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3 years ago

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

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(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