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3 years ago

1) Solve for Side A 2) Solve for Angle B

Mathematics

2 answers:

motikmotik3 years ago

8 0

Answer:

1) 42.51

2) 39.76

algol [13]3 years ago

5 0

Answer:

solution given;

let

AB=a

AC=b=30ft

AB=c=20ft

<A=115°

By using Cosine rule.

a²=b²+c²-2bc cos angle

a²=30²+20²-2*30*20 Cos 115°

a²=1807.1419

a=√[1807.1419]

a=42.51

Side A is 42.51ft.

Again

Cos B= $\frac{a²+c²-b²}{2ac}$

Cos B= $\frac{42.51²+20²-30²}{2*42.51*20}$

Cos B=0.7687

<B=Cos -¹(0.7687)

<B=39.46°

Angle B is 39.46

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3 years ago

Let L be a tangent line to the hyperbola x y = 2 at x = 9 . Find the area of the triangle bounded by L and the coordinate axes.

mafiozo [28]

Answer:

$A = 4$

Step-by-step explanation:

The equation of the slope of the tangent line L is obtained by deriving the equation of the hyperbola:

$y = \frac{2}{x}$

$y'=-2\cdot x^{-2}$

The numerical value of the slope is:

$y' = -2 \cdot (9)^{-2}\\y' = -\frac{2}{81}$

The component of the y-axis is:

$y = \frac{2}{9}$

Now, the tangent line has the following mathematical model:

$y = m \cdot x + b$

The value of the intercept is found by isolating it within the equation and replacing all known variables:

$b = y - m \cdot x$

$b = \frac{2}{9}-(-\frac{2}{81} )\cdot (9)\\b = \frac{4}{9}$

Thus, the tangent line is:

$y = -\frac{2}{81}\cdot x + \frac{4}{9}$

The vertical distance between a point of the tangent line and the origin is given by the intercept.

$d_{y} = \frac{4}{9}$

In order to find horizontal distance between a point of the tangent line and the origin, let equalize y to zero and clear x:

$-\frac{2}{81}\cdot x + \frac{4}{9}=0$

$-\frac{2}{9}\cdot x + 4 = 0$

$x = 18$

$d_{x} = 18$

The area of the triangle is computed by this formula:

$A = \frac{1}{2}\cdot d_{x}\cdot d_{y}$

$A = \frac{1}{2}\cdot (18)\cdot (\frac{4}{9} )$

$A = 4$

4 0

3 years ago

9/11 = ?/22<br> a. 12<br> b. 9<br> c. 4<br> d. 18

svlad2 [7]

18/22 = 2*9/2*11 = 9/11

Therefore, your answer is 18

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