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poizon [28]
3 years ago
7

Find the limits: lim h➡0 (1+h)^3-1/h​need to show work

Mathematics
1 answer:
Fed [463]3 years ago
3 0

9514 1404 393

Answer:

  3

Step-by-step explanation:

The expression can be simplified and then evaluated at h=0.

  ((1 +h)^3 -1)/h = (1 +3h +3h^2 +h^3 -1)/h

  = (3h +3h^2 +h^3)/h

  = 3 +3h +h^2

This evaluates to 3 when h=0.

The limit of interest is 3.

_____

Strictly speaking, the simplified form is only valid when h ≠ 0. However, that form is useful for finding the limit.

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Here is 93 points but answer this please If x = 4, what is the value of x3 + x - 5?
Luda [366]

Answer:

11

Step-by-step explanation:

x3+x-5

3*4=12

12+4-5

16-5=11

pls i REALLY need brainliest

8 0
3 years ago
Everyone please answer one of these my birthday tuesday but i might be in trouble if i fail this :(
Andreas93 [3]

Answer:

11. b

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Step-by-step explanation:

11.

3/x-4=7/x

*cross multiply

3x=7 (x-4)

3x=7x-28

28=7x-3x

28=4x

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10.

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Cos☆=11/14

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9.

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8 0
2 years ago
Solve for k:<br><br> k + 6 &gt; -20
FrozenT [24]

Answer:

k > -26

Step-by-step explanation:

To isolate the k, you subtract 6 from both sides to get k < -26.

4 0
4 years ago
Read 2 more answers
A chemist needs 130 milliliters of a 47% solution but has only 35% and 87% solutions available. Find how many milliliters of eac
choli [55]

Answer: Amount of 35% solution required = 100 milliliters

Amount of 87% solution required  = 30 milliliters

Step-by-step explanation:

Given: Total solution required =  130 milliliters

Let x = Amount of 35% solution (in milliliters).

Then, Amount of 87% solution required = 130 - x

As per given,

Concentration = 0.35x +0.87(130-x)=0.47(130)

0.35x +113.1-0.87x=61.1\\\\\Rightarrow\ -0.87x+0.35x=61.1-113.1\\\\\Rightarrow\ -0.52 x= -52\\\\\Rightarrow\ x=\dfrac{52}{0.52}=100

Hence, Amount of 35% solution required = 100 milliliters

Amount of 87% solution required  = 130-100 milliliters =30 milliliters

3 0
3 years ago
Fire katie shes spam deleting evreything on brainly
Mrrafil [7]

Answer:

Thank you so much. She has even deleted one of my questions before. I'll take care and beware.

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