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Delvig [45]
3 years ago
12

This is probley easy...

Mathematics
2 answers:
vovangra [49]3 years ago
8 0

Answer:

6

Step-by-step explanation:

The other person is correct I did the test

Yakvenalex [24]3 years ago
4 0

Answer:

6

Step-by-step explanation:

divide 450 and 75

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What is the difference between | -3 | and -3
sineoko [7]

Answer:

6

Step-by-step explanation:

l -3 l - (-3)

3 + 3

6

5 0
2 years ago
Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that s
WINSTONCH [101]

Answer:  \bold{c=\dfrac{1+\sqrt{19}}{3}\approx1.8}

<u>Step-by-step explanation:</u>

There are 3 conditions that must be satisfied:

  1. f(x) is continuous on the given interval
  2. f(x) is differentiable
  3. f(a) = f(b)

If ALL of those conditions are satisfied, then there exists a value "c" such that c lies between a and b and f'(c) = 0.

f(x) = x³ - x² - 6x + 2     [0, 3]

1. There are no restrictions on x so the function is continuous \checkmark

2. f'(x) = 3x² - 2x - 6 so the function is differentiable \checkmark

3. f(0) =  0³ - 0² - 6(0) + 2 = 2

   f(3) =  3³ - 3² - 6(3) + 2  = 2

   f(0) = f(3) \checkmark

f'(x) = 3x² - 2x - 6 = 0

This is not factorable so you need to use the quadratic formula:

x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(3)(-6)}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{4+72}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{76}}{2(3)}\\\\\\.\quad =\dfrac{2\pm 2\sqrt{19}}{2(3)}\\\\\\.\quad =\dfrac{1\pm \sqrt{19}}{3}\\\\\\.\quad \approx\dfrac{1+4.4}{3}\quad and\quad \dfrac{1-4.4}{3}\\\\\\.\quad \approx1.8\qquad and\quad -1.1

Only one of these values (1.8) is between 0 and 3.

5 0
2 years ago
You move down 5 units and right 1 unit. You end at (5,-3). Where did you start?
stiv31 [10]

Answer:

ff

Step-by-step explanation:

4 0
2 years ago
the Wright flyer was the first successful powered aircraft a model was made to display in a museum with the length of 35 cm and
Savatey [412]
To compare the displayed Wright flyer in the museum with the actual plane, we have to compute for the ratio of the corresponding parameter. In this item, for both the display and the actual plane, we are given with the length. Computing for the ratio gives us,
    ratio = length of the model / length of the actual plane

Length of the model = 35 cm
We need to calculate for the length of the model in ft.
    length of the model = (35 cm)(1 in/2.54 cm)(1 ft/12 in)
   length of the model = 1.148 ft

Going back to the equation for the ratio,
   ratio = (1.148 ft)/21 ft
   ratio = 0.055

Hence, the dimensions used in the model is equal to 0.055 times the actual dimensions.

Error may occur because of the number of significant figures considered for rounding up or down of the answers after each calculation. 
6 0
3 years ago
Hey I'm confused can someone help
lilavasa [31]
The answer is 95 based on the line
5 0
3 years ago
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