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4 years ago

10

Why does cos(90°) = 0?

1 answer:

Anestetic [448]4 years ago

4 0

Answer:

Step-by-step explanation:

The horizontal leg is 0, since there is no horizontal component of the triangle, so, the cosine is 0. Consider the angle itself and the unit circle. There is no horizontal component of the angle: The horizontal component of the angle is 0, so, the cosine of 90° is 0.

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Is One four-ounce shot of vanilla extract really equal to drinking four shots of vodka??

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Nirmala graphed the relationship between the duration (in hours) of using an oil lamp and the volume (in milliliters) of oil rem

mariarad [96]

Answer:

Nirmala can use the lamp for 31.66 hours before it runs out of oil.

Step-by-step explanation:

From the graph,

Take any two points, let say

(15, 25)

(25, 10)

$\mathrm{Slope\:between\:two\:points}:\quad \mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(15,\:25\right),\:\left(x_2,\:y_2\right)=\left(25,\:10\right)$

$m=\frac{10-25}{25-15}$

$m=-\frac{3}{2}$

The equation of line in slope-intercept form

$y = mx + b$

Putting $m=-\frac{3}{2}$ and any point, let say (25, 10), to find the y-intercept 'b'

$10=-\frac{3}{2}\left(25\right)+b$

$-\frac{3}{2}\left(25\right)+b=10$

$-\frac{75}{2}+b=10$

$\mathrm{Add\:}\frac{75}{2}\mathrm{\:to\:both\:sides}$

$-\frac{75}{2}+b+\frac{75}{2}=10+\frac{75}{2}$

$b=\frac{95}{2}$

$b=47.5$

So the equation of line will be:

$y = mx + b$

$y=-\frac{3}{2}x+47.5$

In order to find how long Nirmala can use the lamp before it runs out of oil, we need to find x-intercept which can be obtained by putting y = 0, and solve for x, as duration lies on x-axis.

so

$y=-\frac{3}{2}x+47.5$

Putting y = 0

$0=-\frac{3}{2}x+47.5$

$-\frac{3}{2}x+47.5=0$

$-\frac{3}{2}x=-47.5$

$\mathrm{Multiply\:both\:sides\:by\:}2$

$2\left(-\frac{3}{2}x\right)=2\left(-47.5\right)$

$-3x=-95$

$\mathrm{Divide\:both\:sides\:by\:}-3$

$\frac{-3x}{-3}=\frac{-95}{-3}$

$x=\frac{95}{3}$

$x=31.66$

Therefore, Nirmala can use the lamp for 31.66 hours before it runs out of oil.

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