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3 years ago

A line has this equation : y = 9/4x + 9 Write an equation for the parallel line that goes through (12, 7) .

Mathematics

1 answer:

faltersainse [42]3 years ago

4 0

Answer:

y=9/4x-20

Step-by-step explanation:

Do the parallel line equation

7=9/4(12)+b

Solve for b

7=9×12÷4+b

7=27+b

subtract 27 to both sides

-20=b

Parallel lines have the same slope

y=9/4x-20

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Need help with trig problem in pic

Sidana [21]

Answer:

a) $cos(\alpha)=-\frac{3}{5}\\$

b) $\sin(\beta)= \frac{\sqrt{3} }{2}$

c) $\frac{4+3\sqrt{3} }{10}\\$

d) $\alpha\approx 53.1^o$

Step-by-step explanation:

a) The problem tells us that angle $\alpha$ is in the second quadrant. We know that in that quadrant the cosine is negative.

We can use the Pythagorean identity:

$tan^2(\alpha)+1=sec^2(\alpha)\\(-\frac{4}{3})^2 +1=sec^2(\alpha)\\sec^2(\alpha)=\frac{16}{9} +1\\sec^2(\alpha)=\frac{25}{9} \\sec(\alpha) =+/- \frac{5}{3}\\cos(\alpha)=+/- \frac{3}{5}$

Where we have used that the secant of an angle is the reciprocal of the cos of the angle.

Since we know that the cosine must be negative because the angle is in the second quadrant, then we take the negative answer:

$cos(\alpha)=-\frac{3}{5}$

b) This angle is in the first quadrant (where the sine function is positive. They give us the value of the cosine of the angle, so we can use the Pythagorean identity to find the value of the sine of that angle:

$cos (\beta)=\frac{1}{2} \\\\sin^2(\beta)=1-cos^2(\beta)\\sin^2(\beta)=1-\frac{1}{4} \\\\sin^2(\beta)=\frac{3}{4} \\sin(\beta)=+/- \frac{\sqrt{3} }{2} \\sin(\beta)= \frac{\sqrt{3} }{2}$

where we took the positive value, since we know that the angle is in the first quadrant.

c) We can now find $sin(\alpha -\beta)$ by using the identity:

$sin(\alpha -\beta)=sin(\alpha)\,cos(\beta)-cos(\alpha)\,sin(\beta)\\$

Notice that we need to find $sin(\alpha)$ , which we do via the Pythagorean identity and knowing the value of the cosine found in part a) above:

$sin(\alpha)=\sqrt{1-cos^2(\alpha)} \\sin(\alpha)=\sqrt{1-\frac{9}{25} )} \\sin(\alpha)=\sqrt{\frac{16}{25} )} \\sin(\alpha)=\frac{4}{5}$

Then:

$sin(\alpha -\beta)=\frac{4}{5}\,\frac{1}{2} -(-\frac{3}{5}) \,\frac{\sqrt{3} }{2} \\sin(\alpha -\beta)=\frac{2}{5}+\frac{3\sqrt{3} }{10}=\frac{4+3\sqrt{3} }{10}$

Since $sin(\alpha)=\frac{4}{5}$

then $\alpha=arcsin(\frac{4}{5} )\approx 53.1^o$

4 0

3 years ago

Please help me I’m really stuck

dalvyx [7]

Answer:

/ 3=/ 5 { alternate anglea}

/ 5+/6 =180° {straight angle}

/ 6=180-26

/ 6=154°

hope it helps

<h3>stay safe healthy and happy.</h3>

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3 years ago

Write an expression that can be used to check the quotient of 646 divided by 3?

Katyanochek1 [597]

It is 3x=646

x is the solution of 646/3. :) hope this helps.

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4 years ago

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miskamm [114]

Answer: 102 degrees

Step-by-step explanation:

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51 + 51 = Angle 4

Angle 4 = 102 degrees

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Doss [256]

Answer:

Step-by-step explanation:

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