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4 years ago

Solve this system of linear equations by substitution method

1 answer:

6 0

If I were you, I would select the first equation to put in y=mx+b form so that you can substitute it in to the second one. What you do is you subtract the x from the left side and bring it over to the right and then you divide everything by 5. The y=mx+b equation is y=-1/5x-7. Then, you substitute this equation into the bottom one so you are now working only with x's. You should distribute and combine like terms. When you distribute and clean it up, at the end you should get 18/5x-14=8. You then add 14 to get 22 on the right side and then multiply that by the reciprocal of the fraction. When you multiply 22 by 5/18, you should get a reduced fraction of 55/9. You then substitute this value into either equation to get the y-value. You should get a y-value of -260/45. Your answer as an ordered pair should be (55/9,-260/45).

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mario62 [17]

Answer:

26.4

Step-by-step explanation:

<u>Law Of Cosines:</u>

$cos(A)=\frac{b^2+c^2-a^2}{2bc}$

This should work for any side. This can generally be thought as:

$cos(\text{angle}) = \frac{\text{sum of squares of two other sides-opposite side squared}}{\text{2 times the product of the other two sides}}$

If this is too confusing here's the formula for the other sides (which is essentially the same, just different variables)

$cos(B)=\frac{a^2+c^2-b^2}{2ac}$

$cos(C) =\frac{a^2+b^2-c^2}{2ab}$

Anyways now just plug in the known values into the equation

$cos(A)=\frac{4^2+6^2-3^2}{2(6)(4)}\\$

Square and multiply values

$cos(A)=\frac{16+36-9}{48}$

Add the values in the numerator

$cos(A)=\frac{43}{48}$

Take the inverse of cosine on both sides

$A=cos^{-1}(\frac{43}{48})$

calculate arccosine (inverse cosine) using a calculator

$A\approx 26.384$

Round to nearest tenth

$A\approx26.4$

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