Help me I don't know what's this is about

90 POINTS!!!!! Given: KLIJ inscr. in k(O),m∠K = 64°, measure of arc LI = 69°, measure of arc IJ = 59°, measure of arc KJ =97°

Kay [80]

In any cyclic quadrilateral, angles opposite one another are supplementary, meaning

$m\angle K+m\angle I=m\angle L+m\angle J=180^\circ$

and given that $\boxed{m\angle K=64^\circ}$ , we have $\boxed{m\angle I=116^\circ}$ .

By the inscribed angle theorem,

$m\angle JLK=\dfrac12m\widehat{KJ}$

$m\angle ILJ=\dfrac12m\widehat{IJ}$

and since

$m\angle L=m\angle JLK+m\angle ILJ$

we have

$m\angle L=\dfrac{97^\circ+59^\circ}2\implies\boxed{m\angle L=78^\circ}$

and it follows that

$m\angle J=180^\circ-m\angle L\implies\boxed{m\angle J=102^\circ}$

5 0

3 years ago

Let f(x)= x(2x-3)/(x+2)x. Find the domain, vertical and horizontal asymptote

Reptile [31]

Step-by-step explanation:

Domain of a rational function is everywhere except where we set vertical asymptotes. or removable discontinues

Here, we have

$\frac{x(2x - 3)}{(x + 2)x}$

First, notice we have x in both the numerator and denomiator so we have a removable discounties at x.

Since, we don't want x to be 0,

We have a removable discontinuity at x=0

Now, we have

$\frac{2x - 3}{x + 2}$

We don't want the denomiator be zero because we can't divide by zero.

so

$x + 2 = 0$

$x = - 2$

So our domain is

All Real Numbers except-2 and 0.

The vertical asymptors is x=-2.

To find the horinzontal asymptote, notice how the numerator and denomator have the same degree. So this mean we will have a horinzontal asymptoe of

The leading coeffixent of the numerator/ the leading coefficent of the denomiator.

So that becomes

$\frac{2}{1} = 2$

So we have a horinzontal asymptofe of 2

7 0

2 years ago

Rachel purchased a prepaid phone card for $30. Long distance calls cost 6 cents a minute using this card. Rachel used her card o

Amiraneli [1.4K]

Answer: her call lasted for 26 minutes.

Step-by-step explanation:

Let x represent the number of minutes for which her call lasted.

Rachel purchased a prepaid phone card for $30. Long distance calls cost 6 cents a minute using this card. Converting 6 cents to dollars, it becomes 6/100 = $0.06

This means that the cost of x minutes of long distance call is

0.06 × x = $0.06x.

If the remaining credit on her card is $28.44, it means that

0.06x + 28.44 = 30

0.06x = 30 - 28.44

0.06x = 1.56

x = 1.56/0.06

x = 26

8 0

3 years ago

Prove that the triangle EDF is isosceles. Give reasons for your answer.

Gekata [30.6K]

Answer:

$\triangle EDF$ is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are given the following things:

$\angle EDF = y$

$\text{External }\angle DFG = 90 +\dfrac{y}{2}$

Let us try to find out $\angle E$ and $\angle DFE$ . After that we will compare them.

Finding  $\angle DFE$ :

Side EG is a straight line so $\angle GFE = 180$

$\angle GFE$ is sum of internal $\angle DFE$ and external $\angle DFG$

$\angle GFE = 180 = \angle DFE + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)$

Finding  $\angle E$ :

Property of external angle: External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external $\angle DFG$ = $\angle E + \angle EDF$

$\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2} -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)$

Comparing equations (1) and (2):

It can be clearly seen that:

$\angle DFE = \angle E =90-\dfrac{y}{2}$

The two angles of $\triangle EDF$ are equal hence $\triangle EDF$ is isosceles.

8 0

3 years ago

A triangle with an angle measuring 104° must be which of these?

Tema [17]

This will be an obtuse angled triangle because one angle is greater than 90 degrees.

8 0

3 years ago