Don’t you guys just hate when you ask a question on here and two different people have different answers???

12(5x + 10x²) = 256 + 60x How do I solve this equation

Serggg [28]

Answer:

30

Step-by-step explanation:

idek

7 0

3 years ago

Read 2 more answers

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 dr

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

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

Solve the following inequality: ALGEBRA 1

Alexxandr [17]

Answer:

$m \leqslant 10 \: or \: m > 4$

Step-by-step explanation:

$\frac{m - 2}{3} \leqslant - 4 \\ \frac{m - 2}{3} \times 3 \leqslant - 4 \times 3 \\ m - 2 \leqslant - 12 \\ m - 2 + 2 \leqslant - 12 + 2 \\ m \leqslant 10$

OR

$3m - 8 > 4 \\ 3m - 8 + 8 > 4 + 8 \\ 3m > 12 \\ \frac{3m}{3} > \frac{12}{3} \\ m > 4$

6 0

2 years ago

How does f(x) = 3 ^x change over the interval x = 4 to x= 5

zloy xaker [14]

Answer:

f(x) = 3^x increases steadily on the interval [4,5].

Step-by-step explanation:

This exponential function f(x) = 3^x has a positive base (3) which is larger than 1. Thus, this function continues to increase as x increases, including the case where x increases from 4 to 5.

6 0

4 years ago

For f(x) = 3x - 1 and g(x) = x+2, find (f – g)(x).

kobusy [5.1K]

Well, I bet you want your answer right away! So here it is.

Given f (x) = 3x + 2 and g(x) = 4 – 5x, find (f + g)(x), (f – g)(x), (f × g)(x), and (f / g)(x).

To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.

(f + g)(x) = f (x) + g(x)

= [3x + 2] + [4 – 5x]

= 3x + 2 + 4 – 5x

= 3x – 5x + 2 + 4

= –2x + 6

(f – g)(x) = f (x) – g(x)

= [3x + 2] – [4 – 5x]

= 3x + 2 – 4 + 5x

= 3x + 5x + 2 – 4

= 8x – 2

(f × g)(x) = [f (x)][g(x)]

= (3x + 2)(4 – 5x)

= 12x + 8 – 15x2 – 10x

= –15x2 + 2x + 8

\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}(gf)(x)=g(x)f(x)= \small{\dfrac{3x+2}{4-5x}}=4−5x3x+2

My answer is the neat listing of each of my results, clearly labelled as to which is which.

( f + g ) (x) = –2x + 6

( f – g ) (x) = 8x – 2

( f × g ) (x) = –15x2 + 2x + 8

\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}(gf)(x)=4−5x3x+2

Hope I helped! :) If I did not help that's okay.

-Duolingo

7 0

3 years ago