All categories

3 years ago

Some group insurance plans cover ordinary doctor visits and dental work. true or false

Mathematics

2 answers:

kherson [118]3 years ago

5 0

<u>Answer:</u>

The correct answer option is: true.

<u>Step-by-step explanation:</u>

It is true that some of the group insurance plans do cover ordinary visits and dental work.

Not all of the insurance plans that are offered cover these ordinary doctor visits and dental work but there are some plans that facilitate people with these benefits.

These plans not only cover the cost of heavy medical care facilities but the ones that are used in a routine.

Ede4ka [16]3 years ago

3 0

True, but not many. Insurance companies typically have a cap off so the client doesn't go in let's say every week for a check up to see if they grew an inch. This is form personal experience in reading into the fine print.

You might be interested in

Debbie buys a tree for the holidays. She would like to determine the amount of space it will take up in her living room. The tre

german

Answer:

Volume of Cone =

$= \frac{1}{3}\pi (2)^2 (4)\\\\=\frac{1}{3}(3.14)(4)(4)\\\\=\frac{1}{3}\times 3.14\times 16 \\\\=\frac{1}{3}\times50.24\\\\=16.746\\\\\approx16.75 \texttt {ft} ^3$

Thus, Volume = 16.75 cubic feet

Step-by-step explanation:

Volume of a cone is given by the formula

$V=\frac{1}{3}\pi r^2 h$

Where r is the radius and

h is the height

Given radius r = 2 and

height is 2 times that.

So height is 2*2 = 4

Plugging these into the formula we get:

Volume of Cone =

$= \frac{1}{3}\pi (2)^2 (4)\\\\=\frac{1}{3}(3.14)(4)(4)\\\\=\frac{1}{3}\times 3.14\times 16 \\\\=\frac{1}{3}\times50.24\\\\=16.746\\\\\approx16.75 \texttt {ft} ^3$

Thus, Volume = 16.75 cubic feet

8 0

3 years ago

Suppose that f: R --> R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR su

Pachacha [2.7K]

<h2>Answer with explanation:</h2>

It is given that:

f: R → R is a continuous function such that:

$f(x+y)=f(x)+f(y)------(1)$ ∀ x,y ∈ R

Now, let us assume f(1)=k

Also,

f(0)=0

( Since,

f(0)=f(0+0)

i.e.

f(0)=f(0)+f(0)

By using property (1)

Also,

f(0)=2f(0)

i.e.

2f(0)-f(0)=0

i.e.

f(0)=0 )

Also,

f(2)=f(1+1)

i.e.

f(2)=f(1)+f(1) ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

Similarly for any m ∈ N

f(m)=f(1+1+1+...+1)

i.e.

f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)

i.e.

f(m)=mf(1)

i.e.

f(m)=mk

Now,

$f(1)=f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=f(\dfrac{1}{n})+f(\dfrac{1}{n})+....+f(\dfrac{1}{n})\\\\\\i.e.\\\\\\f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=nf(\dfrac{1}{n})=f(1)=k\\\\\\i.e.\\\\\\f(\dfrac{1}{n})=k\cdot \dfrac{1}{n}$

Also,

when x∈ Q

i.e. $x=\dfrac{p}{q}$

Then,

$f(\dfrac{p}{q})=f(\dfrac{1}{q})+f(\dfrac{1}{q})+.....+f(\dfrac{1}{q})=pf(\dfrac{1}{q})\\\\i.e.\\\\f(\dfrac{p}{q})=p\dfrac{k}{q}\\\\i.e.\\\\f(\dfrac{p}{q})=k\dfrac{p}{q}\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q$

(

Now, as we know that:

Q is dense in R.

so Э x∈ Q' such that Э a seq belonging to Q such that:

$\to x$ )

Now, we know that: Q'=R

This means that:

Э α ∈ R

such that Э sequence $a_n$ such that:

$a_n\ belongs\ to\ Q$

and

$a_n\to \alpha$

$f(a_n)=ka_n$

( since $a_n$ belongs to Q )

Let f is continuous at x=α

This means that:

$f(a_n)\to f(\alpha)\\\\i.e.\\\\k\cdot a_n\to f(\alpha)\\\\Also\\\\k\cdot a_n\to k\alpha$

This means that:

$f(\alpha)=k\alpha$

This means that:

f(x)=kx for every x∈ R

4 0

3 years ago

Please Answer Correctly - Will Give Branliest & Extra Points. (links or unidentified/wrong answers will be reported and auto

jonny [76]

Answer:

Step-by-step explanation:

11 no

46+4x=90 degree(being perpendicular)

4x=90-46

x=44/4

x=11

12 no

They are complementary angles

6 no

angle B+27=180 degree(being straight line)

angle B=180-27

angle B=153

10 no

angle 38 +angle DFE=180 degree(being supplementary)

angle DFE=180-38

angle DFE=142 degree

5 no

They are vertically opposite angles.

7 no

angle AGF and angle CGD (because they have common arm and vertex)

6 0

3 years ago

Read 2 more answers

your bakery is in a large shopping center. The entrie shopping center is 4,500 square feet. Your bakery takes up only 1/10 of th

Kisachek [45]

So if the bakery is 1/10 of the shopping center's area, and the shopping center's area is 4500 feet, then the bakery's area is 1/10th of 4500 feet. 1/10th of 4500 feet is 450.

A more in-depth explanation:

A fraction is:
x/y
where y would equal how many parts you wanted to split a number into and x would equal how many parts you want to take.

So splitting 4500 into 10 parts and then taking one is basically 1/10 of 4500.

4500 in 10 parts, one would be 450.

5 0

3 years ago

1. Find the area of the shaded area of this figure:

statuscvo [17]

(area of square) - (area of triangle)

We know the are of a square is equal to s^2, where s is the side length.

While the are of a right angle tringle is 1/2*base*height

Then we have,

s = 10
base = 4
height = 4

(area of square) - (area of triangle)

=> (10^2) - (1/2)(4)(4)
=> 100 - 8
=> 92

Finally we get: 92 in^2

8 0

2 years ago