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igomit [66]
3 years ago
6

F (×)=(×H)(×-1) is this polynomial function​

Mathematics
1 answer:
Anastaziya [24]3 years ago
7 0

Answer:

I think so

Step-by-step explanation:

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True or false 1. All parallelograms are squares.2. All squares are parallelograms. 3. All trapezoids are scalene.4. All squares
ruslelena [56]
1-2:<span>Yes. But not all parallelograms are squares</span><span>
3-false ! are </span><span>plane figures.
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4 0
3 years ago
What is a counterexample to this claim ?
MaRussiya [10]
Rectangle is the answer
6 0
3 years ago
1. Check all solutions of the equation: 5x(x - 2) = 0
Elenna [48]
1. A and B are both solutions
2. B
3. C
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7 0
3 years ago
Para ampliar uma sala quadrada, foram acrescentados 4 metros em seu comprimento e 2 metros em sua largura, conforme a figura. Qu
ollegr [7]

Answer:

b) x^2 + 6x + 8

Step-by-step explanation:

Para ampliar la habitación cuadrada, se agregaron 4 metros de largo y 2 metros de ancho, como se muestra en la figura.

Deje que la longitud y el ancho de la habitación antes de la expansión sean x.

La nueva longitud de la sala es (x + 4).

El nuevo ancho de la sala es (x + 2).

La nueva área de la sala será:

(x + 4)(x + 2)\\x^2 + 2x + 4x + 8\\x^2 + 6x + 8\\\\

Esa es la nueva área de la habitación.

3 0
3 years ago
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
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