]3 years ago
5 0

Answer:

You were close; you don’t have to add 3x to the right side though.

We’re given 2y= -3x + 6

Step one: divide both sides by 2

y= (-3/2)x + 3

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What is the HCF of two consecutive numbers?
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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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