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2 years ago

How do I solve 4(x+3)=-6(4y-2)

Mathematics

1 answer:

True [87]2 years ago

4 0

<h2>Answer:</h2><h2>x=-6y</h2>

Step-by-step explanation:

Divide both sides by 4:

$\frac{4\left(x+3\right)}{4}=\frac{-6\left(4y-2\right)}{4}$

Simplify to give:

x+3 = -3(2y-1)

Subtract 3 from both sides:

x+3-3 = -3(2y-1)-3

Simplify to give your answer:

x=-6y

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Given cosα = −3/5, 180 < α < 270, and sinβ = 12/13, 90 < β < 180

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I got

$- \frac{6 3}{65}$

What we know

cos a=-3/5.

sin b=12/13

Angle A interval are between 180 and 270 or third quadrant

Angle B quadrant is between 90 and 180 or second quadrant.

What we need to find

Cos(b)

Cos(a)

What we are going to apply

Sum and Difference Formulas

Basics Sine and Cosines Identies.

1. Let write out the cos(a-b) formula.

$\cos(a - b) = \cos(a) \cos(b) + \sin(a) \sin(b)$

2. Use the interval it gave us.

According to the given, Angle B must between in second quadrant.

Since sin is opposite/hypotenuse and we are given a sin b=12/13. We. are going to set up an equation using the pythagorean theorem.

${12}^{2} + {y}^{2} = {13}^{2}$

$144 + {y}^{2} = 169$

$25 = {y}^{2}$

$y = 5$

so our adjacent side is 5.

Cosine is adjacent/hypotenuse so our cos b=5/13.

Using the interval it gave us, Angle a must be in the third quadrant. Since cos is adjacent/hypotenuse and we are given cos a=-3/5. We are going to set up an equation using pythagorean theorem,

$( - 3) {}^{2} + {x}^{2} = {5}^{2}$

$9 + {x}^{2} = 25$

${x}^{2} = 16$

$x = 4$

so our opposite side is 4. sin =Opposite/Hypotenuse so our sin a =4/5.Sin is negative in the third quadrant so

sin a =-4/5.

Now use cosine difference formula

$- \frac{3}{5} \times \frac{5}{13} + - \frac {4}{5} \times \frac{12}{13}$

$- \frac{15}{65} + ( - \frac{48}{65} )$

$- \frac{63}{65}$

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