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1 year ago

If tan theta = startfraction 11 over 60 endfraction, what is the value of cot theta?

Mathematics

1 answer:

Dennis_Churaev [7]1 year ago

6 0

The value of $cot\theta$ when $tan\theta =\frac{11}{60}$ comes to be $\frac{60}{11}$ .

Given that trigonometric ratio:

$tan\theta = \frac{11}{60}$

<h3>What is the tangent of an angle?</h3>

The tangent of an angle is the ratio of the opposite side(to that angle) to the adjacent side(to that angle).

So, for the given problem

Opposite side to $\theta$ = 11

Adjacent side to $\theta$ = 60

So, $cot \theta =\frac{AdjacentSide }{OppositeSide}$

$cot\theta =\frac{60}{11}$

Therefore, the value of $cot\theta$ when $tan\theta =\frac{11}{60}$ comes to be $\frac{60}{11}$ .

To get more about trigonometric ratios visit:

brainly.com/question/24349828

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What is the distance between the building B (4, 6) and the station K (–6, –2)? Use a calculator to round to the nearest tenth.

Svetllana [295]

Answer:

The answer is

<h2>12.8 units</h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{ ({x1 - x2})^{2} + ({y1 - y2})^{2} } \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

B (4, 6) and K (–6, –2)

The distance is

$|BK| = \sqrt{ ({4 + 6})^{2} + ({6 + 2})^{2} } \\ = \sqrt{ {10}^{2} + {8}^{2} } \: \: \: \: \: \: \: \: \: \\ = \sqrt{100 + 64} \: \: \: \: \: \: \: \: \: \\ = \sqrt{164} \\ = 2 \sqrt{41} \\ = 12.80$

We have the final answer as

<h3>12.8 to the nearest tenth</h3>

Hope this helps you

5 0

3 years ago

What type of graph uses percents to display data?

Tamiku [17]

It depends on the data type. if you want to show an increase or decrease in something, you want a LINE GRAPH. if you want to show the difference between more than one variable, i would use a BAR GRAPH.
*hope this helped

4 0

2 years ago

Pls help sixowiskosixo

erma4kov [3.2K]

The given question is a quadratic equation and we can use several methods to get the solutions to this question. The solution to the equation are 3/4 and -5/6 and the greater of the two solutions is 3/4

<h3>Quadratic Equation</h3>

Quadratic equation are polynomials with a second degree as it's highest power.

An example of a quadratic equation is

$y = ax^2 + bx + c$

The given quadratic equation is $24x^2 + 2x = 15$

Let's rearrange the equation

$24x^2 + 2x = 15\\24x^2 + 2x - 15 = 0$

This implies that

a = 24
b = 2
c = -15

The equation or formula of quadratic formula is given as

$y = \frac{-b +- \sqrt{b^2 - 4ac} }{2a}$

We can substitute the values into the equation and solve

$y = \frac{-b +- \sqrt{b^2 - 4ac} }{2a}\\y = \frac{-2 +- \sqrt{2^2 -4 * 24 * (-15)} }{2*24} \\y = \frac{-2+-\sqrt{4+1440} }{48} \\y = \frac{-2+-\sqrt{1444} }{48} \\y = \frac{-2+- 38}{48} \\y = \frac{-2+38}{48} \\y = \frac{3}{4}\\ \\or\\y = \frac{-2-38}{48} \\y = \frac{-40}{48} \\y = -\frac{5}{6}$

From the calculations above, the solution to the equation are 3/4 and -5/6 and the greater of the two solutions is 3/4

Learn more on quadratic equation here;

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7 0

1 year ago

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$