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n200080 [17]
3 years ago
14

Help with #19 plzzz!!!???

Mathematics
1 answer:
Olegator [25]3 years ago
5 0
Do "42 + 12" Since 7 x 6 is 42 and 4 x 3 is 12.
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Consider the following function. f(x) = 9 − x2/3 Find f(−27) and f(27). f(−27) = f(27) = Find all values c in (−27, 27) such tha
oksano4ka [1.4K]

I guess the function is f(x)=9-x^{2/3}. Then f(-27)=0 and f(27)=0.

The derivative is f'(x)=-\dfrac23 x^{-1/3}, but there is no c such that

-\dfrac23c^{-1/3}=0

This doesn't contradict Rolle's theorem because f'(0) does not exists. In other words, f is not differentiable on (-27, 27), so the conditions of Rolle's theorem are not met. (Looks like that would be the last option, or the second to last option if the last one is "Nothing can be concluded")

6 0
3 years ago
What's the answer to 9,10,11, and 12
nydimaria [60]
Where is the math u are talking about
3 0
3 years ago
What are three consecutive integers whose product is 480 more than their sum?
Natali [406]

Answer:

The three numbers are 7 8 and 9

Step-by-step explanation:

Givens

  • Let the first number be n - 1
  • Let the second number be n
  • Let the third number = n + 1

Equation

(n - 1)(n)(n + 1) - (n-1 + n + n+1) = 480

Solution

Multiply (n - 1) and (n + 1) = (n - 1)*(n + 1) = n^2 - 1

Multiply the second integer by the result of the first and third: n (n^2 - 1)

Add the three integers together: (x - 1) + (n - 1) + n = 3n  Combine these 2 steps

n(n^2 - 1) - 3n = 480 Remove the brackets

n^3 - n - 3n = 480

n^3 - 4n = 480

n^3 - 4n - 480 = 0

Graph

The graph shows that the intercept point is n =8. This is the only way I can see to solve this cubic. There are no other real roots.

Answer

n - 1 = 7

n = 8

n + 1 = 9

Check

Product 7*8*9 = 504

Sum = 7 + 8 + 9 = 24

504 - 24 = 480 Which checks.

4 0
3 years ago
Which of the following equations has only one solution?
Daniel [21]
X^2-6x+9=0 has only one solution
Hope this helps
4 0
3 years ago
Read 2 more answers
Imma need sum help please
MrRa [10]

Answer:

Step 5

Step-by-step explanation:

The mistake is in step 5.

The previous step was

step \: 4 : 2 + 10 \div 2

The order operations, PEDMAS, must be applied:

We have Addition and Division here,

Using PEDMAS, we must divide first to get:

step \: 5 : 2 + 5

We can now add to get:

step \: 6: 7

Therefore the mistake occurred at step 5

8 0
3 years ago
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