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OlgaM077 [116]
2 years ago
12

each of exercises 15–30 gives a function ƒ(x) and numbers l, c, and e 7 0. in each case, find an open interval about c on which

the inequality 0 ƒ(x) - l 0 6 e holds. then give a value for d 7 0 such that for all x satisfying 0 6 0 x - c 0 6 d the inequality 0 ƒ(x) - l 0 6 e hol
Mathematics
1 answer:
3241004551 [841]2 years ago
6 0

The given inequality holds for the open interval (2.97,3.03)

It is given that

f(x)=6x+7

cL=25

c=3

ε=0.18

We have,

|f(x)−L| = |6x+7−25|

          = |6x−18|

          = |6(x−3)|

          = 6|x−3|

Now,

6|x−3| <0.18  then |x−3|<0.03 ----->−0.03<x-3<0.03---->2.97<x<3.03

the given inequality holds for the open interval (2.97,3.03)

For more information on inequality click on the link below:

brainly.com/question/11613554

#SPJ4

Although part of your question is missing, you might be referring to this full question: For the given function f(x) and values of L,c, and ϵ0, find the largest open interval about c on which the inequality |f(x)−L|<ϵ holds. Then determine the largest value for δ>0 such that 0<|x−c|<δ→|f(x)−|<ϵ.

f(x)=6x+7,L=25,c=3,ϵ=0.18

 

.

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(Before you answer I am looking for how I do it. I have the answers but I need to know an equation to do this instead of using l
Tema [17]

Answer:

14 and 16.

Step-by-step explanation:

We know that two consecutive even numbers is 30.

Let's let n be <em>any number</em>.

Then the first <em>even</em> number must be 2n.

This is because n can be any number, either even or odd. However, if we multiply it by 2, this <em>ensures</em> that 2n is even. Remember that any number multiplied by 2 yields an even number.

Therefore, the consecutive even number will be (2n+2). Not (2n+1), because that gives an odd number.

We know that they sum to 30. So, we can write the following equation.

(2n)+(2n+2)=30

And we solve from there:

4n+2=30\\\Rightarrow4n=28\\\Rightarrow n=7

So, the value of n is 7.

Therefore, the first even number is 7(2)=14.

And the consecutive even number is 16.

Edit: Typo

7 0
3 years ago
(b) Out of 200 students in an examination of class IX, 140 passed in Maths, 120 passed in Scien
iren [92.7K]

Answer:

n(U)=200

n(M)=140

n(S)=120

n(MuS)complement=40

n(MuS)=n(U)-n(MuS)complement

=200-40

=160

n(MnS)=n(M)+n(S)-n(MuS)

=140+120-160

=260-160

=100

3 0
3 years ago
Read 2 more answers
WILL GIVE BRAINLIEST!!!!!!! A boy had 20 cents. He bought x pencils for 3 cents each. If y equals the number of pennies left, wr
kumpel [21]

Answer:

Hey there!!

The total number of cents - 20

Cost for each pencil - 3 cents

Number of pencils bought - ' x '

y = number of pennies left

What is the domain ?

Show how y is dependent on x ..

Let's get this into an equation :

... The total cost for x pencils bought = 3x

... Number of pennies left = 20 - 3x

... y = number of pennies left

... y = 20 - 3x

Notice : If the x value changes, the y value changes too

... If x = 1 , then , y = 17; If x = 2 , then , y = 14

Hence, we could say y is dependent upon x

Domain = ?

... Remember - The total number of pennies = 20 ; hence, the total cost cannot go above 20 cents.

Hence, we will have to work with inequalities

The equation :

... 20 ≥ 3x

Divide 3 on both sides

20 / 3 ≥ x

x ≥ 6.667

Let's take this as 6

Hence , the domain will be :

D : { 1 , 2 , 3 , 4 , 5 , 6 }

Hope my answer helps!

5 0
3 years ago
17. El precio de los terrenos en Lima es proporcional al área e inversamente proporcional a su distancia con respecto al centro
grin007 [14]

Answer:

375000

Step-by-step explanation:

We have that in this case the following proportion would be fulfilled:

Price * Distance / area

Now, we have that for a distance of 45 km and an area of 900 m ^ 2 the price is 250,000, now for a distance of 40 km but with an area of 1,200 m ^ 2, how would the price be, we replace:

250000 * 45/900 = P * 40/1200

12500 * 1200/40 = P

P = 375000

Which means that for these conditions the price is 375000.

3 0
3 years ago
A paperweight is shaped like a triangular pyramid. The base is an equilateral triangle. Find the surface area of the paperweight
stich3 [128]

Answer:

The surface area of the paperweight is SA=8.3\ in^2

Step-by-step explanation:

we know that

The surface area of the triangular prism is equal to the area of its triangular base plus the area of its three triangular lateral faces

so

<em>Find the area of its triangular base</em>

SA=\frac{1}{2}(2)(1.7)

SA=1.7\ in^2

<em>Find the area of its three triangular lateral faces</em>

SA=3[\frac{1}{2}(2)(2.2)]

SA=6.6\ in^2

Adds the areas

SA=1.7+6.6

SA=8.3\ in^2

4 0
4 years ago
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