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umka21 [38]
3 years ago
5

7 × 3 = 3 × ?

Mathematics
1 answer:
faust18 [17]3 years ago
5 0

Answer:

1. 7

2. 3

3. 9

Step-by-step explanation:

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Three to the power of three minus seven plus ten
BARSIC [14]

Answer:

30

Step-by-step explanation:

You’re correct

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3 years ago
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Please can someone answer this.
ivanzaharov [21]

Answer:

  •  x = -4, x = 0, x = 1

Step-by-step explanation:

x is a factor of all terms, so x=0 is a zero. (Eliminates choices 1 and 5.)

The sum of coefficients is 0, so x=1 is a zero. (Eliminates choices 3 and 4.)

Reversing the sign of the odd-degree terms gives signs of -++, so there is one sign change, hence one negative real root (by Descartes' rule of signs). This confirms choice 2 as the answer.

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Of course, your graphing calculator can answer this almost as quickly.

7 0
3 years ago
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crimeas [40]

Answer:

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6 0
3 years ago
a climber starts descending from 504 feet above sea level and keeps going until she gets to 40 feet below sea level, how may fee
GenaCL600 [577]

Answer:

544 feet.

Step-by-step explanation:

|504| +|-40|=544

8 0
3 years ago
Verify that the function <img src="https://tex.z-dn.net/?f=g%28x%29%3D2x%5E3-3x%2B1" id="TexFormula1" title="g(x)=2x^3-3x+1" alt
ch4aika [34]

Lets check if the three conditions hold.

<u>1 : Continuity of g on the interval [0,2]</u>

First, g(x) is a continuous function on R, as the sum of a cubic function wich is continuous on R, and a linear polynomial of the form ax + b which is also continuous on R. Finally g is also continuous on the interval [0,2]

<u>2 : Differentiable on the same interval</u>

Since the cubic function and the linear polynomial one are differentiable on R, g also is differentiable and particularly on the interval [0,2]

Also we have g'(x) = 2*3*x² - 3 = 6x² - 3

<u>3 : Do we have g(0) = g(2) ?</u>

Lets compute g(0) = 2*0^3 - 3*0 + 1 = 1

And g(2) = 2*2^3 - 3*2 + 1 = 2 * 8 - 6 + 1 = 16 - 6 + 1 = 11

Since g(0) ≠ g(2), Rolle's theorem is not applicable. Thus unfortunately, we can not conclude that there exist c ∈ (0,2) such that f'(c) = 0

5 0
2 years ago
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