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2 years ago

Help me!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

Ahat [919]2 years ago

6 0

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ahrayia [7]2 years ago

4 0

Answer:

9. x=-6, y=0

10. x=-9, y=-10

11. x=3, y=-1

Step-by-step explanation:

$2x-9y=-12\\3x+18y=-18$

Multiply the first equation by 2.

$4x-18y=-24\\3x+18y=-18$

Add the equations.

$7x=-42$

$x=-6$

We figured out x now we have to figure out y. To figure out y plug in x into any equation.

$2(-6)-9y=-12$

$-12-9y=-12$

$-9y=0$

$y=0$

Your answer is x = -6 and y = 0 for number nine.

$6x+3y=24\\2x+3y=-12$

Change the signs of the second equation becuase 3y and 3y are both positive. We need one of them to be negative.

$6x+3y=24\\-2x-3y=12$

Add them.

$4x = 36$

$x=9$

Now we have to figure out y.

$2(9)+3y=-12$

$18+3y=-12$

$3y=-30$

$y=-10$

x=-9 and y=-10 is your answer to number ten.

$10x+5y=25\\-7x-3y=-18$

$30x+15y=75\\-35x-15y=-90$

$-5x=-15$

$x=3$

$-7(3)-3y=-18$

$-21-3y=-18$

$-3y=3$

$y=-1$

x=3 and y=-1 is the answer to number eleven. I hope this helps! Let me know if I got anything wrong.

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2 years ago

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krok68 [10]

Answer:

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3 years ago

1. if csc β = 7/3 and cot β = - 2√10 / 3, Find sec β

slava [35]

Step-by-step explanation:

$\tan \beta = \frac{1}{ \cot \beta } = - \frac{3}{2 \sqrt{10} } = - \frac{3 \sqrt{10} }{20}$

$\csc \beta \tan \beta = \frac{1}{ \cos \beta } = \sec \beta$

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$\sec \beta = ( \frac{7}{3} )( - \frac{3 \sqrt{10} }{20} ) = - \frac{7 \sqrt{10} }{20}$

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Use the identity

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We chose the negative value of the cosine because of the condition where cot y > 0. Otherwise, choosing the positive root will yield a negative cotangent value. Now that we know the sine and cosine of y, we can now solve for the tangent:

$\tan \beta = \frac{ \sin y}{ \cos y} =( - \frac{ \sqrt{6} }{3} )( - \frac{3}{ \sqrt{3} } ) = \sqrt{2}$

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3 years ago