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Gelneren [198K]
3 years ago
12

F(x)=2x+3/4x+5 find f(-9)pls solve it ​

Mathematics
1 answer:
Marysya12 [62]3 years ago
4 0

Answer:

15/31

Step-by-step explanation:

substitute -9 were there's x in the function

f(-9)=2(-9)+3/4(-9)+5

=-18+3/-36+5

=-15/-31

the negatives cancel giving you

15/31

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Think about the expression (x-8)(x+4)
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Answer:

x*x +4*x-8*x -8*4

x^2+4x-8x-32

x^2-4x-32

4 0
3 years ago
Suppose that P(n) is a propositional function. Determine for which improper subset of the domain of n the statement P(n) must be
Alinara [238K]

Answer:

a) P(n) is true for all 'n' in the set ; { 0,2,4,6,8 ….. }

b) P(n) is true for all 'n' in the set ; { 0,1,2,3,4,5 ............ }

Step-by-step explanation:

a) As P(0) is true

we will assume that

  • P(2) is true
  • P(4) is true
  • P(6) is true

this simply means  that ;  P(n) is true for all 'n' in the set

{ 0,2,4,6,8 ….. }

b) since P(0) and P(1) are true

we will assume that

  • P( 0+2 ) = P(2)  is true

also P(1) and P(2) are true

we will assume that

  • P(1+2) = P(3)  is true

Also from the previous answers it can be seen that P(2) + P(3) is true

we will assume

  • P(2+2) = P(4)  is true

This simply means that P(n) is true for all 'n' in the set

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8 0
3 years ago
The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for
4vir4ik [10]
A) g=f' is continuous, so f is also continuous. This means if we were to integrate g, the same constant of integration would apply across its entire domain. Over 0, we have g(x)=2x. This means that


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\displaystyle\lim_{x\to1}x^2+C=1+C=3\implies C=2


Now, over x, we have g(x)=-3, so f_{x, which means f(-5)=17.


b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,


\displaystyle\int_1^6g(x)=4+\int_3^62(x-4)^2\,\mathrm dx=4+6=10


c) f is increasing when f'=g>0, and concave upward when f''=g'>0, i.e. when g is also increasing.

We have g>0 over the intervals 0 and x>4. We can additionally see that g'>0 only on 0 and x>4.


d) Inflection points occur when f''=g'=0, and at such a point, to either side the sign of the second derivative f''=g' changes. We see this happening at x=4, for which g'=0, and to the left of x=4 we have g decreasing, then increasing along the other side.


We also have g'=0 along the interval -1, but even if we were to allow an entire interval as a "site of inflection", we can see that g'>0 to either side, so concavity would not change.
5 0
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kolbaska11 [484]

Answer:

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Step-by-step explanation:

Please see the attached pictures for full solution.

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Your proof is correct and very well done
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