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ElenaW [278]
3 years ago
14

Simplify 4х+ 2-

Mathematics
1 answer:
exis [7]3 years ago
7 0

Answer:

3x - 2

Step-by-step explanation:

4x + 2 -  \frac{ {x}^{2}  + x - 12}{x - 3}  \\  \\  = 4x + 2 -  \frac{ {x}^{2}  + 4x  - 3x- 12}{x - 3}  \\  \\  =  4x + 2 -  \frac{ {x}(x  + 4)  - 3(x  + 4)}{x - 3}  \\  \\  = 4x + 2 -  \frac{ (x  + 4)   \cancel{(x- 3)}}{\cancel{(x- 3)}}  \\  \\  = 4x + 2 - (x + 4) \\  \\  = 4x + 2 - x - 4 \\  \\  = 4x - x + 2 - 4 \\  \\  = 3x - 2

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Subtract 210 by the total number of students which you didn't give

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The answer is A. x < 78 because the dot isn’t filled in and it’s pointing to the left indicating that it’s less than 78.
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Julie checked the prices of her five favorite CDs at Music Madness. What is the mode for these prices? $14, $12, $15, $12, $17
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The force exerted by an electric charge at the origin on a charged particle at a point (x, y, z) with position vector r = x, y,
suter [353]

Answer:

\frac{25k}{34}[\frac{1}{4}-\frac{1}{50^{1/2}}] \\

Step-by-step explanation:

The straight line path between point (a,b,c) and (l,m,n) is parametric by the expression

r(t)=(1-t)(a,b,c) +t(l,m,n)\\

since we are giving point (4,0,0) and (4,3,4), the parametric equation is giving below

r(t)=(1-t)(4,0,0) +t(4,3,4)\\

using the dot product system of multiplication, we have

r(t)=(4,3t,4t) \\

t is between 0,1.

Next we define the line integral for work done which is express as

\int\limits^a_b {F.} \,dr\\

First we define the general expression for the force

f(x,y,z)=\frac{K(x,y,z)}{(x^{2}+y^{2}+z^{2})^{3/2}} \\

If we substitute our parametric equation we arrive at

F(r(t))=\frac{K(4,3t,4t)}{(4^{2}+(3t)^{2}+(4t)^{2})^{3/2}}\\F(r(t))=\frac{K(4,3t,4t)}{(16+34t^{2})^{3/2}}\\

also we need to find the expression for <em>dr</em>

r(t)=(4,3t,4t)\\dr=(0,3,4)dt\\

Now we substitute into the integral expression

\int\limits^1_0 {\frac{k(4,3t,4t)}{(16+34t^{2})^{3/2}}} \, .(0,3,4)dt

using dot product we arrive at

\int\limits^1_0 {\frac{25kt}{{(16+34t^{2})^{3/2}} } \,

let make a simple substitution so we can simplify the integral,

let assume u=16+34t^{2}\\

\frac{du}{dt}=68t\\ dt=\frac{du}{68t} \\

and changing setting the new upper and lower limit, we have

\frac{25k}{68} \int\limits^a_b {\frac{1}{u\frac{3}{2} } } \, du\\a=50\\b=16\\

by simple integral we arrive at

-\frac{25k}{34}[\frac{1}{u^{1/2}}] ^{50}_{16} \\\frac{25k}{34}[\frac{1}{4}-\frac{1}{50^{1/2}}] \\

Hence the workdone is

\frac{25k}{34}[\frac{1}{4}-\frac{1}{50^{1/2}}] \\

3 0
3 years ago
What is 26633646376+63761876784
Lemur [1.5K]

Answer:63,761,876,784

Step-by-step explanation:

6 0
3 years ago
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