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mina [271]
3 years ago
10

In the number 14,685,000,000, the digit 8 is in what place

Mathematics
2 answers:
Ghella [55]3 years ago
7 0
I believe it is in the ten millionth place.
scZoUnD [109]3 years ago
4 0
Ten millionth place
You might be interested in
(06.01)
MAVERICK [17]

After solving I have come up with 6 2/9 or 6.22222.

4 0
3 years ago
How do you solve it?
Elis [28]

At heart we're being asked for a line through two points,


(40^\circ \textrm{ C}, 355 \textrm { m/s}) \quad \textrm{and} \quad (49^\circ \textrm{ C}, 360 \textrm { m/s})


In general the line through (a,b) and (c,d) is


(y-b)(c-a)=(x-a)(d-b)


Check that you understand why both (a,b) and (c,d) are on this line.


Here our indepedent variable, instead of x, is T, temperature. Our dependent variable is v, velocity. Substituting,


(v - 355)(49 - 40) = (T - 40)(360 - 355)


9(v - 355) = 5(T - 40)


v-355 = \frac 5 9 T - \frac{200}{9}


v= \frac 5 9 T - \frac{200}{9} + 355


v= \frac 5 9 T + \frac{2995}{9}


That's our answer; let's check it.


When T=40, v = (5/9)40 + (2995/9) = 355 good


When T=49, v= (5/9)49 + (2995/9) = 360 good




7 0
3 years ago
Use a table of values to estimate the value of the limit. Confirm your result graphically by graphing the function with a graphi
il63 [147K]
Hello,

\lim_{x \to 0}\dfrac{ \sqrt{x+25}-5}{x}\\\\

=\lim_{x \to 0}\dfrac{(\sqrt{x+25}-5)*(\sqrt{x+25}+5)}{x*(\sqrt{x+25}+5)}\\\\

=\lim_{x \to 0}\dfrac{x+25-25)}{x*(\sqrt{x+25}+5)}\\\\


=\dfrac{1}{10}

















4 0
3 years ago
3m-5p=12 solve for p
umka21 [38]
3m - 5p = 12

3m(-3m) -5p = (-3m)+ 12
-5p = -3m + 12
(-5)/-5p = (-5)/ (-3m + 12)
p = 3/5m + 12/-5
5 0
3 years ago
Read 2 more answers
Video Example EXAMPLE 1 Find the linearization of the function f(x) = x + 1 at a = 3 and use it to approximate the numbers 3.98
Blizzard [7]

Answer:

  the linearization is y = 1/4x +5/4

  the linearization will produce <em>overestimates</em>

  the values computed from this linearization are ...

     f(3.98) ≈ 2.245

     f(4.05) ≈ 2.2625

Step-by-step explanation:

Apparently, you have ...

  f(x)=\sqrt{x+1}

from which you have correctly determined that ...

  f'(x)=\dfrac{1}{2\sqrt{x+1}}

so that f(3) = 2 and f'(3) = 1/4. Putting these values into the point-slope form of the equation of a line, we get the linearization ...

  g(x) = (1/4)(x -3) +2

  g(x) = (1/4)x +5/4

__

The values from this linearization will be overestimates, as the curve f(x) is concave downward everywhere. The tangent (linearization) is necessarily above the curve everywhere.

__

At the given values, we find ...

  g(3.98) = 2.245

  g(4.05) = 2.2625

4 0
4 years ago
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