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kiruha [24]
3 years ago
8

Andre's phone is 11.5 cm long. How many millimeters is this?

Mathematics
1 answer:
Eduardwww [97]3 years ago
6 0

Answer:

about 115 millimeters

Step-by-step explanation:

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ValentinkaMS [17]
4+4+6+6=12+8=20

hope this helps
7 0
2 years ago
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On a coordinate plane, a curved line with a minimum value of (negative 2.5, negative 12) and a maximum value of (0, negative 3)
goldfiish [28.3K]

Answer:

f(x) > 0 over the interval (-\infty,-4)

Step-by-step explanation:

If  f(x)  is a continuous function,  and that all the critical points of behavior change are described by the given information, then we can say that the function crossed the x axis to reach a minimum value of -12 at the point x=-2.5, then as x increases it ascends to a maximum value of -3 for x = 0 (which is also its y-axis crossing) and therefore probably a local maximum.

Then the function was above the x axis (larger than zero) from - \infty, until it crossed the x axis (becoming then negative) at the point x = -4. So the function was positive (larger than zero) in such interval.

There is no such type of unique assertion regarding the positive or negative value of the function when one extends the interval from - \infty to -3, since between the  values -4 and -3 the function adopts negative values.

6 0
3 years ago
Read 2 more answers
last year my math book cost $126,and i was told the price would increase 9% each year. If the cost continues to increase at this
mrs_skeptik [129]
2,268 because u have to add the 2 years with 126
3 0
3 years ago
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I need help with this plzzz
Advocard [28]
I think it is BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
3 0
3 years ago
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What is the distance covered by a train while slowing down from
alukav5142 [94]

Answer:

The distance covered is 113.75 m

Step-by-step explanation:

As per the question:

The initial velocity of the train, v = 20 m/s

The final velocity of the train, v' = 6 m/s

Uniform deceleration, a = 1.6m/s^{2}

Or uniform acceleration, a = - 1.6m/s^{2}

<em>Here, the body decelerates, i.e., slows down at a uniform rate thus we take acceleration with negative sign.</em>

Now, to find the distance covered, s:

Using the eqn of Kinemetics:

v'^{2} = v^{2} + 2as

6^{2} = 20^{2} + 2(- 1.6)s

36 - 400 = - 3.6s

s = \frac{- 364}{- 3.6} = 113.75\ m

5 0
3 years ago
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