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larisa [96]
3 years ago
6

Use side lengths to evaluate. tan (π/3)

Mathematics
1 answer:
ch4aika [34]3 years ago
5 0

Tan (pi/3) = Sin (pi/3) / Cos (pi/3) or y/x

y= (squareroot of 3)/2

x= 1/2

sqrt 3/2 × 2/1 (the reciprocal) = sqrt 3

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Determine the value of a so that the line whose equation is ax+y-4=0 is perpendicular to the line containing the points (2,-5) a
soldi70 [24.7K]
First, write the equation of the line containing the points <span>(2,-5) and (-3,2).

We can use 2 point form, or point-slope form.

Let's use </span>point-slope form.

the slope m is \frac{-5-2}{2-(-3)}= \frac{-7}{5}, then use any of the points to write the equation. (ex, pick (2, -5))

y-(-5)=(-7/5)(x-2)

y+5=(-7/5)x+14/5

y= (-7/5)x+14/5 - 5 =(-7/5)x+14/5 - 25/5 =(-7/5)x-11/5


Thus, the lines are 

i) y=-ax+4      and  ii) y=(-7/5)x-11/5

the slopes are the coefficients of x: -a and (-7/5),

the product of the slopes of 2 perpendicular lines is -1, 

so 

(-a)(-7/5)=-1

7/5a=-1

a=-1/(7/5)=-5/7


Answer: -5/7
6 0
3 years ago
1/2 - 1/5=<br> Please answer
Veronika [31]

Answer: 3/10

Step-by-step explanation:

Subtract them

5 0
3 years ago
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Help plz plzzzzzzz i need heelllp
emmainna [20.7K]

Answer:

I can't see the picture right, it's backward sorry

Step-by-step explanation:

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3 0
3 years ago
Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x &gt; 0)
vfiekz [6]

Answer:

Step-by-step explanation:

given a point (x_0,y_0) the equation of a line with slope m that passes through the  given point is

y-y_0 = m(x-x_0) or equivalently

y = mx+(y_0-mx_0).

Recall that a line of the form y=mx+b, the y intercept is b and the x intercept is \frac{-b}{m}.

So, in our case, the y intercept is (y_0-mx_0) and the x  intercept is \frac{mx_0-y_0}{m}.

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph (x_0,\frac{1}{x_0}). Which means that y_0=\frac{1}{x_0}

The slope of the tangent line is given by the derivative of the function evaluated at x_0. Using the properties of derivatives, we get

y' = \frac{-1}{x^2}. So evaluated at x_0 we get m = \frac{-1}{x_0^2}

Replacing the values in our previous findings we get that the y intercept is

(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}

The x intercept is

\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0

The triangle in consideration has height \frac{2}{x_0} and base 2x_0. So the area is

\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2

So regardless of the point we take on the graph, the area of the triangle is always 2.

6 0
3 years ago
1/3x² + x+5/3x² = 1/9<br><br>HAHAHAHA......​
mr_godi [17]

Answer:

Now, 3x 2+x+5⩾0

This is because b² −4ac=1−4×5×3

=−59 (roots are imaginary)

3x²+x+5=(x−3)² =x²+9−6x and x−3⩾0

2x²+7x−4=0

2x²+8x−x−4=0

2x(x+4)−1(x+4)=0

(2x−1)(x+4)=0

x= 1/2 and−4 but x≥3

∴ No solution.

lol hehehe

7 0
2 years ago
Read 2 more answers
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