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2 years ago

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A line has a slope of 2. It passes through the points (1, 2) and (3, y) . What is the value of y?

2 answers:

Alexus [3.1K]2 years ago

7 0

Answer:

Step-byThe equation of the line with slope 2 is y=2x+b, b is a constant which should be determined. The line passes via through points (1, 2) and (3,y)

(1, 2): 2=2+b --> b=0

So, the equation of the line is y=2x. Now, when x=3, y=6. So the line y=2x passes through points (1, 2) and (3, 6). (3,y)=(3, 6).-step explanation:

yuradex [85]2 years ago

7 0

Answer of y is 6.

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<em> </em> $\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha =0$ <em></em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the definite integration</em>

<em> </em> $\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha$ <em></em>

<em>we know that the trigonometric formula</em>

<em> sin²∝+cos²∝ = 1</em>

<em> cos²∝ = 1-sin²∝</em>

<u><em>step(ii):-</em></u>

<em>Now the integration</em>

<em> </em> $\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha = \int\limits^\pi _0 {(\sqrt{cos^{2} \alpha } } \, )d\alpha$ <em></em>

<em> = </em> $\int\limits^\pi _0 {cos\alpha } \, dx$ <em></em>

<em>Now, Integrating </em>

<em> </em> $= ( sin\alpha )_{0} ^{\pi }$ <em></em>

<em> = sin π - sin 0</em>

<em> = 0-0</em>

<em> = 0</em>

<u><em>Final answer:-</em></u>

<em> </em> $\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha =0$ <em></em>

<em></em>

<em></em>

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