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Dmitry_Shevchenko [17]

2 years ago

7

Find the measure of each missing angle.

1 answer:

nadya68 [22]2 years ago

5 0

Answer:

Step-by-step explanation:

Angle sum property of triangle: Sum of all the angles of a triangle is 180

Alternate interior angles: When two parallel lines are intersected by a transversal, the pair of angles on the inner side of each of these lines but on the opposite side of the transversal are called alternate interior angles

In ΔABC,

∠1 + 90 + 38 = 180 {angle sum property of triangle}

∠1 + 128 = 180

∠1 = 180 - 128

$\sf \boxed{\bf \angle 1 = 52^ \°}$

AB // CD and AC is transversal.

$\sf \boxed{\bf \angle 3 = 38^\circ}$ {Alternate interior angles are equal}

In ΔACD,

∠2 + ∠3 + 63 = 180 {angle sum property of triangle}

∠2 + 38 +63 = 180

∠2 + 101 =180

∠2 = 180 - 101

$\sf \boxed{\bf \angle 2 = 79^ \°}$

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Let f(x)=2/x-3. Find f^-1(x)

KonstantinChe [14]

Answer:

Two questions:

Question 1: $f^{-1}(x)=?$ given $f(x)=\frac{2}{x}-3$ .

Answer 1: $f^{-1}(x)=\frac{2}{x+3}$

Question 2: $f^{-1}(x)=?$ given $f(x)=\frac{2}{x-3}$ .

Answer 2: $f^{-1}(x)=\frac{2}{x}+3$

Step-by-step explanation:

So $f^{-1}$ is used in most classes to represent the inverse function of $f$ .

The inverse when graphed is a reflection through the y=x line. The ordered pairs $(a,b)$ on $f$ implies $(b,a)$ are on $f^{-1}$ .

This means we really just need to swap x and y.

Since we want to write as a function of x we will need to solve for y again.

Question 1:

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

Swap x and y:

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

We want to solve for y.

Add 3 on both sides:

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

Make the left hand side a fraction so we can cross-multiply:

$\frac{x+3}{1}=\frac{2}{y}$

Cross multiply:

$y(x+3)=1(2)$

Simplify right hand side:

$y(x+3)=2$

Divide both sides by (x+3):

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

So $f^{-1}(x)=\frac{2}{x+3}$ .

Question 2:

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

Swap x and y:

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

Make left hand side a fraction so we can cross multiply:

$\frac{x}{1}=\frac{2}{y-3}$

Cross multiply:

$(y-3)x=1(2)$

We have to distribute here:

$yx-3x=2$

Add 3x on both sides:

$yx=2+3x$

Divide boht sides by x:

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

You could probably stop here but you could also simplify a little.

Separate the fraction into two terms since you have 2 terms on top bottom being dividing by x:

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

Simplify second fraction x/x=1:

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

So $f^{-1}(x)=\frac{2}{x}+3$ .

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d1i1m1o1n [39]

Yaaaaay
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